HC Verma Physics Part 1 Solutions for Chapter 7 Circular Motion

5/21/2012

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Solution of HC Verma Concept of Physics -1- Chapter 7 Circular Motion

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142 Comments

Ankur

10/20/2012 12:47:11 pm

please explain question 23 of chapter 7

Admin

10/20/2012 10:15:41 pm

in the question it is given that the child is rotating around the pivot on rod. the child is rotating, there will be only centrifugal force acting on it which will try to push it outwards. he is holding the rod so friction force is acting between the rod and his hands. so in order to rotate
friction force = centrifugal force
= m*r*w*w
where m = mass of the child
r = radius
w = angular velocity

Trisha

8/28/2015 05:42:09 pm

Value of omega is already given. Why we have not used it

ishita

8/30/2015 11:59:30 pm

its given 20 revolutions per minute...
so we will take 2#n/60*60.......right

ANKIT YADAV

10/8/2015 12:13:14 am

CIRCULAR MOTION

ankur

10/20/2012 01:02:40 pm

some explain question 24,ch-7

Admin

10/20/2012 10:49:08 pm

this is very simple question. you just need to check the direction of forces. you may see the free body diagram in the solutions that at the maximum angular velocity the block will try to move upward and friction force will act in the opposite direction in order to stop it.
also at minimum angular velocity the block will try to move downward and the friction force will move upward to maintain equillibrium. now you resolve forces in x and y direction and get the form the equation

sunil kumar

8/17/2013 09:50:10 pm

not getting answer to this problem plz provide complete solution with explanation.

Trisha

8/28/2015 05:53:51 pm

why the maximum angular velocity will try to move it upward

Karen kumar

10/11/2015 10:57:04 pm

Admin

11/3/2012 03:41:14 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

Sharad

11/5/2012 10:46:56 pm

All question of chapter circular motion starting from question no 18 to question 30(except 24) ...{book hc verma}

Admin

11/6/2012 09:13:46 am

Explanation for question - 18
In this question we are asked to find the two extreme velocity i.e the minimum and the maximum velocity.
please see the diagram given in the solutions. At minimum velocity the velocity will try to slip down.so friction force will act up the inclined plane in order the resist the vehicle from slipping down.
Similarly at Maximum Velocity, the vehicle will try to skid up and friction force will act down the inclined plane.
Though this we form free body diagram as given in question and solve question by resolving forces.

Admin

11/6/2012 09:23:10 am

Explanation for Question 19
In this question it is given that motorcycle is moving on a circular track.
so here two forces will be acting. first is centrifugal force which will try to lift the motorcycle and second is gravitational force which will act downward.
part a - here we simply equate centrifugal force = gravitational force
part b - same as above
part c - Vehicle will loose contact at the end of the bridge. so we calculate at that point.

Asmin

11/6/2012 09:28:24 am

Explanation for question 20
in this question a car is moving on a circular track having radial as well as tangential accelaration. so here we calculate the resultant force as given in solutions and then we keep
resultant force = frictional force for equillibrium

Admin

11/6/2012 09:39:22 am

Explanation for question 21
Statement-
it is given there is a ruler fixed at one end and a block is kept on it
we are asked to find out speed at which block slips.
part a - friction force will act as a centripetal force and will balance centrifugal force(see diagram in solutions).
we simply equate them.
part b - it is same as that of above 20th question.
there are two accelerations in radial and tangential direction. we first calculate the resultant and then equate it will frictional force.

Admin

11/6/2012 09:54:51 am

explanation for question - 22
first see the free body diagram as given in solutions.
part a - normal force is the difference of weight and centrifugal force.

part b - friction force = 0 at B and D because at highest and lowest points it has no tendency to slip.
But at point C , frictional force = mgsin(@) where @ = 45 is the angle which mg will make with normal.

part C - normal force is the difference of weight and centrifugal force.

part D - To find out the minimum desired coeff. of friction, we have to consider a point just before C. because N is minimum at that point due to which friction force may not balance the component of weight.

