HC Verma Physics Part 1 Solutions for Chapter 10 Rotational Mechanics

5/21/2012

Download HC Verma Solutions for HC Verma Physics Part 1 Solutions Chapter 10 Rotational Mechanics solved by our expert teachers. We have curated solutions for all questions of chapter 10. You can download the Solution of HC Verma - Rotational Mechanics and prepare for your upcoming competitive examinations.

HC Verma Physics Part 1 Solutions for Chapter 10 Rotational Mechanics

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253 Comments

Debraj Saha

6/22/2012 08:28:55 pm

Can i 've H C verma answers for chapter 10 "Questions for short answer" no. 21

Admin

6/23/2012 03:12:29 pm

Hi Debraj, you can download the solution by clicking on Download bottom.
If you have problem in solving question you can ask us any time.

karthik link

10/26/2014 01:46:37 am

hcv in chapter no 10 rotational mechanics problemm no 33 how do we get (T1-T2)=ia\r2

irfan khan

10/3/2015 09:35:40 pm

sir;please i need to this chapter solution so please sir not forget you

Reviewer

12/13/2015 02:31:46 am

Please check what you upload, there are tons of mistakes in many questions.... I have reviewed this earlier as well.. Suckers

abhishek

9/21/2013 10:24:28 pm

sir please help me to solve question number 10 of the exercises..!!!

abhishek

9/21/2013 10:25:22 pm

sir please help me to solve question number 10 of the exercises..i have not got its solution by you!!!

abhishek

9/21/2013 10:25:40 pm

sir please help me to solve question number 10 of the exercise

Brijesh Yadav link

10/24/2013 12:18:57 pm

harshit

4/30/2015 06:14:53 pm

Sir,please explain que.60 thoroughly

shubham

12/23/2013 03:16:19 pm

q.no22 nhi ban raha hai

gaua

3/8/2016 12:06:22 am

Aman

10/29/2012 10:50:18 pm

what is in Q-44, ch- 10

Admin Answer

10/30/2012 11:35:55 am

Hi Aman,
for this question you need to understand the free body diagram as given in the solutions
In this, hinges will apply two forces F1 and N1 for hinge1 and F2 and N2 for Hinge2
F1 and F2 are originated to balance the weight of door and N1, N2 are the normal forces which will cancel each other.
so through we may easily calculate.

Harsh

9/15/2013 03:43:25 am

but sir why are the normal contact forces opposite to each other... how can we find it out? in ques. 44 chap. 10

vanshi

1/15/2016 03:42:11 am

Sir,how did u te ak torque about point B.what is .75

Dinesh Yadav link

10/30/2012 07:20:08 pm

HC VERMA Vol 1 Page: 191 ;Rotational Mechanics
Example:29 The sphere lies..... from the centre of the sphere.
Can somebody tell me that in its (b) part solution,why does they have Conserved Angular Momentum about the centre of mass? there is Toruqe by friction force that we can't neglect.
According to Me, instead of abt the centre of mass, we should do the same about the point of contact because there is torue only by mg that's very small that we can neglect. So by this method, answer should be: w=(mV*h)/[(2/5)M+m]R*R

Please explain me with Reason if I am wrong or not? and the reason for using this about C.O.M. in book?

Admin Answer

10/30/2012 10:02:32 pm

Dear Dinesh,
In this question we have used Conservation of angular momentum because it is given in the question that particle and sphere are part of the system that is combined system. so there is no external torque acting on the system. so angular momentum remains constant.
Also friction force occurs when there is any relative motion. But in the question there is no relative motion between particle and sphere nor between sphere and ground because it is given that sphere starts pure rolling(friction = 0) after collision.

faculty

4/14/2013 12:53:35 am

dear ,when it is said 'just after collision' that means before friction work.its not practical actually but here meaning is this only.and in b) part pure rolling is not given.even it is given there will be static friction so ext torque n force.

abhishek choudhary

9/20/2013 01:42:45 pm

sir, i am stuck on question no. 10 of exercises, please help

Admin

11/3/2012 03:08:52 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Prashant

11/15/2012 11:06:53 pm

HC VERMA Vol 1 Page: 192 ;Rotational Mechanics
Questions for short answer
Q12
if the resultant torque.......on the body...any other point?
Doubt:is the answer NO..because the sum of torques about any line is zero iff net force on the body is zero...but here it is asking about all points..I am getting confused between points and line..please help by giving explanation of answer

Admin

11/16/2012 11:49:24 am

Your answer will be uploaded on 18th Nov.
Please wait for few hours because there is some technical problem in the Website. we will trying to serve you as fast as possible.

Admin

11/17/2012 07:13:03 pm

Dear Prashant,
This is just a conceptual question. Do not get confused with the question. Yes the answer is No. We all know that torque depends on the product of two factors i.e force and distance from the point. It may be possble that product becomes zero at any point but it will not be zero at every point.Generally torque becomes zero whenever there is symmetry about any point. We can see symmetry may not be there about every point.
Now taking about torque about the line. In this Torque, we take the distance as the perpendicular distance from the line and everything remains same.

satish

11/29/2012 07:28:59 pm

can u explain Q 50

Admin

11/29/2012 08:30:44 pm

Dear Satish,
In this question it is given that two balls of 0.5 kg is attached to the end of rods of length 50 cm. An impulsive force of 5N is acting on one of the ball. so this force will exert external torque on the system. we know that
Torque(T) = Iα
where I = moment of inertia of particle
and α = angular acceleration
so we know that
T = F x r
I = mr²+mr²
T = Iα
through this we will find the value of angular acceleration(α).
ω = ω₁+αt
t = time and ω₁ = intial velocity
through this we may find ω.

Raman

12/4/2012 10:41:03 pm

Sir in question 63 I'am not able to understand the reasoning for the velocity of ball B that it has the same velocity.Please explain explicitly.And sir i wanted to ask ,can momentum conservation equation be written for only one part of the system like say only for one ball dealing only with its initial angular and final momentum and not the other ball .

Admin

12/5/2012 02:22:06 pm

Dear Raman,
Here I am going to explain why the Velocity of the ball B remains same after the collision. In order to explain this let us take an example that rod (AB) is moving on a friction less surface with some velocity v. After some time we hinged the point A, then what will happen the point B will start rotating about A with the same velocity because the force is acting on point B only along its length which is providing it the centrifugal force.
Similarly here the point A is not hinged but atleast its velocity is decreased which means that the rod is moving as well as rotating. But this is the case when the frame of reference is outside the rod.
Definitely the velocities will change if we change the frame of reference.
Now I am going to explain “can momentum conservation equation be written for only one part of the system like say only for one ball dealing only with its initial angular and final momentum and not the other ball .”

See do not get confused with the answer given in the solutions
Let us again solve part a of question number 63 –
After and before collision velocity of B = v₀
After collision velocity of (A+P) = v
Before collision velocity of A = v₀
Now applying linear momentum equation. I told you that before applying the conservation of momentum equation you just have to check that all the bodies of the system should be non acceleration. Here A, B, P are moving with constant velocity or are at rest. Now applying linear momentum equation.
Momentum due to B + Momentum due to A = Momentum due to B + Momentum due to (A+P)
Momentum due to A = Momentum due to (A+P)
So while solving question you should take all the bodies.

Harshit

4/30/2015 07:07:37 pm

Sir,what are the conditions of applying law of conservation of angular momentum,law of conservation of energy,conservation of angular momentum.like in que.67 can we do it by applying law of conservation of angular momentum. Why or why not?

Raman

12/4/2012 10:47:29 pm

Sir I'am also a little confused about when is torque 0 in a question and when is it not 0.For your reply to dinesh you mentioned considering thye ball and the sphere as a system the momentum remains conserved but friction is acting on it .If we say system what do we mean by it no force would be considered on it if this is so then everything will noit have a torque on it if considered as a system.Sir please explain this to me distinguishing the cases when will torque not act and when will it act with some good examples.

