Loading....
Coupon Accepted Successfully!

 

Question-1

Give the location centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform density. Does the centre of mass of a body necessarily lie inside the body?

Solution:

(i) at its geometric centre
(ii) at the centre of its axis of symmetry
(iii) at the geometric centre (centre of the ring)
(iv) at the geometric centre i.e, where the three diagonals intersect.
No, the centre of mass need not necessarily lie inside the body. It may lie outside the body as in case of a ring, a hollow cube, a hollow sphere, a hollow cylinder.

Question-2

In the HCl molecule, the separation between the nuclei of two atoms is about 1.27 Ao (1 Ao = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Solution:

Let the mass of the hydrogen atom be x

Let the hydrogen atom be located at x1 = 0

Let the chlorine atom of mass 35.5 m be located at x2 = 1.27 x 10-10 m

Let x be the distance of the centre of mass of HCL molecule

x = (m1x1 + m2x2) / (m1 + m2)
=  {0 + (35.5 x 1.27 x 10-10) / (1 +35.5) m
= 1.235 x 10-10 m = 1.235 Ao

Therefore the centre of mass of HCl molecule is located on the line joining hydrogen and chlorine nuclei at a distance of 1.235 Ao from the H2 nucleus.

Question-3

A child sits stationary at one end of a long trolley moving uniformly with a speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?

Solution:

When the child runs about on the trolley in any manner, the child exerts only an internal force on the trolley- child system.

Since no external force acts on the system and hence the speed of the centre of mass remains the same , ie v remains the same.

Question-4

Three mass points m1, m2, and m3 are located at the vertices of an equilateral triangle of length a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m1?

Solution:
 

M. I of a particle is defined as the product of the mass and the square of the distance from the axis of rotation.

Since the particle of mass m1 is situated on the axis of rotation (XY) itself, its distance from the axis is zero.

If r1 r2 r3 are the distances of m1 m2 m3 respectively from the axis of rotation,

M. I of the system = m1 r12 + m2 r22 + m3 r32
= m1 (0)+ m2[ a/2 ] 2 + m3a/2]2
since r1 = 0, r2 = a/2 and r3= a/2
= a2/4 [ (m1 + m2 )]

Question-5

What is the moment of inertia of a uniform circular disc of radius R and mass M about an axis
(i) passing through its centre and normal to the disc;
(ii) passing through a point on its edge and normal to the disc? 
The moment of inertia of the disc about any of its diameters is given to be ( ¼ M R2).

Solution:
M. I of a disc about an axis passing through its centre and normal to the disc can be calculated using the perpendicular axis theorem.

The theorem states that M. I of a body about an axis perpendicular to it and passing through a point is equal to the sum of the moments of inertia of the body about the two mutually perpendicular axes in the plane of the body and passing through that point.

(i) Let the moment of inertia of the disc about its centre = M. I about one diameter + M. I about the perpendicular diameter.

= ¼ MR2 + ¼ MR2
= ½ MR2

(ii) M. I of the disc about an axis passing through a point on its edge and normal to the disc is determined using the parallel axis theorem.

M. I = M. I about an axis through the center + M x square of the distance between the axes

M. I = ½ MR2 + MR2 = ( 3/2) x MR2.

Question-6

A solid cylinder of mass 20 kg rotates about its axis with angular speed of 100 radians. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Solution:
Mass of the cylinder = M = 20 kg
Angular speed =
ω = 100 rad/s
Radius of the cylinder = R = 0.25 m
Moment of inertia of the cylinder about its axis = I = ½ M R2 = ½ x 20 x (0.25)2 kg m2 = 0.625 kg m2
Kinetic energy of rotation = ½ I
ω2 = ½ x 0.625 x (100)2 J = 3125 J

Angular momentum about the axis L = I
= ½ x 0.625 x (100)2 J = 3125 J

Angular momentum about the axis L = I
ω  = 0.625 x 100 Js = 62.5 Js .