Explanation for question 23
in the question it is given that the child is rotating around the pivot on rod. the child is rotating, there will be only centrifugal force acting on it which will try to push it outwards. he is holding the rod so friction force is acting between the rod and his hands. so in order to rotate
friction force = centrifugal force
= m*r*w*w
where m = mass of the child
r = radius
w = angular velocity

Admin

11/6/2012 07:57:08 pm

Explanation for Question 25
In this Question we have to calculate the radius of curvature of the circular arc at the highest point.
Particle is projected with speed ‘u’ at an angle @. At the highest pt. the vertical component of velocity is
‘0’
So, at that point, velocity = u cos @
so consider this velocity as the tangential velocity, we calculate its centripetal force. we can see centripetal force is only provided by weight mg.
mg = centripetal force

Explanation for question 26

it is same as that of question no. 25.
See the free body diagram.
we can see that there is no force in the x-direction. so velocity will remain constant.
Through this we calculate the velocity at that point.
Rest is same we know that centripetal force is perpendicular to the velocity. so we resolve our mg in direction perpendicular to the velocity.
and every thing is same as question 25.

Explanation for question 27

In this question a block is moving on a circular track and touching against the walls of cylinder.
And there is a friction between block and cylindrical walls
for parts A,B,C see the solution you will be able to understand.
For part D, we have just done the integration that you will study in 12th standard.

Explanation for question 28

It is very simple question
In this question we are given that a cabin is rotating with angular w and there is a groove AB and a particle is kept at point A.
Here we will see that centripetal force will act on the particle in the horizontal direction and particle will move along AB due to centrifugal force.
so we resolve the centrifugal force along AB and found the particle's acceleration.
Now applying kinemtic equation we calculate time.

Explanation for Question 29

In this there is car moving on a circular Road and there is a wooden plate fixed vertically on a car's seat. Now a block is kept against the support of wodden
block
part A - normal force is simly equal to centrifugal force.

part B - Now wooden plate is turned at an angle @ with the radius.
Now Normal force = centrifugal force* [email protected]
frictional force = centrifugal force * [email protected]
Through this we may calculate the angle.

Explanation for Question 30

In this question there is a cabin in which there is a table and pulley is fixed in it with two masses m and 2m on each side.When the cabin will rotate the two masses
will start accelerating.
See free body diagram to see forces acting on it and form the equation.

Trisha

8/28/2015 07:38:59 pm

Here normal reaction is providing centripetal or friction?

vallabh

12/20/2012 06:44:32 pm

explanation for q.no 20

Admin

12/21/2012 02:24:22 pm

Dear Vallabh,
In this question, it is given that a car is moving on a circular road and also it is accelerating with acceleration a. Now here two forces are acting on car. These are
1. centrifugal force- It will try to take car away from the center. acceleration due to centrifugal force = v²/r.
Here friction force is balancing and restricting the car to remain in the circle.
centrifugal force = mv²/r

2. tangential force - it will try to move the car along the tangent. Again friction force will restrict the car to move in circle.
Tangential force = ma
Now we can see that friction force is balancing the above two forces.
Now we will calculate the resultant of above two forces and equate it with the friction force.
resultant force = √ ((mv²/r) ²+ (ma)²) = µN
where N = mg

Robin

7/8/2017 04:20:40 am

Sir but why to equate as they had said that it will skid so friction force should not provide it

1/11/2013 02:53:43 pm

shalu

1/11/2013 03:02:18 pm

please explain me question no 22

Admin

1/11/2013 08:43:47 pm

Dear Shalu,
In this question a person sitting on a bicycle is moving along the specified path as given in the question.
part A-
in this part we have to find the normal force between cycle and road at point B and D.
Here at the maximum and the minimum point there will be no slipping due to which the friction force is zero.
so, at point b we can see that Normal and centrifugal force are acting in the upward direction. so,we get
mg = N+mv²/r
mv²/r = centrifugal force and it always act away from the center.
and find Normal force(N)
at point D, weight and centrifugal force are acting in the downward direction, so we get
N = mg+mv²/r
solution for part b -

we know that the cycle is moving at constant speed which means net force acting is zero.
here there is two components of mg. these are-
1. mg cos Ɵ which is balanced by normal force
where Ɵ = angle between vertical(weight)and the normal force acting at a point on the circle

