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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.

Reply

Harsh

15/09/2013 10:43am

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

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?

Reply

Admin Answer

31/10/2012 5:02am

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.

Reply

faculty

14/04/2013 7:53am

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.

Reply

abhishek choudhary

20/09/2013 8:42pm

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

Reply

Admin

03/11/2012 10:08pm

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

Reply

Prashant

16/11/2012 7:06am

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

Reply

Admin

16/11/2012 7:49pm

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.

Reply

Admin

18/11/2012 3:13am

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.

Reply

satish

30/11/2012 3:28am

can u explain Q 50

Reply

Admin

30/11/2012 4:30am

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.

Reply

Raman

05/12/2012 6:41am

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 .

Reply

Admin

05/12/2012 10:22pm

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.

Reply

Raman

05/12/2012 6:47am

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.

Reply

Admin

06/12/2012 6:24pm

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.

Reply

Raman

06/12/2012 8:09am

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.

Reply

Admin

07/12/2012 8:47am

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.

Reply

Raman

07/12/2012 5:57pm

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

Reply

Admin

07/12/2012 7:18pm

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.

Reply

Raman

08/12/2012 6:32am

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

Reply

Admin

08/12/2012 6:45pm

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.

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.

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.

Reply

Admin

26/12/2012 9:44pm

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

Reply

Shreyas

26/12/2012 11:34pm

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

09/11/2013 9:49am

is this book..helful fr clearing medical examzzz

Shreyas

26/12/2012 5:52am

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.

Reply

Admin

26/12/2012 9:51pm

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.

Reply

Shreyas

26/12/2012 11:43pm

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

27/12/2012 10:14pm

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.

Reply

Shreyas

27/12/2012 10:50pm

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???

Reply

shreyas

27/12/2012 10:54pm

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

28/12/2012 11:34pm

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

01/01/2013 4:43am

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

02/01/2013 1:12am

thank you for the answer

Reply

Arshdeep Bajwa

11/01/2013 10:46am

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

Reply

Admin

12/01/2013 3:24am

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).

Reply

Admin

12/01/2013 3:53am

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.

Reply

himanshu

20/01/2013 10:48am

sir can u help me in question no 40 and 41

Reply

Admin

21/01/2013 3:24am

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.

Reply

Admin

21/01/2013 3:52am

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.

Reply

himanshu

21/01/2013 9:18am

In qustion no 41 R1 is coming wrong

Reply

Admin

22/01/2013 4:35am

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)

Reply

himanshu

22/01/2013 10:57am

thanx sir but i am finding problem in qustion no 45

Reply

Admin

24/01/2013 2:02am

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.

Reply

Rajat

23/01/2013 11:06am

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??

Reply

Admin

24/01/2013 2:12am

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.

Reply

Rajat

24/01/2013 3:13am

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...

Reply

Admin

25/01/2013 9:42pm

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)

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

Reply

Prateek

12/02/2013 12:58pm

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

13/02/2013 4:52am

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

12/02/2013 12:55pm

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..!!

Reply

Admin

13/02/2013 5:03am

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.

Reply

manju

13/02/2013 12:54am

plz expln Q: 15

Reply

Admin

13/02/2013 5:31am

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

Reply

Amrish Rai

25/03/2013 12:41am

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

Reply

theyatin

25/03/2013 2:35am

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.

Reply

Amrish Rai

27/03/2013 8:34pm

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

27/03/2013 10:32pm

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

29/03/2013 1:53am

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...?

Reply

theyatin

29/03/2013 7:36am

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

Reply

theyatin

29/03/2013 7:40am

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.

Reply

DJ

01/04/2013 5:56am

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 ??

Reply

theyatin

01/04/2013 10:26pm

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

Reply

Monika

05/04/2013 3:23am

sir please explain me short question 18, 19, 20

Reply

thayatin

06/04/2013 8:00am

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

Reply

theyatin

06/04/2013 8:07am

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

Reply

theyatin

06/04/2013 8:12am

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.