Question-7

(a) A child stands at the centre of a turntable with its two arms outstretched. The turn table is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hand back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.
(b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Solution:

(a) Let I1  be the moment of inertia of the child with outstretched arms and I2 be the corresponding moment of inertia with folded arms.

If ω1and and ω2be the corresponding angular speeds, then
 
By the principle of conservation of momentum, I1
be the corresponding angular speeds, then
By the principle of conservation of momentum, I1
ω1 = I2 = I2 ω2
                                                       ω2 = (I1 = (I1 ω1) / I2  
                                                            = (I1 x 40) / (2/5 x I1)
                                                            = 100 rev/min
                                                            = 2.5
) / I2  
                                                            = (I1 x 40) / (2/5 x I1)
                                                            = 100 rev/min
                                                            = 2.5
ω1


(b) The new initial kinetic energy of rotation E1 = ½ I1 ω12
      The final kinetic energy of rotation   E2 = ½ I2
2
        The final kinetic energy of rotation   E2 = ½ I2
ω22
                                                               = ½ (2/5 I1 x (5/2
2
                                                               = ½ (2/5 I1 x (5/2
ω2)2 = 5/4  I1 )2  
                                                              
  = 5/4  I1 ω12 
                                                   
Thus, the final (new)  kinetic energy is greater than the initial K. E of rotation by a factor of 5/2.

The child uses his internal muscular energy to increase the rotational kinetic energy, by way of folding the arms.

Question-8

A rope negligible mass is wound round a hollow cylinder of mass 3 Kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope?
Solution:
Mass of the cylinder m = 3 kg
Radius of action of the force r = 0.4 m
Force acting on the cylinder (pull) F = 30 N
Applied torque
τ = F x r
Therefore the torque acting on the cylinder, 
t = Fr
Moment of inertia of the hollow cylinder about the axis of rotation = I = m x r2
I = m x r2
= 3x (0.4 )2        
= 0.48 kgm2

Torque = τ = I x α
Angular acceleration of the cylinder = α = τ / I
                                                       = 12/ 0.48
                                                       = 25 rad/s2
Linear acceleration of the rope a= F / m = 30 / 3 =10 m/s2

Question-9

A cylinder of mass 5 kg and radius 30 cm and free to rotate about its axis receives an angular impulse of 3 kg m2/s initially, followed by a similar impulse after every 4s. What is the angular speed of the cylinder 30s after the initial impulse? The cylinder is at rest initially.

Solution:
Moment of inertia of the cylinder about its axis = ½ MR2 = ½ x 5 x (0.3)2 kg m2 = 0.225 kg m2

Since angular impulse is defined as change in angular momentum, an impulse of 3 kg m2 s-1 will mean that

Final angular momentum- initial angular momentum = 3
Angular momentum = I
ω 
impulse = change in angular momentum

                  = final momentum - initial momentum
            = I
ω - I x 0 = I ω

Angular speed ω = (Angular impulse) / (I)
                        = 3 / 0.225 = 13.33 rad/s = speed after first impulse.

Because the angular impulse of same magnitude is given after every 4 seconds producing equal change in angular speed. The angular speed remains a constant till another impulse is given, i. e for values of t lying between, 28 and 32 seconds, the angular speed will still have the same value as t= 28, the value being 106.64/s

Therefore, the angular speed after 30 s = 8 x 13.33 rad/s =106.64 rad/s.

Question-10

To maintain a rotor at a uniform angular speed of 200 rad/s, an engine needs to transmit a torque of 180 Nm. What is the power of the engine required? (Note: Uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque? Assume that the engine is 100% efficiency.

Solution:

Power = Torque x angular speed
         = 180 x 200 Nm/s
         = 36,000 Nm/s
         = 36,000 W
         = 36 kW

Question-11

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Solution:
Moment of inertia of the hollow cylinder about its axis = I1 = M R2
Moment of inertia of the solid sphere about its diameter= I2 =2/5 M R2
Angular acceleration of hollow cylinder =
α1 
Angular acceleration of the solid sphere =
α2
Torque = τ = I1 α1 = I2 = I2 α2 [Given that the torques are of equal magnitude]
α1 /α2 = I2 / I1
         = I2 / I1

α1 = = α2 I2 / I1

= α2 { MR2 / MR2}

= 2.5 α2
Angular speed of the cylinder after a time t is given byω1 = = ω0 + + α1 t

Angular speed of the sphere after a time t is given byω2 = = ω0 + + α2 t = ω0 + 2.5 + 2.5 α1 t

as (ω2) is greater than (ω1), clearly the angular speed of the sphere (ω2) will be greater than that of the cylinder (ω1) under the given conditions.