Admin

12/6/2012 10:24:27 am

Dear Raman,
In order to use the momentum equation. Just take all the bodies which are colliding. Then check that whether they are carrying any acceleration or not that may be linear or angular acceleration. If they are carrying then momentum conservation equation is not applied because the velocity is changing every instant.
Now taking about the friction. Forces like friction and weight are the properties of body that's why they are taken as internal forces.
Now may see that while deriving the collisions equations for non elastic collision we apply momentum conservation equations where friction is acting.

Raman

12/6/2012 12:09:35 am

Sir please i have doubts in the 85th and 86th questions.Sir in the 85th question ho wto take the directions of the linear and angular momentum ,i mean that when will they be opposite to each other because in ques 86th thefinal linear and ang momentum being opp to each other are taken as opposite.I just didnt understand hpw will one decide the direction .Sir please explain me the 85th question in detail,i got to the answer but now some things are confusing me and please write the full equation.

Admin

12/7/2012 12:47:42 am

Dear Raman,
Solution of question number 85 -
In this question we are taking moment about A,
It is given that the velocity of center of mass is V which can be taken as the tangential velocity. so at that instant it is having moment((MVR) moving in the clockwise direction with respect to A.
Also it is having angular velocity(ω) in the opposite direction which is in the anti-clock wise direction.
Before angular momentum = MVR - Iω
where I = moment of inertia of sphere.
ω = angular velocity
Now when sphere stops rotating, then is having only linear velocity which is taken as the tangential velocity(V₀) with respect to A and we angular momentum = MV₀R
Now equating both values we get V₀
After some time when the sphere starts pure rolling then all the angular momentum are in same direction. We can calculate new velocity.

Solution for Question 86 -
here do not get confused. we are taking the situation after the collision. when the ball has just collided then due to collision the velocity of center of mass of sphere is reversed but the ball is still having angular momentum in the clockwise direction. Now the angular momentum due to center of mass velocity is opposite to the angular momentum due to ω.
Now again taking moment about the point of contact.
Angular momentum = mvR – (2/5) mR² × v/R
Now after some time the sphere will start pure rolling
angular momentum = mv₀R + (2/5) mR²× (v₀/R)
Now equating both we get value of v₀.

Raman

12/7/2012 09:57:11 am

in youe explaination for 85 and 86 question the things i wanted to be explained are that how is momentum and ang momentum opposite and the language that is first it stops rotating then it starts pure rolloing and how in pure rolling the direction of linear and ang mom becomes same

Admin

12/7/2012 11:18:45 am

Dear Raman,
In question number 85 - The velocity of center of mass is moving with the linear velocity towards the right direction. Now try to analyse that if a particle moving in the circle has tangential velocity towards right then it means that it is moving in the clockwise direction. but In the question it is also given that there is a angular velocity which is in the anticlockwise direction so it is opposite to the above momentum.

Now i am explaining your another doubt that for example you have a ball and you pushed it with linear velocity V towards right . you may see that initially it will slide and after some time it will start rolling clockwise. Here also is the same case initially sphere will slide and as the surface is rough so sphere is having the tendency to roll . But due to the angular velocity in the opposite direction it will not allow the ball to have pure rolling motion. but after some time the opposite angular velocity will come to zero because linear velocity which will try to rotate it in the clockwise direction and when the opposite angular velocity will become zero then it will start rolling purely.

Solution for question 86 -
In this question there is also the same case as above.
Initially the ball was rolling in the clockwise direction but when the rolling ball hits the wall, due to elastic collision its linear velocity reverses but the ball is still having angular velocity in the clockwise direction. then same thing will happen it will resist the motion for sometime then it will become zero and ball will rolling purely having linear and angular momentum in the same direction.

Raman

12/7/2012 10:32:10 pm

Sir can you please help me to give the moment of inertia of various objects and also objects like box,plates etc.....or sir provide me the details of some site from where i can download all these stuff

Admin

12/8/2012 10:45:43 am

Dear Raman,
if you are preparing for entrance. Then you don't need learn moment of inertia for different objects. Just learn the moment of inertia for the standard objects which are given in your text books or reference books for engineering entrance examinations.

theyatin link

3/27/2013 03:15:54 pm

dear Raman, curiosity has no limits apart from test there is a real world where you face many question but answer for most of them lies in books you read so give more stress on study.
apart from this,
moment of inertia only depends upon center of mass and velocity or angular velocit and shape size density doesnt matters.
here i am giving you a link for standard objects whose M.I. ca be calculated easily.

basically plate is a disc
and box is cube so these are very close to standard objects.

theyatin

3/27/2013 03:20:01 pm

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

shreyas

12/25/2012 09:46:58 pm

Sir i have an doubt.A general one.If a rod of lenght l and mass M is lying horizontally and no friction is present.if a ball of mass m and velocity v hits it at one end and comes to rest.Say that end to be pt A.now if we conserve angular momemtum about pt A.we will get
0+0(angular momentum of ball+rod)=Iw+v(rod after collision)M(l/2)
now in the above equation why do we get w negative.If i have made any mistake in conserving the angular momemtum please tell me.Thank You.

Admin

12/26/2012 01:44:32 pm

Dear shreyas,
when the ball is striking the end of rod, then there are two motions that the rod will experience. these are -
1. linear motion
2. circular motion about center of mass
so, you have to apply the momentum conservation for both the motions to solve the question
applying momentum conservation for linear motion
MV1 = mv
where M = mass of rod
V1 = velocity of rod
m = mass of ball
v = velocity of rod
applying momentum conservation for circular motion about center of mass, we get
mvr/2 = Iω
where r = length of rod

Shreyas

12/26/2012 03:34:41 pm

sir but i have conserved momemtum from point A itself.then
intial angular momemtum of ball is 0 as the velocity vectir passes through it.then final angular momentum would be
Iw+mvr/2
where v is the linear velocity gained by te rod after collision.then why do we get w negative.

doctor

11/9/2013 01:49:51 am

is this book..helful fr clearing medical examzzz

akash link

1/4/2017 08:58:52 am

Hc verma solution of chapter 10 of class11 .Q54

Shreyas

12/25/2012 09:52:48 pm

Sir one more doubt.If a rod length L and mass M is hanging horizontally with the help of two springs at both ends A and B.if we cut the spring corresponding to B.how can we get its angular acceleration.i get it wrong if i find torque about A and then equate to Ia.but i get it right when i equate torque about the other end.Please tell me why do i get it wrong.

Admin

12/26/2012 01:51:51 pm

Dear shreyas,
Please explain your question in depth.
according to me, you should get the right answer about A. if this is the question of HC verma, then tell me question number, I will see it.
if it is not the question of HC verma, then explain me in depth.

Shreyas

12/26/2012 03:43:34 pm

Sir the rod is hanging horizontally in gravity with the help of two springs at both ends.That is tention in both springs will be mg/2.now if we cut one spring ie from end B,then we have tp find angular accn immediately after cutting the spring.
now if we look from point A.
we get torque as,
mgxl/2.this the only torqueacting from point A
but if we look from other end B.
the torque will be,
mgxL/2+mg/2xL
here the latter torque is due to spring which is mg/2xL.as the spring is at a distance L from the other end.
I also might have done any mistake.please check.

admin

12/27/2012 02:14:22 pm

Dear shreyas,
first in am going to explain you about your first question i.e rod of length L collided with ball of mass m.
If we apply conservation of angular momentum about point A, then we are getting negative ω.
yes it depends upon the frame of reference. here our frame of reference is A. now imagine that you are standing at point A.
you know that after colliding, the rod will start rotating about center of mass. Now as you seeing the rod from point A, it is rotating in the anti clock direction about point A. But center of mass will try to rotate in the clock wise direction because velocity of center of mass is in the forward direction. that's why we are getting negative ω.
here torques are opposite to each other.

explanation for other question -

here we will calculate the torques from both point A and B.
first we calculate torque about point A.
here there is only one torque i.e. mgl/2.

now we are calculating torque about B -
now when the spring is cut, the whole tension is transferred to spring at A. Now tension at the spring A is mg.
force at the centroid i.e weight will try to rotate the rod in the anti clock direction about B.
force at spring A will try to rotate the rod in the clock wise direction
now calculating torques about point B, we get
mgl-mgl/2 = mgl/2

so, we are getting same torques about both the points.