2. mg sin Ɵ which will be balanced by frictional force
so, we get frictional force = mg sin Ɵ where Ɵ = angle between vertical(weight)and the normal force acting at a point on the circle

Now at highest and lowest point that B and D, Ɵ = 0 and we get sin 0 = 0. so, friction force = 0 at point B and D.
but at point C,
Ɵ = 45° and we get frictional force at point C.

solution for part C
in this part we have to find normal force at just before and after C.
There is just one difference in the two situations. The difference is that of the direction of centrifugal force.
Just before Cthe cycle is moving in the circle ABC so the centrifugal force is acting upwards.
but just after C the cycle is moving in the circle CDE and centrifugal force is acting downwards.
Through this you may calculate Normal force.

solution for part d
here we have to find the minimum value of cofficient of friction(µ).
we have considered the point just before C because at that point Normal force is zero.

katyani

1/11/2013 03:22:38 pm

explain the chapter of work power and energy

Admin

1/11/2013 06:15:48 pm

Dear Katyani,
Please start reading the chapter from a good book like HC verma or your text book and ask the doubts and topics that you are not able to understand. In this way you will be able understand better by yourself.

katyani

1/11/2013 03:32:10 pm

please give explation now

Admin

2/22/2013 05:28:06 pm

The doubt solving service for HC verma has been disabled for few days. We are very happy with the faith the students have shown in our service. We are hiring more and more teachers and upgrading our features like audio video techniques to solve your doubts.

arpit

3/24/2013 10:41:06 pm

plz explain q-13

theyatin

3/25/2013 12:40:05 am

dear arpit,
as you may know law of conservation every thing inside car is at the same momentum as of the car which is linear momentum.
when car turns the pendulum starts leaning away from the turn just to maintain momentum (like wise we feel some force when car urns suddenly)
thus we need to calculate the angle upto which the pendulum leans away from the turn. as pendulum moves away from mean position the thread of pendulum must feel some tension. . .
as from the change in momentum we can find force so we can find tension
thus comparing both components of tension we can find angle up to which the pendulum would lean.

Akhil krishnan

5/23/2013 02:22:35 am

OBJECTIVE II QUESTION NO.7 (last one) !!

theyatin

5/23/2013 06:23:53 pm

dear,
it is possible that a paricle is in circular motion still it can have uniform velocity
in case when whole system is moving in some particular direction.
for example a yo yo inside a moving car as seen from outside of car.
thus it implies that there are other forces also resultant of other forces is varying in magnitude as well as direction. because to keep circular motion in linear as seen from inertial frame, the system must be moving in different directions at different time so as with different velocity.
thus it may have b) and d) as its properties.

anjireddy

7/3/2013 11:56:39 am

piz explain the answer for q no 6 from objective II

theyatin

7/4/2013 12:36:51 am

dear,
a circular road is made at some angle so that for a particular speed car wont slip away from the road. if speed is greater the car would slip away.
if speed is slow then car would slip downward. (in case friction is negligible.
hence 1 is wrong.
2 is right .
c is again wrong in case road exerts same force as mg or same angular force as of car it will only neutralize its motion but not be able to turn that car.
d) is right as greater force will make it turn otherwise it will keep moving in straigh tline.