Question-12

Two discs of moment of inertia I1 and I2 about their respective axes ( normal to the disc and passing through the centre), and rotating with angular speeds ω1 and and ω2 are brought into contact face to face with their axes of rotation coincident. 
(a) What is the angular speed of the two disc system?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take
are brought into contact face to face with their axes of rotation coincident. 
(a) What is the angular speed of the two disc system?
(b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take
ω1 ω2 .

Solution:

Let the moments of inertia of the two discs before contact be I1 and I2 about their respective axes.

Let their respective angular speeds be ω1 and and ω2.

When they are brought together, the moment of inertia of the combination is I1 + I2 .

Let the corresponding angular speed of the combination be .

When they are brought together, the moment of inertia of the combination is I1 + I2 .

Let the corresponding angular speed of the combination be ω
Since there is no other external forces involved, according to the principle of conservation of angular momentum

(a) I1 ω1 + I2 + I2 ω2 = ( I1 + I2 ) = ( I1 + I2 ) ω
                           ω = (I1 ω1 + I2 + I2 ω2) / (I1 + I2)
(b) Total kinetic energy of the discs before contact = E1
                          E1 = ½ I1
) / (I1 + I2)
(b) Total kinetic energy of the discs before contact = E1
                          E1 = ½ I1
ω12 + ½ I2 2 + ½ I2 ω22
Kinetic energy of the combination E2 = ½ ( I1 + I2)
2
Kinetic energy of the combination E2 = ½ ( I1 + I2)
ω2
                                      = ½ ( I1 + I2) / {(I1
                                      = ½ ( I1 + I2) / {(I1 ω1 + I2 + I2 ω2) / (I1 + I2 )}2
E1 - E2 = ½ I1 
) / (I1 + I2 )}2
E1 - E2 = ½ I1 
ω12 + ½ I2  2 + ½ I2  ω22 - ½ (I1  2 - ½ (I1  ω1 + I2  + I2  ω2)2 /(I1 + I2 )
          = {I1 I2 / 2 (I1 + I2)} (
)2 / ( I1 + I2 )
          = {I1 I2 / 2 (I1 + I2)} (
ω1 ω2)2
(
ω1 ω2)2 is positive.
E1 - E2 is positive.

hence E1 > E2
The kinetic energy of the combined system is less than that of kinetic energies of the two discs.   

There is a loss of kinetic energy in the system when the discs are joined together.
The loss in energy is due to the energy being used up to overcome frictional forces between the surfaces of the two discs while in contact,

These forces, in fact, bring about a common angular speed (ω) of the two discs on combining.

Question-13

A disc rotating about its axis with angular speed ω0 is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B, and C on the disc shown in the figure? Will the disc roll in the direction indicated?

Solution:

Since there is no friction, the table does not offer any force to the rotating disc. Since there is no external force the rotary motion of the disc is not altered and it continues to rotate about its centre with the same angular speed ω. The disc will not roll on the frictionless surface.

Linear velocity of point A =
ω0 R in the direction of the arrow R in the direction of the arrowshown in the figure.
Linear velocity of B =
ω0 R in the direction opposite to the arrow R in the direction opposite to the arrow shown in the figure.
Linear velocity of C =
ω0 R/2 = ½ R/2 = ½ ω0 R in the direction of the arrow R in the direction of the arrow shown in the figure.

Question-14

Explain why friction is necessary to make the disc in the figure under question 13, to roll in the direction indicated?
(a) Give the direction of friction force at B and the sense of frictional torque, before perfect rolling begins.
(b) what is the force of friction after perfect rolling begins ?