Reply

Rama

05/04/2013 3:25am

Objective 1 - q11, 12

Reply

theyatin

06/04/2013 8:17am

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

Reply

theyatin

06/04/2013 8:23am

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

Reply

Rana Vishwajeet

29/04/2013 5:27am

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

Reply

theyatin

29/04/2013 7:50pm

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

Reply

Rajat

29/04/2013 8:43pm

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..

Reply

theyatin

30/04/2013 6:58am

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.

Reply

Rajat

30/04/2013 10:22am

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...

Reply

theyatin

02/05/2013 7:18am

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.

Reply

aman sangal

04/05/2013 7:56am

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

Reply

theyatin

04/05/2013 9:31pm

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)

Reply

aman sangal

05/05/2013 5:42am

what is point A

theyatin

05/05/2013 6:38am

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.

Reply

aman sangal

05/05/2013 8:50am

thank you so much

jai

06/05/2013 8:10pm

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

Reply

theyatin

10/05/2013 10:22am

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. . .

Reply

rohit

23/05/2013 1:14am

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.

Reply

arjun

31/05/2013 1:20am

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

Reply

theyatin

31/05/2013 5:59am

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

Reply

theyatin

31/05/2013 6:15am

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.

Reply

arjun

04/06/2013 7:48am

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

Reply

theyatin

04/06/2013 8:09am

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

rakesh

08/06/2013 9:48am

sir can u explain me Q15 from question for short answers

Reply

theyatin

09/06/2013 11:20am

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.

Reply

Swarnabha Das

09/07/2013 11:53am

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.

Reply

Sanghati Sussama

14/08/2013 5:43pm

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?

Reply

abhishek

22/09/2013 5:28am

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

Reply

theyatin

22/09/2013 10:44am

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.

Reply

abhishek

22/09/2013 5:28am

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

Reply

abhishek

22/09/2013 5:28am

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

Reply

divesh

01/10/2013 8:56pm

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

Reply

divesh

01/10/2013 8:58pm

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

Reply

abhishek

11/10/2013 7:54am

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

Reply

theyatin

12/10/2013 10:48am

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.

Reply

utkarsh

13/10/2013 9:48pm

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.

Reply

theyatin

15/10/2013 10:15am

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

Reply

Aymanzoor

25/10/2013 6:13am

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

Reply

theyatin

26/10/2013 10:02am

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

Reply

Aymanzoor

26/10/2013 10:29am

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

26/10/2013 9:18pm

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

Reply

Aymanzoor

27/10/2013 5:30am

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

Reply

Aymanzoor

27/10/2013 5:37am

Thanks but i got it

Reply

SHIVANI SHARMA

28/10/2013 10:48am

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?

Reply

theyatin

30/10/2013 12:11pm

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

Reply

SHIVANI SHARMA

30/10/2013 12:30pm

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

29/10/2013 3:30am

sir please help me and please reply me .

Reply

theyatin

31/10/2013 9:51am

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

Reply

SHIVANI SHARMA

30/10/2013 12:34pm

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?

Reply

theyatin

31/10/2013 9:45am

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

Reply

pravesh

05/11/2013 6:10am

i want question also can i get it

Reply

theyatin

05/11/2013 7:25am

dear,
which questions???

Reply

sunny singh

13/11/2013 6:10pm

i can't understand Q.no 33

Reply

theyatin

17/11/2013 10:44am

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.

Reply

Avith

15/11/2013 11:21pm

can you plz explain Q no 81

Reply

theyatin

16/11/2013 1:42am

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

Reply

kunal

20/11/2013 4:15am

can u help me in q 56

Reply

theyatin

22/11/2013 8:58am

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

Reply

Divyanshu Sahu

21/11/2013 1:38am

I am not able to understand question no. 37

Reply

theyatin

22/11/2013 8:26am

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

Reply

Sidharth

23/01/2014 10:14pm

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

darshan

21/11/2013 5:41am

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

Reply

theyatin

22/11/2013 8:30am

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

Reply

darshan

21/11/2013 5:48am

Please explain question 10 of rotational mechanics

Reply

theyatin

22/11/2013 8:38am

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.

Reply

prahaasa

21/11/2013 7:19am

can you please explain Q63 and 64 in detail.