Pritesh Kwatra

7/4/2013 02:10:23 am

please explain question 23 ( exercise ) of ch- circular motion part 1

theyatin

7/5/2013 01:53:05 am

dear,
Q23. d = 3m  R = 1.5m
R = distance from the centre to one of the kids
N = 20 rev per min = 20/60 = 1/3 rev per sec
w= 2pir = 2pi/3
m = 15kg
now frictional force is equal to mrw^2
thus we have all these values thus we can have answer.
which part you are having difficulty?

shivam

7/23/2013 02:13:48 pm

plz explain ques15

theyatin

7/25/2013 12:11:04 am

dear,
we need to calculate tension on the string where angle of deflection and mass and the velocity is given
hence angular momentum is = mv^2 /r
that is equal to tension T- resultant weight of the bob (mg cos theta)
hence T-mg cos theta =mv^2 /r
you can check this formula too physically.
now solving to get answer.

Rakesh

8/2/2013 03:34:38 am

sir can you explain me all the objectives of part 2 from this lesson .

theyatin

8/3/2013 03:51:31 am

dear,
you must ask question one by one. . .

chetan link

8/11/2013 01:26:14 am

ch 7 objective 1 q.no.12 hc verma

theyatin

8/18/2013 05:38:03 am

dear,
Q21
in this case angular velocity of the earth may be additive.
as the earth rotates from west to east,
so by relativity theorem the seed of train A may be added up thus more force is applied on the tracks.
this is most probable explanation.

anish

8/12/2015 02:25:06 pm

but i feel the explanation itself is not very clear sir...can you further discuss about this....
thanks and regards

madhavi

8/18/2013 05:24:15 am

sir can you please explain me objective 1 ques. no.3 ,5 ,11

theyatin

8/18/2013 05:30:38 am

dear,
Q3
as the angle of turn is sharper at A than B so the normal force applied by the road on the car to turn it would be larger at B than A. because the car would turn easier on point b than point a .
so NA <NB

theyatin

8/18/2013 05:34:01 am

Q11
C) is the answer
because car C would exert maximum speed as it is due to gravity hence more force is exerted as it in middle oh the bridge.

madhavi

8/19/2013 03:58:01 pm

sir please give my question 5 answer of objective 1 and also give detailed ans of ques no. 7,10,11

theyatin

8/20/2013 07:36:45 pm

dear,
Q7
in case when mg > mv^2 / r the water would fall as its weight is greater than angular force
but when mg is not greater than mv^2/r the water would not fall.

theyatin

8/20/2013 07:39:18 pm

dear,
Q14
as the distance from the pivoted end increases
so lesser the weight associated to that point. thus lesser the weight lesser the tension associated with that point.
so T1>T2

tehyatin

8/20/2013 07:41:45 pm

dear,
Q10
as over bridge is curved at ascending to maintain speed constant the motorcycle must have increasing acceleration and thus the normal force also increases.

madhavi

8/19/2013 04:07:21 pm

sir ,l have already ask you ques 11 so plz explain me ques 14 of obj 1

Moumita

9/1/2013 02:34:07 pm

I am having problem with the following sums, so could you just help me out?
A particle is projected with a speed u at an angle x with the horizontal. Consider a small part of its oath near the highest position and take it approx. to be a circular arc. What is the radius of the circle, where the radius is called the radius of curvature of the curve at the point.

theyatin

9/4/2013 04:00:17 am

dear,
as formula for maximum height is
h= v^2 sin^2 theta / 2g
if we take very small curve near maximum height then radius of curve would be similar to maximum height of object. . .

Alisha

9/7/2013 03:39:10 pm

Sir i have 3 doubts.
1. in q19 how can we say that it has a maximum tendency to lose contact at the end of the bridge??
2. in q20 and q21 why did we take the resultant of accelerations? how exactly is the friction acting on the car and block?
3. in q23 is the friction on the boys by rod acting towards the centre or away from the centre? pls also give the reason..

theyatin

9/11/2013 01:29:06 am

dear,
1. in q19 how can we say that it has a maximum tendency to lose contact at the end of the bridge??

as its the end point of bridge so its the point where the bike has maximum probability of losing contact.

theyatin

9/11/2013 01:32:41 am

dear,
in q20 and q21 why did we take the resultant of accelerations? how exactly is the friction acting on the car and block?