Solution:
Since the force of friction is necessary to provide the centripetal force, in order that the disc may, roll in the direction of the arrow, it is absolutely necessary that the frictional force must be present. The disc will not roll on a frictionless horizontal surface,

(a) Frictional force opposes the relative motion between the two surfaces involved, It opposes the direction of velocity of the disc at that point, i. e, in the direction of the arrow shown in the figure. This generates a torque about the centre of mass The velocity of B is in opposite direction of the arrow. The frictional force will act in a way so as to oppose angular motion. Both ω0 and and τ are normal to the plane of the paper.ω0 is inside the paper and the sense of torque will be directed out of the paper.

(b) The condition of perfect rolling is that the force of friction at that point B should be zero. Hence after perfect rolling begins, the force of friction is zero.

Question-15

A solid disc and a ring both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed = 10 π rad/s. Which of the two will start to roll earlier? The coefficient of kinetic friction is μ k = 0.2.

Solution:

Let mass = m, 
radius = r = 0.1 m, 

initial angular speed = ω0 = 10 = 10π rad/s
The force due to Friction =
μk x N= ( x N= (μk mg) where N is the normal reaction of the table.

The centre of mass of a body will move with an acceleration ‘a’ given by Newton’s second law of motion.

                     mg) where N is the normal reaction of the table.

The centre of mass of a body will move with an acceleration ‘a’ given by Newton’s second law of motion.

                     μ k m g = m a       or
a= 
μk g                  ................. (1)
The torque due to friction( =
g                  ................. (1)
The torque due to friction( =
μ k mgR) will decrease the initial angular speed ω0 and hence will produce angular retardation, about the centre of mass. This acts counter to the rotation imparting a deceleration ( and hence will produce angular retardation, about the centre of mass. This acts counter to the rotation imparting a deceleration (α) to the rotation.
                  
μ km g R = - I α
                              α = - (μ  k mg r/ I)          .................. (2)
The acceleration generated by the frictional force imparts a linear acceleration to the body at its centre of mass and increases the velocity from zero to say v in a time say t. The centre of mass will have a speed v ( initial speed u=0) given by the relation,

                    v = 0+μ kgt
                    v =
μ k g t                      .................. (3)
The friction force, during the time t reduces the angular speed from
ω0 to to ω.
Then                      
ω = ω0 - - α τ
                                 = ω0 - ( - (μ kmg R / I) t    ................. (4)          

When the body starts rolling without slipping, the linear speed = angular speed x radius.
                                  v =
ωr
Using the values from (3) and (4) above,
                        
μ kg t = {(ω0 - ( - (μ k mg r )/ I) t} r   .............. (5)
Moment of inertia for disc = m r2
Using this value in ..(5) above, the disc starts rolling in a time
                                
t1 =
ω0 r / 2 r / 2μ kg        ............... (A)
Moment of inertia for ring = ½ m r2
Time for ring to start rolling t2 = 1/3
ω0 r / r / μkg  .. ............ (B)
Disc starts rolling earlier since t1 is smaller.

Question-16

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30o. The coefficient of static friction μs = 0.25. 
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination
= 0.25. 

(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination
θ of the plane is increased, at what value of θ does the cylinder begin to skid and not roll perfectly?

Solution:
 

Since the cylinder is rolling without slipping, static frictional coefficient is relevant here.
The forces acting on the object rolling are
1. weight acting downwards = mg.

This is resolved in to two rectangular components mg cos θ normal to the plane and mg Sin θ along the plane downwards
2. Reaction force normal to the plane and = mg cos
θ
3. Frictional force F acting parallel to the plane in upward direction and F = μ s R = μs mg cos mg cos θ

a) Force of Friction on the cylinder   F = 1/3 mg sinθ
                                                  F = 1/3 x 10 x 9.8 x sin 30o 
                                                     = 16.33 N

b) Work done against friction during rolling is zero.

c) The condition for no slipping is  

(F/N) = (1/3) tan θ = μs

where N is the normal reaction on the plane of the cylinder.