Shreyas

12/27/2012 02:50:19 pm

Yup.Got the first one.Thanks a lot.
Even got the second one.
Now according to the equations we will get the angular acceleration as 3g/2l.But the answer given is 3g/l.Can you please explain why???

shreyas

12/27/2012 02:54:31 pm

The second question also further has sub question
With what acceleration will the end A go up?
and the answer given to this is g.Please explain.

admin

12/28/2012 03:34:56 pm

Dear Shreyas,
I am also getting the same answer 3g/2l. i am trying to get it. I will upload your answer when i will get. i am sorry for your inconvenience.

Shreyas

12/31/2012 08:43:10 pm

Hey nywaz got dat ques...
the tension in the spring A wudnt become mg...as it is a spring the tension would remain same ie mg/2.Now the spring at end A wud go upward.Therefore we cannot equate torque about that point.But we can equate torque abou COM.therefore from COM we get,
mg/2*L/2=mL^2/12*a
where m is mass of the rod,L is lenght and 'a' is angular acceleration of the rod.

admin

1/1/2013 05:12:59 pm

thank you for the answer

Arshdeep Bajwa

1/11/2013 02:46:44 am

Sir can u explain question 37 n 38 of exercise.....

Admin

1/11/2013 07:24:52 pm

Dear Arshdeep,
solution for question 37
In this question there is a block which is connected to the pulley as shown in the diagram.
Let us assume that
T1 = tension in left side string of the pulley.
T2 = tension in the right side string of the pulley
Also if the block is moving with acceleration (a) then the pulley will move down with (a/2) acceleration.
so, here we will just form three equations according to the given conditions -

we can see that tension T1 is acting on the block.
T1 = ma .............(1)

we can also see that due to difference in tension T1 and T2, the pulley will also rotate about its own axis.
(T2 - T1) = Iɑ where ɑ = angular acceleration

Now we are going to solve the third equation. we can see that the pulley is going down due to its weight(m´g) with acceleration a/2
where m´ = mass of the pulley. so, we get
m´g - m´a/2 = T1+T2
where m´a/2 = pseudo force and we know that pseudo force acts in the opposite direction of the motion. That's its sign is negative.
now we have just solved the three equations and calculated the value of acceleration (a).

vishesh

12/4/2015 06:21:39 am

Why u have taken the concept of pseudo force ?we can write m'g-(T1+T2)=m'a/2..

Admin

1/11/2013 07:53:08 pm

Dear Arshdeep,
Solution for Question 38
In this is question two blocks which are attached to each other with a string through pulley. these blocks are resting on an isosceles block of angle 45°.
Block of 4 kg is heavier. so, it will move down with acceleration (a)
and block of 2 kg will move up with acceleration (a).
T1 = tension acting in the string attached with 4 kg block
T2 = tension acting in the string attached with 2 kg block

Here also we will form three equations according to the given conditions -
mg sin Ɵ - ma = T1 .............(1)

where ma = pseudo force
Ɵ = 45°
m = 4 kg

Now 2 kg block is moving up with acceleration a. so, we get
m´ g sin 45° + m´a = T2 .............(2)
where m´a = pseudo force
m´ = 2 kg

Now we are going to solve third equation. we know that the pulley is also rotating about its own axis due to difference in the tension of strings on both sides. so, we get
(T1 - T2)r = Iɑ where ɑ = angular acceleration = a/r
where a = linear acceleration
so, we get
(T1-T2) = Ia/r² ...........(3)

so, solving these three equation we can calculate the value of acceleration a.

himanshu

1/20/2013 02:48:56 am

sir can u help me in question no 40 and 41

Admin

1/20/2013 07:24:41 pm

Dear Himanshu,
In question 40,
there is a rod of 200 g connected to the ceiling through the strings attached to the ends of the rod. A mass of 20 g placed on the rod at point 70 cm from the left end.
Let that point be G and the weight (.2g) is acting at the middle of the rod at .5 m from the ends, let this point is C.
Let
T1 = tension in the left string
T2 = tension in the right string
we will have to two equations to calculate T1 and T2
First equation

T1 + T2 = mg + (m1)g
where m = 200g
m1 = 20 g
This is so because weight is acting downwards which will be balanced by the tension T1 and T2 which are acting upwards.

Second equation

A-----------C------G----------B
Here AB = rod = 1 m
AC = 0.5 m
CG = .2 m
GB = 0.3 m
AG = 0.7 m

To form this equation we will make the use of torque equation.
we know that the body is not rotating which means that sum of all the torques will be zero.
so, let us calculate the torque about point G(the point where 20 g is kept).
T1 x AG - T2 x GB - .2g x CG
T1x 0.7 - T2 x 0.3 - 0.2g x .2 = 0
so, we have two equations now and through this we will calculate T1 and T2.

Admin

1/20/2013 07:52:52 pm

Dear Himanshu,
In question number 41
it is given that a ladder is resting against the wall at an angle of 37˚ with the vertical wall.
The electrician is resting at a distance of 8m from one end.
Now there are two weights acting on the ladder.
first is the weight of electrician( 60g) at 8m from the downward end.
second is the weight of the rod( 16g) at the center at 5 m from the ends.
Now due to these forces there are reaction forces on the ladder. These are
R1 = this is the reaction force from the vertical wall
R2 = this is the reaction force from the horizontal wall
These forces are always perpendicular to the contacting surface i.e walls.
Friction force = µR2 as shown in the diagram
due to the wight of the electrician the ladder will try to slip in the outward direction or right direction.
so, friction force is acting in the left direction.
By the diagram given in the solution we can say that
R2 = (16+60)g
Now calculating R1
we know that ladder is not rotating and the sum of torque about any point on the rod will be zero.
so, we calculate torque about R2 or the end of rod where R2 is acting.
through this we have calculated R1 as given in the solution.

Now to calculate coefficient of friction.
we have kept friction force = R1 as shown in the diagram and calculated µ.

himanshu

1/21/2013 01:18:11 am

In qustion no 41 R1 is coming wrong

Admin

1/21/2013 08:35:16 pm

R1 = µR2 = force of friction by ladder
and the answer is coming correct in the above solutions,
R1 = 336g/8 = 42g = 42x9.8 = 411.6 N(approx)

himanshu

1/22/2013 02:57:56 am

thanx sir but i am finding problem in qustion no 45

Admin

1/23/2013 06:02:15 pm

Dear Himanshu,
In question number 45 it is given that the rod is resting on a smooth roller and making as angle Ɵ
with the horizontal.
Now here you need to see the direction of the reaction forces on the rod.
Here two reactions are there. These are-
R1 = this is the reaction force acting due to the roller on the rod.
R2 = this reaction force is acting at the end of rod due to base.
Now direction R2 is simply perpendicular to the base. But you will face problem in calculating the direction of R1.
Earlier i told you that the reaction force is always perpendicular the contacting surface. we will apply this here also.

We know that the rod is resting on the rod. Here rod is acting as a tangent to the roller.
And we also know that the tangent passes through the point on a circle. So, here a point on a roller is contacting surface. Force on that point should pass through the radius and we know the fact that the line passing though the radius is always perpendicular to the tangent. That’s why the reaction R1 is perpendicular to the rod.
After this we resolved R1 in two directions. These are R1 cos Ɵ and R1 sin Ɵ as shown in the diagram.
Now we simply keep R1 cos Ɵ = R2
R1 sin Ɵ = friction force
Now solving these equation we get the answer.
We are sorry for delaying your answer.

Rajat

1/23/2013 03:06:27 am

A solid sphere rolls on a rough surface. It rolls a distance 'x' metre then what is the loss in its energy?
Mass: m
radius: r
coeff of friction: k
is it right away kmgx??
In hcv it is given that sphere slows down because the point of contact is not perfectly spherical..but if it was pefectly spherical will it roll forever? There would not be any loss in its energy no mattter how large the friction is??