resultant of the acceleration is the force applied by the road on the car.

friction acts in opposite direction of motion at every point of the circular motion on car so as to turn it
and on box it acts in direction of motion so at to move it along the motion.

theyatin

9/11/2013 01:34:47 am

dear,
3. in q23 is the friction on the boys by rod acting towards the center or away from the center? pls also give the reason..

friction is acting toward the center so as to cancel force acting outward from the road and is equal so that the boy wouldn't slip.

shubhi

9/20/2013 06:36:44 pm

Sir, please explain question 24

theyatin

9/22/2013 04:02:06 am

dear,
force of rotating bowl would be F=mw^2 r
thus we have balanced equation as.
now we can find w1 and w2 by calculating tensions.

Siddhant

9/21/2013 12:20:09 pm

Sir can u also tell me some q from dc pandey?

theyatin

9/22/2013 03:54:34 am

dear,
you can write whole question here with all detail
and we will try to answer it.

Abhay

9/22/2013 06:36:01 am

Sir plz xplain me ques10 obj level 1 ....i didn't understand this concept and plz relate it to question 3 of obj level 1 (hcv)....

theyatin

9/23/2013 01:47:18 am

dear,
in this part as the bike has to maintain its speed constant it must increase its acceleration thus the normal force acting on it is increasing.

shubhi

9/22/2013 11:01:51 am

sir please explain question 24

Jay

9/24/2013 04:17:22 pm

Sir please explain this,
If we tie a stone using string of constant length and rotate it in a circle why it's speed increases
As we increase the tension in the string although it does not have a component tenge tangentially to increase the speed of the stone.

theyatin

9/25/2013 04:46:32 am

dear,
actually its other way around
when speed of stone increases thus centrifugal force increases thus tension on the string increases.
you cant increase tension on string wile its rotating.

mehtab singh

10/5/2013 06:50:09 pm

sir plrase explain q no. 19
how angle taken is 2a

theyatin

10/8/2013 04:56:30 pm

dear,
as we have normal in the center of bridge passing through angle and bisecting it
for further simplified calculations we assume that angle as 2a
so that if in case of either end we can have bisected angle that is a

tripti

10/11/2013 04:39:42 am

can u please provide the solution of Q.9 of objective 1 ch 7.

theyatin

10/12/2013 04:17:38 am

dear,
centripetal force must be exceeded so as coin to slip away
that is F=m r w^2
or W^2 is directly proportional to 1/r
thus for w^2 d.p. 4
then (2w)^ is tha 4w^2
thus if w is 4 times then r becomes 1/4 th
so it is r=1

adi

10/18/2013 02:34:18 am

how to do question no. 30 of hcv circular motion??

shilpi

10/20/2013 06:09:04 pm

Sir...plz xplain clearly the situations in which linear momentum is conserved or energy or both....when should we conserve momentum nd when the mechanical energy.??

theyatin

10/23/2013 04:51:46 pm

dear,
when a body is moving in a straight line with uniform velocity has both linear momentum and energy conserved.
it depends upon problem which quantity you need to get conserved if you need momentum in result you need to conserve momentum
if you need either of kinetic or potential energy or any of the component of these then you would conserve mechanical energy.

shilpi

10/27/2013 12:28:39 am

sir...but both conservations require absence of any external force on the body.....then how to decide the right assumption in case of absence of external force...bt in inelastic collision momentum is conserved bt not energy...inspite of absence of external forces.on tthe system...y.??

theyatin

10/28/2013 03:20:41 am

Dear,
when you have bodies in motion then you need to conserve momentum.
as you may have change in state of similar body so as from dynamic to steady state or to transfer momentum to other body you need to conserve momentum
when you have change energies you need to conserve energies.
but it varies as per demand of the numerical.
you need to calculate energies due to position you need to conserve mechanical energy.

komal

11/15/2013 09:57:04 am

would u pls give me the explanation of questiin no. 17 of ch7

theyatin

11/17/2013 02:50:46 am

dear,
a) Net force on the spring balance.