Reply

kunal

25/11/2013 1:03am

can you explain q 67

Reply

ekansh gupta

25/11/2013 7:28am

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

Reply

ekansh

25/11/2013 7:30am

please

Reply

shivam

13/12/2013 9:07pm

please explain q.no 2 ,14&20

Reply

shubhu

16/12/2013 2:38am

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 !!

Reply

theyatin

17/12/2013 9:48am

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

theyatin

17/12/2013 9:43am

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. . .

Reply

shivam

17/12/2013 9:08am

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

Reply

theyatin

17/12/2013 9:45am

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

Reply

Alok Kumar Sharma

18/12/2013 5:42pm

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

Reply

theyatin

20/12/2013 7:00am

dear,
the question is not much clear. . .

Reply

shivam

24/12/2013 6:22pm

please explain q.no 28

Reply

theyatin

25/12/2013 7:39am

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.

Reply

shivam

24/12/2013 6:57pm

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

Reply

shubhu

24/12/2013 8:41pm

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

Reply

shivam

05/01/2014 7:17am

please explain how v=rw in pyre roll

Reply

theyatin

08/01/2014 10:31am

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

Reply

shivam

05/01/2014 7:19am

sir its a mistake not pyre its a pure roll

Reply

saurav

20/01/2014 5:56am

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

Reply

theyatin

20/01/2014 10:12pm

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.

Reply

vish

21/01/2014 7:28pm

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

Reply

theyatin

23/01/2014 9:13am

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

Reply

Ansh Arora

24/01/2014 1:22am

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

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

mglsinθ-(mglsinθ/5) = KE

Reply

Harish

27/01/2014 10:36pm

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?

Reply

theyatin

29/01/2014 10:51am

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. . .

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

Reply

theyatin

25/02/2014 11:28pm

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

Reply

waka

09/03/2014 12:41am

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.....?

Reply

manish

10/03/2014 5:46am

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 ....

Reply

theya

10/03/2014 6:52am

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

Reply

kumar shivam

10/03/2014 11:00am

Reply

manish

12/03/2014 4:10am

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

Reply

THEYATIN

19/03/2014 7:52am

dear please write complete question here for particular solution.

Reply

airpit aggarwal

16/03/2014 7:49am

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...

Reply

theyatin

19/03/2014 8:00am

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

Reply

shubhu

20/03/2014 7:29am

how u arrived at at the answer ??? explain

Reply

waka

25/03/2014 1:27am

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

Reply

waka

25/03/2014 1:32am

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

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.

Reply

shivani sharma

08/04/2014 10:12am

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

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

10/04/2014 5:53am

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

10/04/2014 6:00am

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

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

04/04/2014 1:31am

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

Reply

Anshul Bansod

19/04/2014 3:17am

Q-37

Reply

ER.MIRJAIK

09/05/2014 9:21pm

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

Reply

Archie

08/05/2014 10:43am

Q 54

Reply

ER.MIRJAIK

09/05/2014 11:05pm

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

Reply

shreyansh

03/06/2014 8:18am

qus. 32(b)

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Pratyush

11/06/2014 9:50am

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

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admin

22/06/2014 8:44pm

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

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iod

02/07/2014 4:53am

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

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Admin

02/07/2014 11:35pm

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

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Satya

16/07/2014 6:12am

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

Reply

Satya

16/07/2014 6:14am

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

Reply

Top Gun

27/07/2014 2:08am

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.

Reply

ashish

29/08/2014 2:46pm

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

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

Reply

SHIVANI SHARMA

14/09/2014 3:00am

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

Reply

Brock

25/09/2014 5:51am

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?

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

Reply

shivani sharma

29/11/2014 7:03am

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

Reply

Mike

03/12/2014 8:52am

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

Reply

shashank

06/12/2014 9:10am

Sir plz xplain ques no 43

Reply

shivani sharma

24/12/2014 11:54am

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

Reply

shivani sharma

24/12/2014 11:57am

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

08/12/2014 2:43am

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

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shalu

08/12/2014 9:43pm

sir, please explain me question no 78

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pinki

12/12/2014 10:15pm

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

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mrunal

16/12/2014 10:12pm

how to solve objective 10,15 in objective 1

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konark

09/01/2015 12:22pm

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!

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yash parkar

22/01/2015 9:39am

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