Admin

1/23/2013 06:12:12 pm

Dear Rajat,
Let us again see the concept of friction. we know that friction friction opposes the relative motion between the two surface. Here try to understand the meaning of relative motion. In simple words we can say that relative motion only occurs when there is skidding of one surface over the other.
In rolling there is no skidding so, if the body is rolling purely on the surface then there will be no friction force and the object will move forever. But if the object is moving on a rough surface then there will skidding as well as some rolling due to which friction force will start acting and object will slow down.
Ideally there is no concept like pure rolling.

Rajat

1/23/2013 07:13:03 pm

A sphere of radius R, mass M is placed on an inclined plane of angle theta. What is the minimum height of an obstacle to be placed on the inclined plane so that sphere remain in equibrium?
thank you whole heartedly...

Admin

1/25/2013 01:42:36 pm

Dear Rajat,
In this question you need to visualize the diagram. please visit the link "http://padhaichannel.com/main/rolling-motion-along-an-inclined-plane/". I am going to explain this question according to this diagram only. There is an inclined plane and on that we have kept a sphere. we know that due to its weight it will try to come down. so, we have kept an obstacle of height h in front of it to stop it from moving down. Now you need to imagine that the ball will not slide but it has the tendency to rotate about the upper most point of the obstacle.Let that uppermost point be G. Now we resolve the weight in two directions. These are
1. mg cos θ - its direction is along the inclined plane.
2. mg sin θ - its direction is perpendicular to the inclined plane.
The other forces which are acting are -
1. Normal force(N) = this force is acting opposite to mg sin θ and is balancing it.
2. friction force(f) = you need to imagine the direction of this force. if there is an obstacle and if the sphere needs to overcome that obstacle then sphere will rotate about the uppermost point and will over come the obstacle. now according to the given diagram when the sphere starts rotating then at that instant its lowermost point which is touching the plane is moving towards right. so, to resist that motion the friction force will act towards left. Here f = mg cos θ
3. reaction force due to obstacle (R)= this force is resisting the sphere to move down and it is acting at the uppermost point of the obstacle about which the sphere is rotating.
Now we will apply the torque equation about the uppermost point G.
to stop the sphere from rotating,
f x h - mg sin θ(R - h) > 0
(mg cos θ)h > mg sin θ(R - h)
solving this we get h > mg sin θ/(mg sin θ + mg cos θ)
so. minimum h = mg sin θ/(mg sin θ + mg cos θ)

pramod

1/27/2016 06:23:44 am

Hey , given answer is (R-costheta) please explain

Sridhar

7/4/2017 10:46:09 am

how is the perpendicular distance between the force mg sin θ and the point G is (R-h)?

Himanshi link

1/30/2013 01:49:34 am

Sir can u help me in d ques 49

Admin

1/30/2013 08:02:46 pm

Dear HImanshi,
In this question we have just applied the formula
angular momentum = Iw
where I = moment of inertia
w =angular velocity
Here there are two masses m1 and m2 having distance r between them.
so, the center of mass of the system will lie at
r1 = (m2)r/(m1 +m2)
where r1 = distance between m1 and center of mass
r2 = (m1)r/(m1 +m2)
where r2 = distance between m2 and center of mass
I1 = moment of inertia of m1 = m1(r1)²
I2 = moment of inertia of m2 = m2(r2)²
total angular momentum = I1 w + I2 w ..........1

Now we put the values in eq 1 and rest is done in the solutions

Prateek

2/12/2013 04:58:35 am

sir..in the ques we are askd to take the axis as the centre of rod or system..but why are we considering centre of mass in our solutions??..if the masses are unequal the centre of rod and centre of mass qont coincide..

Admin

2/12/2013 08:52:57 pm

Dear Prateek,
First of all the rod is considered light weight whose mass is neglected.Also in the question it is given that the system rotates about the center of mass of the system. so, in the system we need to consider the masses m1 and m2. as the masses are unequal so, the center of mass will not lie at the center.

Prateek

2/12/2013 04:55:04 am

Sir isnt the ques 83 conceptually incorrect..the sphere cant possible move linearly on a smooth surface..as body moves forward only due to friction..!!

Admin

2/12/2013 09:03:22 pm

Dear Prateek,
Its not always the case. You would be thinking that concept of walking that we walk with the help of friction. Yes friction helps in the movement but its not always the case. You just need to see that if the force is acting in any direction then the object will move in that direction until and unless any other force is canceling the above force.

manju

2/12/2013 04:54:52 pm

plz expln Q: 15

Admin

2/12/2013 09:31:54 pm

Dear Manju,
In this question we have to find the moment of inertia of square about the diagonal.
So, please the diagram given in the solution. I will explain according to that diagram.
First we have calculated the moment of inertia about xx = ma²/12
Now we know that I(zz) = I(xx) + I(yy)
perpendicular axis theorem.
Now I(xx) = I(yy)
I(zz) = 2 I(xx) = ma²/6
Now we can also write I(zz) = I(x'x') + I(y'y')
where x'x' and y'y' are the diagonals and are perpendicular to each other and z axis.
I(x'x') = I(y'y')
Now I(zz) = 2 I(x'x')
I(x'x') = ma²/12

Amrish Rai

3/24/2013 05:41:38 pm

sir,can we do Qno. 66 by Torque=I*(alpha)? if can,pls provide the solution.

theyatin

3/24/2013 07:35:18 pm

dear as per my perception you cant solve this question by using torque
if you were asked to find angular acceleration or angular velocity only then you may use torque in solution.

Amrish Rai

3/27/2013 01:34:09 pm

we have been asked velocity and radius is also provided even the moment of inertia of the disc or pulley rotating is given. so, why we can't use Torque? plz explain.

theyatin

3/27/2013 03:32:50 pm

dear amrish.
because the velocity of box is to be determined and beacause motion of box is linear so we have to convert all the given data in linear values rather than angular.
if you were asked to calculate angular velocity of pulley then you may convert all the equations into angular values and thereafter use torque instead of force likewise we have done here. . .

sreerag k v

3/28/2013 06:53:53 pm

sir,in question 80 why friction is avoided...?? it is a spherical shell so i think friction will be necessary to roll without slipping....am i correct sir..? sir i think if we are considering an axis passing through c.o.m perpendicular to plane. how net torque became zero..sir what hapened to torque due to friction...plz.. answer me...sir how can i do this question by considering instantaneous axis of rotation...?

theyatin

3/29/2013 12:36:38 am

yes dear, you are right to roll a spherical object you need friction but if sphere is completely round and mass is equally distributed so that center of mass lies exactly at the center of sphere
the friction required to roll the sphere is very very small as it can be neglected.
also in this case,before strike of cue the torque is zero
and torque does not relate with friction

theyatin

3/29/2013 12:40:51 am

yes dear, you are right to roll a spherical object you need friction but if sphere is completely round and mass is equally distributed so that center of mass lies exactly at the center of sphere
the friction required to roll the sphere is very very small as it can be neglected.
also in this case,before the strike of cue the torque is zero
and torque does not relate with friction.

3/31/2013 10:56:19 pm

Sir in the last worked out example of rotational motion, how can we conserve linear momentum even though friction is present between sphere and ground...can we solve it by conserving angular momentum along the point of contact bt. sphere and ground ??

theyatin

4/1/2013 03:26:15 pm

dear
i can't understand what do you want to ask. . .
if the friction was prominent enough then the ball would never retain its rotation in clockwise direction
so friction is negligible

Monika

4/4/2013 08:23:30 pm

sir please explain me short question 18, 19, 20

thayatin

4/6/2013 01:00:47 am

dear
Q18
as density increases from A to B then mass would also increase from A to B
thus as rotation needs lesser force if more mass is near center of mass
so at b more mass though more weight will be concentrated thus clamping at B and applying force at A would be easier. . . that means lesser force would be needed

theyatin

4/6/2013 01:07:08 am

dear in Q19
well in case tal buildings are in order of Km like burj khalifa or much more than that as radius of earth is 3600km
then mass of earth will be shifted outwards slightly so rotation of earth will slow down very slightly
so as it will take slightly more time for one rotation so time duration of day and night both will increase a very slightly

theyatin

4/6/2013 01:12:31 am

dear in Q20
as radius of earth is more at equator and lesser at poles by melting ice and water flowing towards equator similarly in Q19 duration will increase very very very slightly increased.