R = mg – mw^2r
So, fraction less than the true weight (3mg) is
mg-R/mg
mg-(mg-mw^2r) / mg
solving this we can have ans
2)
but when mg-(mg-mw^2r) / mg =1/2
then value of w can be calculated
then Time period T=2 pi/ w
so we can caculate T

sidharth

11/22/2013 01:11:14 am

I have a question. In questions like pendulum at an angle @ or object on an incline of angle, sometimes we make the cos & sin components of mg and sometimes of tension or normal. How exactly do I get to know whose components do I have to make in which question?

Samrat

12/21/2013 03:12:58 am

Why the vehicle has maximum tendency to llose contact at end of bridge and why angle is 2'alpha'

theyatin

12/22/2013 12:04:33 am

dear,
at any instance car is applying force on bridge as well bridge on car
at the end of bridge probability is max that it looses contact in case its not adhesive. . .
which angle??

faisal ahmad link

1/6/2014 04:38:19 pm

sir in Q19 where is the concept of normal reaction applied, why dont we consider normal reaction, and is centripetal force equal to mv2/R or centrifugal, it is confusing me a lot.
thanx in advance

theyatin

1/8/2014 02:30:34 am

dear,
centrifugal force is the apparent force that draws a rotating body away from the center of rotation
centripetal force is a force that makes a body follow a curved path
so here mv^2/R=mg that means it balances the weight of the object thus it does not falls so it is centrifugal acting opposite to weight.

abinaya

1/12/2014 05:56:14 pm

how to solve circular motion 9th ,10th sum hcv

theyatin

1/14/2014 01:58:54 pm

dear in case of Q9
torque=force of attraction by coulomb's law
Q10 torque=weight of body thus we balance both equations to find ans

Shreya

2/11/2014 04:14:13 am

In q 28 shudn't our net acceleration be mw^2rcos - mgsin? (can't find theta so didn't place)

theyatin

2/19/2014 01:46:49 am

dear,
we used formula S=ut+1/2at^2
and why are you subtracting verticle component of weight from force??

Aaron

2/14/2014 10:33:41 pm

Can u please explain in detail how we get equations in q18?

theyatin

2/19/2014 01:42:00 am

dear,
we use horizontal and vertical component of every force
for example mg is weight downward then mg co theta is towards right and downward that is opposite to R1 thus it is taken as negative.
similarly mg sin theta is towards friction resultant thus is positive.
you can now try yourself.

manish

3/12/2014 10:06:19 pm

if a particle is kept over a fan blade and surface is friction less then blade passes from below it and particle falls down . my doubt is in q. 30 . the same way above blocks should also get tilted as surface is friction less (Coriolis force) hence some friction is needed to retain the blocks in position . but involving friction will affect our answer. so how is it ??

theyatin

3/17/2014 02:25:22 pm

dear,
in question 30 blocks are connected with pulley plz confirm your query again

tipu sultan

3/15/2014 02:19:06 am

why a flags flutter in a windy day

theyatin

3/17/2014 02:12:47 pm

dear
the cloth of flag turns into directionn of wind so as to offer minimum friction to air.
due to different pressure areas created at different parts of cloths the cloth bends at many points when it gets straighten after bending it makes sound thus flattering sound is heard

tipu sultan

3/15/2014 02:21:58 am

why pressure of a tire increases when a car moves

theyatin

3/17/2014 02:06:45 pm

dear,
when car moves its tire gets heat due to friction and as the air in it is heated it expands.
so its pressure increases as the air expands

tipu sultan

3/15/2014 02:24:42 am

why the buildings are damage due to earthquake?

theyatin

3/17/2014 02:16:25 pm

dear,
earthquake makes makes vibrations of huge amplitude and intensity as building material is hard and stiff its elasticity is none.
the vibration causes the particles of material move in different directions thus the walls of building cracks as the particles are detached from each other at some point