Rama

4/4/2013 08:25:32 pm

Objective 1 - q11, 12

theyatin

4/6/2013 01:17:30 am

dear in Q11
as moment of inertia is directly proportional to radius^2 and doesnot depend upon thickness
so Ia<Ib
as radius of B=4A

theyatin

4/6/2013 01:23:07 am

dear in Q12
if equal torque is appplied on both discs
as M.I. of A is lesser so for same torque applied disc A will have faster rotation.
as M.I. of B is 16 times as of A
so A will have 16 times more rotation
thus even though parameter of A is 1/4th of B velocity at the rim wouldbe stil approximately 4 times as of B
hence Va>Vb

Rana Vishwajeet

4/28/2013 10:27:41 pm

please explain me 3rd paragraph of page no.169 of hc verma.how can torque about any point of axis of rotation is same(since,r is changing). Please explain

theyatin

4/29/2013 12:50:42 pm

please read the line that "the torque of force about a line is independent
of the choice of origin as long as it is chosen on the line". so, in order to solve, choose the distance which is shortest and perpendicular to the force. it is not compulsary that it should start from origin. but it should not go beyond origin

Rajat

4/29/2013 01:43:41 pm

Please explain the direction of friction acting on a rolling sphere when:
1) it rolls down the inclined plane
2) it rolls up the inclined plane

& consider a case when a sphere is rolling on a horizontal surface . We make a point on it using a pen. Then:
1) acceleration of that point ( centipedal, tangential, its direction, it remains constant or changes with time, is its magnitude constant?)
2) velocity of that point ( similar arguments)

i have seen these type of questions in almost every jee adv sample paper but could not find a satisfactory explaination....u can only help me..

theyatin

4/29/2013 11:58:07 pm

Dear Rajat,
whenever we try to find to direction of friction. Always try to see the tendency of the object to slide and skid because sliding and skiding are the form of relative motion.
now when the sphere is rolling down the plane so it has the tendency to skid down the plane due to its weight. so, friction is acting upward.

Now when the sphere is going upward again the friction will act upward because still it has the tendency to skid backward or you can also understand by the example of walking. you know that when we walk we apply force in the backward direction and friction force pushes us forward. so, when the sphere will move up the plane you will see that at the point of contact the sphere is trying to move the plane in the backward direction. so the plane will push the sphere in the forward direction by applying friction as a reaction force.

also i am able to understand ur 2nd question. please explain in detail.

Rajat

4/30/2013 03:22:33 am

Thank you sir..
in the second question we are observing a particular point on the rolling sphere...as the sphere rolls (on horizontal surface) how acceleration & the velocity of "that" point change with time...
some questions mention centipedal acceleration & tangential acceleration of the point on the rolling sphere which i am not able to visualise...

theyatin

5/2/2013 12:18:23 am

Dear Raman,

let us take an example that we are rotating a stone which is tied with a string. now at some instant let the string breaks. Now you will see that it will go in the tangential direction. this means it was having some force which was having tangential direction. so this is all about tangential acceleration.

now lets come to centripetal acceleration. we know that if any object is moving in a circle than it must be having centripetal acceleration which always acts towards the center along the radius.

Now we get the net acceleration which the vector sum of centripetal and tangential acceleration.

aman sangal

5/4/2013 12:56:40 am

can you please explain question no. 80. in that question you have taken initial angular momentum equal to MVcH.shouldn't MVcH be the the angular momentum if it is a ring. with a spherical shell shouldn't it be 2/3MVcH

theyatin

5/4/2013 02:31:00 pm

dear,
read line carefully it says angular momentum at point a. . .
if you can imagine at point a there is a ring .
(cut sphere as point a)

aman sangal

5/4/2013 10:42:57 pm

what is point A

theyatin

5/4/2013 11:38:04 pm

dear,
point a is where cue hits the ball as it is away from mid point consider it as a ring of breath h (from mid point to a).
cut sphere from mid point and from a you will get a ring.

aman sangal

5/5/2013 01:50:01 am

thank you so much

jai

5/6/2013 01:10:19 pm

in question 41(Rotational mechanics), why is the normal force exerted by the electrician on the ladder not perpendicular to the ladder surface?

theyatin

5/10/2013 03:22:16 am

dear. . .
because ladder is slanted and the force of electrician is downward due to his weight thus the normal force may be in upward direction with some angle depending upon the angle of ladder to the horizon. . .

rohit

5/22/2013 06:14:21 pm

hi, i dont have any queries . I just want to thank the site for such a great work. This has helped me a lot in physics.

arjun

5/30/2013 06:20:40 pm

why the tension is different in the string having two different block at its free end when the string is passes through pully having mass whether
the accelaration of the blocks is same

pls expln eg no 8. & why in eg no 5 the block deaccelarates

theyatin

5/30/2013 10:59:11 pm

dear,
your question is not clear to me please ask again with more information.

theyatin

5/30/2013 11:15:18 pm

dear,
there are three main forces acting on ladder,
on due to wall N1
one due to floor N2
and one due to friction.
N1 is equal to frictional force of ladder thus it doensot slides down.
N2 is equal to weight.
that is N1=f
N2=W
as torque at both ends are equal thus ladder doesnt slides
hense N1(hieght)=N2=(hieght of weight))
N1*(AO)=N2*(CB)
where W=10*9.8=98
so you can solve rest of the question.

arjun

6/4/2013 12:48:18 am

why the tension is different in the string passing over a pully having inertia(having mass) refer to eg no 6

theyatin

6/4/2013 01:09:05 am

dear,
as mass attached to both strings is different thus tension on both strings are different.

rakesh

6/8/2013 02:48:27 am

sir can u explain me Q15 from question for short answers

theyatin

6/9/2013 04:20:24 am

dear,
as for a fat person most of fat is on upper body
when the person tries to touch his toes his upper body leans forward
so the center of mass of body
thus the person tends to fall forward.

Swarnabha Das

7/9/2013 04:53:39 am

Sir,
In Chapter 7, I am having a doubt in Question number 18.
The Question is : "A turn of radius 20m is bankedfor the vehicles going at a speed of 36km/h. If the coefficient of the static friction between the road and the tyre is 0.4, what are the possible speeds of vehicle so that it neither slips down nor skid up".

In this question it is asked to Find out the possible speeds of vehicle so that it neither slips down nor skid up. Here it is possible to find out The maximum velocity by using the formula (√(rg (μ+tanθ)/(1-μtanθ)). But in the HC verma Solve to find out the minimum velocity this formula has been used : √(rg (tanθ-μ)/(1+μtanθ)).
Will You plz tell me that how the second formula has been derived?
Before Seeing the solve I could not understand the way of finding the minimum velocity.
Plz Explain The derivation of the second formula.

Thank You.

Sanghati Sussama

8/14/2013 10:43:15 am

dear sir,
Could you please explain Q.26,Q.5,Q.9 and Q.10 of Objective 1 in chapter 10?
In Q.26,I cant understand that how can the length of the string passing through the hands of the man while the cylinder reaches his hands be 2l instead of l?

abhishek

9/21/2013 10:28:07 pm

sir,
please help me in solving question number 10 of the exercises,i am stuck there from long time..

theyatin

9/22/2013 03:44:18 am

dear,
for first two particles
49*1^2+51*1^2
for seconda adjacent particles
48*2^2 + 52*2^2= 100 (2^)
force third adjacent particles
47*3^2 + 53*3^2= 100 (3^2)
and doing it so on
...
1*49^2+ 99 *49^2= 100 (49^2)
and finally 100*50^2 (last remaining particle)
now
100 (1^2+2^2+3^2.... +49^2) + 100 * 50^2
hence solving this we can get answer.

abhishek

9/21/2013 10:28:18 pm

sir,
please help me in solving question number 10 of the exercises,i am stuck there from long time..

abhishek

9/21/2013 10:28:31 pm

sir,
please help me in solving question number 10 of the exercises,i am stuck there from long time..

divesh

10/1/2013 01:56:31 pm

sir in ques 53 my answer is coming 12 pi but answer is given only 12. how????

divesh

10/1/2013 01:58:23 pm

sorry, i got the answer. I just didnt converted it.

abhishek

10/11/2013 12:54:17 am

if a hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane, the which sphere will have a greater kinetic energy on reaching the bottom ?,how? Please explain

theyatin

10/12/2013 03:48:21 am

dear,
as the kinetic energy depends upon the mass of the object
and angular velocity depends upon radius of the object here both are equal,
thus kinetic energy is same.

utkarsh

10/13/2013 02:48:45 pm

sir, please explain example 15
its given that loss of energy of the system is=Mgh-mgh
but it should be mgh-Mgh since the M is going down so there is loss o potential energy while m is going up so gain in potential energy.

theyatin

10/15/2013 03:15:03 am

dear,
as M is going downward and m is going upward
g is negative for m thus Mgh-mgh is right term

Aymanzoor

10/24/2013 11:13:39 pm

Sir in q17 why do we take r = range/2
instead of taking sqrt {(range/2)^2 + (Height)^2}
which is the distance between the particle and the point of projection

theyatin

10/26/2013 03:02:49 am

dear,
if we consider highest point in a trajectory it is at half of range of the particle horizontally that is why range/2

Aymanzoor

10/26/2013 03:29:40 am

Yes
but torque=r x F
where r is the distance between the tail of the vector notation of force
and the point about which torque is calculated
so why do we use range/2

Aymanzoor

10/26/2013 02:18:10 pm

sir
in q61 part d
why do we take for the rod
L=mvr
isnt this applicable only to smalll masses over large radii
how is this correct

Aymanzoor

10/26/2013 10:30:11 pm

Sir
in q84
why are friction and the force taken in the same direction shouldnt they both produce torques in the same direction
and hence be applied in the opposite directions

Aymanzoor

10/26/2013 10:37:05 pm

Thanks but i got it

SHIVANI SHARMA

10/28/2013 03:48:26 am

sir,in ques. 86,we are calculating angular momentum,then first of all about which point we are calculating it and if we are calculating it about centre then the v and r are in the same direction so angular momentum should be zero and second thing sir,initially there was no pure rolling,so why w=v/r?

theyatin

10/30/2013 05:11:27 am

dear,
In this question there is also the same case as above.
Initially the ball was rolling in the clockwise direction but when the rolling ball hits the wall, due to elastic collision its linear velocity reverses but the ball is still having angular velocity in the clockwise direction. then same thing will happen it will resist the motion for sometime then it will become zero and ball will rolling purely having linear and angular momentum in the same direction

SHIVANI SHARMA

10/30/2013 05:30:16 am

ok sir,that means this question is related to the previous question,but sir what about the angular momentum which was mvr,but v and r were in the same direction?

SHIVANI SHARMA

10/28/2013 08:30:10 pm

sir please help me and please reply me .

theyatin

10/31/2013 02:51:23 am

dear,
r dont have direction
but its true that velocity is in forward direction
thus at some point angular momentum would be zero

SHIVANI SHARMA

10/30/2013 05:34:04 am

sir this is a general question-when a ball goes up,it falls towards earth due to gravitational pull but then why does moon revolve around earth as on moon there is also gravittional pull,wh this serves as centripetal force?

theyatin

10/31/2013 02:45:53 am

dear,
it is case when centripetal force balances centrifugal force. . .
thus it only revolve around earth but not fall on it

pravesh

11/4/2013 10:10:49 pm

i want question also can i get it

theyatin

11/4/2013 11:25:05 pm

dear,
which questions???

sunny singh

11/13/2013 10:10:56 am

i can't understand Q.no 33

theyatin

11/17/2013 02:44:45 am

dear,
first we need to find tension in both parts of systems
then we need to see that the change in tension is due to torque acting on the pulley
so T1-T2= Ia/r^2
so comparing first two equations and putting value of T1-T2 we can have value of a of the system.

Avith

11/15/2013 03:21:31 pm

can you plz explain Q no 81

theyatin

11/15/2013 05:42:18 pm

dear,
in this question a wheel roles first freely on its axis then i lands on rough surface it starts moving ahead ad decelerates
angular momentum in free rotation= 1/2 *MV^2 w
then rotation with forward velocity= 1/2 *MV^2 (V/r) + M(V*r)
=3/2 mVr
equating both as its contatnt
1/2 *MV^2 w=3/2 mVr
V=w r/3

kunal

11/19/2013 08:15:25 pm

can u help me in q 56

theyatin

11/22/2013 12:58:09 am

dear,
A kid of mass M stands at the edge of a platform of radius R which has a moment of inertia I. A ball of m
thrown to him and horizontal velocity of the ball v when he catches it.
Therefore if we take the total bodies as a system
Therefore mvR = Iw + mR^2w + MR^2w
={I + (M + m)R^2w
now we can find , w

Divyanshu Sahu

11/20/2013 05:38:22 pm

I am not able to understand question no. 37

theyatin

11/22/2013 12:26:54 am

dear,
que 37
Let us assume that
T1 = tension in left side string of the pulley.
T2 = tension in the right side string of the pulley
Also if the block is moving with acceleration (a) then the pulley will move down with (a/2) acceleration.
so, here we will just form three equations according to the given conditions -

we can see that tension T1 is acting on the block.
T1 = ma .............(1)

we can also see that due to difference in tension T1 and T2, the pulley will also rotate about its own axis.
(T2 - T1) = Iɑ where ɑ = angular acceleration

Now we are going to solve the third equation. we can see that the pulley is going down due to its weight(m´g) with acceleration a/2
where m´ = mass of the pulley. so, we get
m´g - m´a/2 = T1+T2
where m´a/2 = pseudo force and we know that pseudo force acts in the opposite direction of the motion. That's its sign is negative.
now we have just solved the three equations and calculated the value of acceleration

Sidharth

1/23/2014 02:14:09 pm

why did we take acceleration of 2nd block as a/2?

darshan

11/20/2013 09:41:14 pm

In question 8 of rotational mechanics I can't understand how 20 multiplied by 20 is 200

theyatin

11/22/2013 12:30:56 am

dear,
its typing mistake its rw where r-10 and w-20
that is 10*20=200

darshan

11/20/2013 09:48:14 pm

Please explain question 10 of rotational mechanics

theyatin

11/22/2013 12:38:48 am

prahaasa

11/20/2013 11:19:02 pm

can you please explain Q63 and 64 in detail.

kunal

11/24/2013 05:03:45 pm

can you explain q 67

ekansh gupta

11/24/2013 11:28:47 pm

can u help me to prove that the collision is elastic in Q60

ekansh

11/24/2013 11:30:25 pm

please

shivam

12/13/2013 01:07:47 pm

please explain q.no 2 ,14&20

shubhu

12/15/2013 06:38:19 pm

In que-2 it says 50 rev. Which means 100pi of total angle covered (1rev.=2pi) use motion equations . And. convert the quantities in required form
Q14- using I=MK² .
Use parallel axes theorem . Let x be the required distance then MR²/2 +Mx²= MK²[ k= R] solve !!

theyatin

12/17/2013 01:48:57 am

nice spirit shubhu.
keep helping other friends this will also add up your knowledge

theyatin

12/17/2013 01:43:56 am

dear,
Q20
you need to find torque=force *distance
now you can compute torque for each point from central point O
for this you need to find perpendicular distance using right angle
and now find resultant torque using negative torque for anticlockwise torque and positive for clockwise torque. . .

shivam

12/17/2013 01:08:40 am

sir please tell me some other websites for chemistry similar to yours where can i clear my doubts!!!

theyatin

12/17/2013 01:45:47 am

dear,
i will look for on website good enough to study online
till then you can ask us we can help somehow.