NEEL

3/30/2014 07:21:23 pm

Ex Q 22d please

ER. MIRJAIK

5/4/2014 03:24:28 am

In this question a person sitting on a bicycle is moving along the specified path as given in the question.
part A-
in this part we have to find the normal force between cycle and road at point B and D.
Here at the maximum and the minimum point there will be no slipping due to which the friction force is zero.
so, at point b we can see that Normal and centrifugal force are acting in the upward direction. so,we get
mg = N+mv²/r
mv²/r = centrifugal force and it always act away from the center.
and find Normal force(N)
at point D, weight and centrifugal force are acting in the downward direction, so we get
N = mg+mv²/r
solution for part b -

we know that the cycle is moving at constant speed which means net force acting is zero.
here there is two components of mg. these are-
1. mg cos Ɵ which is balanced by normal force
where Ɵ = angle between vertical(weight)and the normal force acting at a point on the circle

2. mg sin Ɵ which will be balanced by frictional force
so, we get frictional force = mg sin Ɵ where Ɵ = angle between vertical(weight)and the normal force acting at a point on the circle

Now at highest and lowest point that B and D, Ɵ = 0 and we get sin 0 = 0. so, friction force = 0 at point B and D.
but at point C,
Ɵ = 45° and we get frictional force at point C.

solution for part C
in this part we have to find normal force at just before and after C.
There is just one difference in the two situations. The difference is that of the direction of centrifugal force.
Just before Cthe cycle is moving in the circle ABC so the centrifugal force is acting upwards.
but just after C the cycle is moving in the circle CDE and centrifugal force is acting downwards.
Through this you may calculate Normal force.

solution for part d
here we have to find the minimum value of cofficient of friction(µ).
we have considered the point just before C because at that point Normal force is zero.

krupa

5/4/2014 05:08:27 pm

sir,will u plz give me full explaination for que no. 24???

er. mirjaik

5/5/2014 11:15:58 pm

dear,
in this queshtion you just need to check the direction of forces. you may see the free body diagram in the solutions that at the maximum angular velocity the block will try to move upward and friction force will act in the opposite direction in order to stop it.
also at minimum angular velocity the block will try to move downward and the friction force will move upward to maintain equillibrium. now you resolve forces in x and y direction and get the form the equation
force of rotating bowl would be F=mw^2 r
thus we have balanced equation as.
now we can find w1 and w2 by calculating tensions.

admin

6/22/2014 01:44:16 pm

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chetan

7/15/2014 01:00:00 pm

A block of mass 1 kg is released from P on a frictionless track. What is the magnitude of radial acceleration of the block when it arrives at Q?
Track is inclined at H=6m, After travelling inclined surface it follows a vertical circular path till mid way position Q.

chetan

7/15/2014 02:05:42 pm

and total resultant acceleration for same problem

Shivank

7/22/2014 01:57:13 am

What does skid means in question no 20?

Kaundi

8/22/2014 02:41:59 am

Is there any mistake in unsolved questions of chapter 7 circular motion please tell me

Gagan kalra

8/23/2014 08:03:53 pm

please explain qestion no 20

Ankit link

9/12/2014 01:00:37 pm

sir if we solve ques 18 by resolving normal reaction and friction instad of centripital force and weight then why we are not gettinng right ans???

jasmine swain

9/16/2014 04:00:46 am

in ques no. 22 .
why frictional force is zero at point B & D will u elaborate plzz..
thanxx..

Asadullah Aman

9/16/2014 02:07:07 pm

which book is best for medical (physics)????
plz suggest the name of book....

yash singh

10/26/2014 01:03:14 am

A sphere of mass 200gm is attached to an inextensible string of length 130cm whose upper end is fixed to the ceiling . The sphere is made to describe a horizontal circle of radius 50cm . Calculate the period time of this conical pendulum and tension in the string.