Alok Kumar Sharma

12/18/2013 09:42:32 am

Sir, I am not able to solve a problem......please help!!!!!!. It is not from H C Verma, it's from D C pandey-Mechanics Part 2. The question is somewhat like this:-
A uniform rod of mass"m" and length "l" is piovted at point "O". The rod is initially in vertical position and touching a block of mass "M" whcih is at rest on a horizontal surface. The rod is give a slight jerk and it starts rotating about point O. This causes the block to move forward. Thr rod loses contact with the block when it make an angle of 30(degrees) with horizontal. All surfaces are smooth. Then:-
The value of ration M/m is

theyatin

12/19/2013 11:00:46 pm

dear,
the question is not much clear. . .

shivam

12/24/2013 10:22:18 am

please explain q.no 28

theyatin

12/24/2013 11:39:21 pm

dear,if T is torque by which wheel stops then it is equal to work done on the wheel
so F*r= T= I*a (where a is angular acceleration.
solving this we can have force F.

shivam

12/24/2013 10:57:22 am

inq.no 57 please explain how w(omega) of ball and platform be same

shubhu

12/24/2013 12:41:29 pm

Dear when ball is thrown by man instantaneously after that man +platform rotates with common angular velocity. as it is not said that platform is frictionless

shivam

1/4/2014 11:17:26 pm

please explain how v=rw in pyre roll

theyatin

1/8/2014 02:31:26 am

dear,
please ask question in detail. it is not clear for me what you want to know

shivam

1/4/2014 11:19:24 pm

sir its a mistake not pyre its a pure roll

saurav

1/19/2014 09:56:37 pm

sir, i am not able to understand q.10 of rotational mechanics ,please help.

theyatin

1/20/2014 02:12:28 pm

vish

1/21/2014 11:28:27 am

sir i want questions for short answers of ch 10. but it is not found on this website plz upload it

theyatin

1/23/2014 01:13:48 am

dear,
we will upload answers soon please wait or ask questions one by one

Ansh Arora

1/23/2014 05:22:52 pm

In Q 76(b) can we use work energy theorum as stated below -

W(gravity) + W(friction) = Change in KE

mglsinθ-(mglsinθ/5) = KE

Harish

1/27/2014 02:36:53 pm

Sir i am not able to understand the solution to the 20th question in the Exercises section. How did you take the torques of each the forces and their components but not the one with the force of 5N which is at 30 degrees?

theyatin

1/29/2014 02:51:07 am

Manshi Tomar link

2/21/2014 11:31:49 pm

sir, wht v hav to do with Q. no.78 of ch.10...?
should i aply energy concept there.

theyatin

2/25/2014 03:28:37 pm

dear,
at top most point weight acts as centripetal force thus
mv^/R=mg
now balance conservation of energy.
we can get v'

waka

3/8/2014 04:41:04 pm

A ring of radius R is rolling purely on the outer surface of a pipe of radius 4R. At some instant the center of the ring has a constant speed =v. Then the acceleration of the point on the ring which is in contact with the surface of the pipe is.....?

manish

3/9/2014 10:46:06 pm

i want to ask the explanation for the single string bicycle wheel experiment that it rotates in the plane as well as rotates around the string ....

theya

3/9/2014 11:52:48 pm

Manish could u elaborate what u actually wanna ask that is which wheel and how is it connect ...so next time do clearly ask..

kumar shivam

3/10/2014 04:00:21 am

manish

3/11/2014 09:10:18 pm

sorry for that.... for the wheel experiment refer class xi ncert physics part - one , chapter no. 7 , page no. 156......

THEYATIN

3/19/2014 12:52:26 am

dear please write complete question here for particular solution.

airpit aggarwal

3/16/2014 12:49:04 am

Sir plz explain this question from ....waka09/03/2014 1:41pm
A ring of radius R is rolling purely on the outer surface of a pipe of radius 4R. At some instant the center of the ring has a constant speed =v. Then the acceleration of the point on the ring which is in contact with the surface of the pipe is.....? Actually i solved it and my answer is 4v^2/5R...is it correct....plz give the detail solution ...also..i thank for ur good work...

theyatin

3/19/2014 01:00:27 am

dear,
your answer is correct.
would you mind explaining how did you get your answer..

shubhu

3/20/2014 12:29:03 am

how u arrived at at the answer ??? explain

waka

3/24/2014 06:27:54 pm

Sir, How did u arrive to the conclusion that his answer is correct..if this answer is correct . I thought at it ...and realized that he must have used the concept of centripetal force..and using concept of no slipping...also angular momentum conservation can not be done....in this case

waka

3/24/2014 06:32:21 pm

Sir do reply ...is any above mentioned concept r wrong...

swadhin agrawal link

4/1/2014 03:52:11 pm

sir,
In a rotating body, a=alpha x r &v=wr. thus a divided by alpha=v/w. can you use the theorems of ratio & proportions to write (a + cx) /(a - cx) = (v+w)/(v - w). & if yes then why. if not then why?
where cx means alpha.

shivani sharma

4/8/2014 03:12:55 am

MY ANSWER IS NO BECAUSE WHENEVER WE ADD OR SUBTRACT 2 QUANTITIES ,THE UNITS OF BOTH THE QUANTITIES SHOULD BE SAME BUT HERE THE UNITS ARE NOT SAME

swadhin agrawal link

4/8/2014 08:18:53 pm

Thankyou friend I hsve some more questions would you help me?

shivani sharma

4/9/2014 04:12:55 am

yes,sure I will definitely try.

swadhin agrawal link

4/9/2014 06:26:28 pm

Thankyou,
Plz explain me the following questions-
Chapter 10- short questions-
No.-5,8,9,19,20
Acc. To me in que. No. 5
It is possible from both the places as both of them revolve as eell as they rotate.
In ques.no 8
If there will be some part of that body 50cm apart then only it is possible.
In no 9
I have done the first part but I want to recheck it through yours way, & second part I dont know.
In no.19
Since the total mas of earth is same & the change in radius is also negligible hence I hope it will not be affected as moment of inertia will be same
& in ques.20
After the ice melts I got two conditions
1- if we consider earth as a disc then it will be affected.
2-if we consider mass & change in radius as negligible then wil not be affected.
& I have many more doubts. I hope you will enjoy my doubts.

shivani sharma

4/9/2014 10:53:10 pm

in quest.5,you are right,we cannot see all the faces.
quest.-8, angular acceleration is produced only when the torque is acting along the axis,but here the torque is perpendicular to the axis,so there is no torque produced along the axis.(SEE PAGE NO.-169,CASE 3)
QUES.19,I think that it will affect it because the moment of inertia of buildings get also added and the angular momentum is conserved,so it will decrease w(omega) slightly.
quest.20,it will not bring any change since there is no change in moment of inertia as earth is a sphere.

shivani sharma

4/9/2014 11:00:50 pm

if we consider earth not as a sphere in quest.20 ,then answer will be different as earth is actually oval-like

swadhin agrawal link

4/10/2014 02:30:14 am

But in ques 5 why we cannot see all the faces.
And in 19 why we are addin up the moment of inertia but not in 20. I think as the whole mass of earth is same so moment of inertia should not change.think upon it .
And thanks gor rest of the ans.

kushal khanna

4/3/2014 06:31:55 pm

sir kindly explain question no.11 and 7 of objective 2
in question 11 i think that the sphere can not roll on the smooth horizontal and inclined surface since there is no friction to counter any type of motion...plzz explain

Anshul Bansod

4/18/2014 08:17:00 pm

Q-37

ER.MIRJAIK

5/9/2014 02:21:33 pm

que 37
Let us assume that
T1 = tension in left side string of the pulley.
T2 = tension in the right side string of the pulley
Also if the block is moving with acceleration (a) then the pulley will move down with (a/2) acceleration.
so, here we will just form three equations according to the given conditions -

we can see that tension T1 is acting on the block.
T1 = ma .............(1)

we can also see that due to difference in tension T1 and T2, the pulley will also rotate about its own axis.
(T2 - T1) = Iɑ where ɑ = angular acceleration