Rachana

10/26/2014 10:26:06 pm

can you plz help me out in Q19,I didn't understand why the horizontal velocity of the man is considered when he is falling dist. x,so that he can fall on the ground???since he is not only going horizontal but also vertically downward??

tarun rai link

10/27/2014 01:01:14 pm

i want theory of rotational kinetic energy

neha

11/10/2014 01:25:31 pm

pls explain obj 2 ques 5

Rishab choudhury

11/21/2014 12:32:11 am

I have problem with q.no. 14,15,16.can you please solve them with explanation?

Shivam Anand

11/22/2014 10:13:15 pm

Why weight is not considered in question 29?? N=[(mv2cos(@))/R]+[mgsin(@)] ????

shalu

11/25/2014 01:45:34 pm

please explain me ques. 3 of objective 1

Shantanu Roy

1/9/2015 12:13:56 pm

Explain me The objective II of circular motion - Q 5,6,7

deeksha link

1/17/2015 05:41:30 pm

Please explain question 17 of objectives 1 hcverma chapter 18

Atul

3/21/2015 04:21:43 am

Sir I am not able to understand the (b) of question no 22
The frictional force at c is [email protected] but how.. Friction [email protected] should be equal to centripetal force

Hamra

3/27/2015 05:12:14 pm

Sir, Why is normal force inversely proportional to centripetal force?

aditya

4/24/2015 11:59:34 pm

Not understanding que no 20

shashank kapoor

5/1/2015 10:21:03 pm

sir,
In question 15 when we do component of T we get Tcos0 which balance by mg + mv^2/r cos0 and Tsin0 is balance by mv^2/r sin0 . So what is wrong in this statement ? please explain

charu link

5/5/2015 12:44:13 pm

In ques 18 ...formula obtained by equations is differnt frm that given in solution...+ - signs are interchanged in numerator and denominator

Smit

6/1/2015 01:55:20 am

Please explain question 15 of objective 1 and questions 2,3,4,5,6,7 from objective 2

Saanika

8/9/2015 04:55:50 pm

Experts I've a doubt in question 20 of pg 7.4. Why we consider centrifugal force instead of centripetal force? In which direction must friction act and why? Why friction will be equal to net acceleration x mass. Thank you so much!

ambika

8/15/2015 01:22:03 pm

sir i would like to know the answer to question no.6 of the objective 1 circular motion

lakshay link

8/30/2015 10:02:31 pm

In q20 components of friction should be a) towards centre and b)opposite to tangent.why have we taken radial component outward

Amit

9/7/2015 08:01:25 pm

Please give solution of question no. 29 also explain it.

Neel

9/21/2015 10:16:09 am

Can anyone please explain me chapter 7 questions 26 and 27

Jack link

10/3/2015 06:58:25 pm

Circular obj 1 question 6,9,12,14
Obj 2 question 5
Exercise questions 19 24 26

anjali patel

10/16/2015 04:01:01 am

please explain ques 11 2nd part of que.

Atharva Ranade

11/8/2015 10:00:11 pm

why is the true weight in question 17 given as 3mg?

omkar

11/19/2015 10:48:01 pm

suvit

11/23/2015 10:43:52 pm

in question 18
V1=under root rg(tan(thetha)+nu) / 1- nu*tan(thetha)
according to equations

but you mentioned
V1=under root rg(tan(thetha)-nu) / 1+ nu*tan(thetha)

how can this be possible ???
literally iam soo much confused please help me ASAP(as soon as possible).

student123

12/6/2015 01:21:59 am

In the diagrams why is it that the centripetal force acts outwards i e towards the body?the centripetal force should act towards the centre right?

M.syed muheer

7/5/2016 07:39:17 am

Sir...I want to knw about the question no=17....

divya

9/1/2016 08:10:45 pm

raushan

11/25/2016 01:26:56 am

anshu

12/12/2016 07:08:32 pm

anshu

12/12/2016 07:12:23 pm

Sir please explain me Q 16 17 and 24.

Yash

2/8/2017 01:31:39 am

why are we using centrifugal force in an inertial frame in Q.18

Chapterwise Solutions of HC Verma