Now we are going to solve the third equation. we can see that the pulley is going down due to its weight(m´g) with acceleration a/2
where m´ = mass of the pulley. so, we get
m´g - m´a/2 = T1+T2
where m´a/2 = pseudo force and we know that pseudo force acts in the opposite direction of the motion. That's its sign is negative.
now we have just solved the three equations and calculated the value of acceleration

Archie

5/8/2014 03:43:53 am

Q 54

ER.MIRJAIK

5/9/2014 04:05:18 pm

DEAR ARCHIE
From the surface of earth reference the umbrella has a angular velocity
greater than that of the angular velocity of the earth reference on which a man is standing because the angular motion to the umberla
is caused by man so equating the two angular vel. we can get angular vel. imparted to the man

shreyansh

6/3/2014 01:18:15 am

qus. 32(b)

Pratyush

6/11/2014 02:50:00 am

In q21 they have asked about the torque of normal force about centre shouldn't it be zero and if not plz explain why not

admin

6/22/2014 01:44:58 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

iod

7/1/2014 09:53:31 pm

sir ques n. 61 part (b) please explain.........

Admin

7/2/2014 04:35:47 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

Satya

7/15/2014 11:12:47 pm

When a fat person tries to touch his toes, keeping the legs straight, he generally falls, Explain.

Satya

7/15/2014 11:14:11 pm

When a fat person tries to touch his toes, keeping the legs straight, he generally falls. Explain.

Top Gun

7/26/2014 07:08:13 pm

This happens because when the fat person tries to touch his toes , there is a change in the position of the COM of his body.
There is now an unbalanced torque due to his own weiight acting on him due to which he develops a tendency to fall.

ashish

8/29/2014 07:46:48 am

why is force acting on dm due to gravity i.e its weight dm x g not considered in problem no 69

Sumedha Ghoshal

9/11/2014 12:03:32 am

i cant understd part b c f of ques 61

tanmay shukla link

9/13/2014 01:58:51 am

in question no. 21 i cannot understand why there wont be any torque because of mgcos theta

SHIVANI SHARMA

9/13/2014 08:00:51 pm

torque is r x f x sintheta,but angle between r and mg cos theta is 180,so no torque is there.

Rutvik Patel

1/1/2016 08:27:56 am

yaa that's true shivani.
i found your comments nice
i think you are intelligent.
if you agree then give me your phone number..
i wanna talk abount some concepts of physics

Brock

9/24/2014 10:51:44 pm

in a chain-plate beam balance that we usually see in markets , when an iteam of mass say M kg is placed in one plate , the beam balance falls on the side of weight. to check the weight the salesman put stone of mass M kg n this balances the plates at the same level.
The question is-what is the restoring force which brings the other plate to the level of the other plate?
all torques are balanced and there is no force couple.
what is the reason of this?

unnamed link

10/25/2014 12:45:44 am

Sayali

11/8/2014 03:29:40 am

Sir, can you please explain Q. 78.

Amy

11/12/2014 08:21:20 pm

Sir plz explain me ques no. 65 in ch 10

Kanad

11/28/2014 02:05:09 am

A person sitting firmly over a rotating stool has his arms streched.If he folds his arms, his angular momentum about the axis of rotation decreases.Explain

shivani sharma

11/28/2014 11:03:00 pm

angular momentum is L=I x ANGULAR VELOCITY
so when the person folds his arms, the angular velocity decreases but moment of inertia remains same
thus the angular momentum decreases

Mike

12/3/2014 12:52:55 am

Could you explain the 2nd part of Q.no 76 in chapter 10 ????

shashank

12/6/2014 01:10:18 am

Sir plz xplain ques no 43

shivani sharma

12/24/2014 03:54:01 am

in the 1st part of this question,we have calculated the torque due to the weight of the body
torque= r x f
=r x f x sin(theta)
here,r is taken from the centre where mg is acting,so r=6.5/2
f=mg
sin(theta)={1- (6/6.5)*2} *1/2
so calculate torque from this

shivani sharma

12/24/2014 03:57:46 am

the ladder is in translational as well as in rotational equilibrium,so equate the vertical forces
mg=R2
similarly equate the horizontal forces also

prajwal

12/7/2014 06:43:10 pm

sir when i am not get the answer of question then what can i do

shalu

12/8/2014 01:43:18 pm

sir, please explain me question no 78

pinki

12/12/2014 02:15:08 pm

respected sir/madam,
could you help me in the 6th and 7th question of objective-1??
thanks

mrunal

12/16/2014 02:12:06 pm

how to solve objective 10,15 in objective 1

konark

1/9/2015 04:22:49 am

Acc to me q no 37 ans should be (5g)/7.
i did it by assuming vel. a1 and a2 of pulley and block after which i could relate a1 and a2.
later i assumed tensions T1 and T2 and wrote eqns. for translational motion of block and pulley and also torque=I(alpha).
I got the mass of the pulley=5kg by I=m×rsquare
Please help me!

yash parkar

1/22/2015 01:39:41 am

Sir can u help me to solve question number 21 of Chapter 10-Rotational Mechanics . Of Exercise .

Shantanu

3/7/2015 11:31:25 pm

Sir can u explain Qno 16

wafdff

4/24/2015 11:39:29 am

sdSd sfFsd

Sir plz explain q. 72

9/18/2015 10:59:20 am

pulkit

10/13/2015 08:45:12 am

Explain Q77 of rotational mechanics why
mg+mv^2/r and not mg -mv^2/r because mg is downward and mv^2/r is upward??

tanya

11/9/2015 08:34:13 pm

what are the conditions for applying conservation of angular momentum?

Rutvik patel

12/16/2015 06:50:08 am

hcv is an easy book
not much hard
instead of hcv try B.M. Sharma published bt cyngage publication.
if u solve B.M. Sharma u r definitely going to crack JEE advance paper. and u will become an IITian too

sayli

12/29/2015 09:56:05 pm

Sir, in Q 82 HCV rotational mech. Iam confused about how can we conserve angular momentum about the lowest point as frictional force is acting at that point?? and also why is the direction of linear momentum shown in direction of the velocity of com

vanshi

1/15/2016 01:53:20 am

please can someone help me with ques 40-45.I am unalbe to understand any of them

VANSHI

1/15/2016 09:33:44 am

Can u help me with ques 69 too.

2/6/2016 08:36:06 am

Rutvik Patel want to try difficult book try
Ie irodov's general physics and krotov's application of...

pallavi

3/7/2016 06:42:33 am

Sir can u tell the solution of Q17,18,19

Koulick Sadhu link

10/28/2016 10:23:11 am

I am stuck in Question no.44 of Rotational Mechanics Of chapter 10..plz gave a detailed explanation of the question...

Aaron Goyal

12/29/2016 09:16:35 am

HC VERMA Vol 1 Page: 192 ;Rotational Mechanics
Questions for short answer
Q9,18

Abhijeet

2/2/2017 07:39:50 am

Sir ,in HCV ch 10 worked out example no 25 , can't the solution be found by conservation of angular momentum about the axis of any one cylinder?I tried but got a different answer than the one given in book.Thank you

Raunak

2/23/2017 11:11:39 pm

Sir
Please tell question

9th. 17th. And 20th
Of hcv rotation chapter
Question for short answers

Austin

3/9/2017 06:39:08 am

Sir.,,In chapter 10 ques. no. 10 the MOI of the lone 50th particle should be 50gm x (50)^2 cm^2

Harsh Agarwal

9/29/2017 03:10:05 am

Sir there is a mistake in solution of hc verma chapter 10 question 36:
When we added equation 2 and 3 we will get 2T (2)-T(3)-T (1)=2I/R^2

Chapterwise Solutions of HC Verma