The value of

where N is the normal reaction on the plane of the cylinder.

The value of θ of the plane at which the cylinder begins to skid is given by tan θ=3μs=0.75

or θ =36.52’
= 370

At the maximum inclination, when the cylinder starts to skid the coefficient of friction reaches the limiting value of static friction μ s

Question-17

A ring, a disc, and a sphere, all of the same radius and mass, roll down an inclined plane from the same height h. Which of the three reaches the bottom (i) earliest, (ii) latest?

Solution:
I

The forces acting on the rolling body are: 
The forces acting on the object rolling down are
1. weight acting vertically downwards = mg.

This is resolved in to two rectangular components mg cos θ normal to the plane and mg Sin θ along the plane downwards

2. Reaction force normal to the plane and = mg cos θ

3. Frictional force F acting parallel to the plane in upward direction and F = μ s R = μs mg cos mg cos θ
    Now, force down the plane imparts an acceleration to the body at its centre of mass.
    i.e.                  m a = mg sin
θ - F                   ...............(1) 
    [a = acceleration of centre of mass]
    A torque is generated by the friction force about the centre of mass

The linear acceleration of a body rolling down an inclined plane making an angle θ with the horizontal is given by,

   a = g sinθ / (1 + I / Mr2)

where, I is the M. I of the body ( of mass M and radius r) about an axis passing through the point of contact of the body with the plane.

(a) Ring:
Moment of inertia about axis of rotation i = I = M r2
                                                      a1 = g sin
θ / (1 + I / Mr2)
                                                      a1 = g sin
θ / (1 + Mr2 / Mr2)
                                                          = ½ g sin
θ
(b) Disc:
Moment of inertia about axis of rotation = I = ½ M r2
                                                     a2= g sin
θ / (1 + I / Mr2)
                                                    a2 = g sin
θ / (1 + ½ Mr2 / Mr2)
                                                        = (2/3) g sin
θ 
(c) Sphere: 
Moment of inertia about rolling axis I = (2/5) M r2
                                                 a3 = g sin
θ / (1 + I / Mr2)
                                                     = g sin
θ / (1 + (2/5)Mr2 / mr2) = (5/7) g sinθ 
Since S and u are the same for all the three objects, t is least when a is greatest
The sphere reaches the the bottom first, the disc next and the ring the last.

Question-18

A solid cylinder of mass 20 kg and radius 0.12 m, rotating with initial angular speed of 125 rad/s is placed lightly (i.e, without any translational push) on a horizontal table with coefficient of kinetic friction μk = 0.15 between the cylinder and the table. 

(a) After how long does the cylinder start rolling?

(b) What is the initial

(i) translational energy, 

(ii) rotational energy, and

(iii) total energy of the cylinder?

(c) What is the final (i.e., after rolling begins) 

(i) translational energy, 

(ii) rotational energy, and

(iii) total energy of the cylinder?

(d) Is the final total energy equal to the initial total energy of motion of the cylinder? If not where does the difference of energy disappear?

(e) Account for the loss of total energy of motion in the following way; find the work done by the friction by the body as regards its rotational motion and the work done against friction by the body as regards its rotational motion. Show that the net work done by friction on the body is negative, equal in magnitude to the loss of total energy computed in (d) above.

Solution:

Mass = m = 20 kg radius = r = 0.12 m
Initial angular speed =
ω0 = 125 rad/s
Coefficient of kinetic friction =
= 125 rad/s
Coefficient of kinetic friction =
μk = 0.15
(a) Time to start rolling = t =
= 0.15
(a) Time to start rolling = t =
ω0 r / 3 r / 3 μkg
                                 = 125 x 0.12 / 3 x 0.15 x 9.8 s
                                 = 3.4 s
(b) Initial translational energy = 0 [ since there is no translational movement]
     Initial rotational energy Er1 = ½ I
g
                                 = 125 x 0.12 / 3 x 0.15 x 9.8 s
                                 = 3.4 s
(b) Initial translational energy = 0 [ since there is no translational movement]
     Initial rotational energy Er1 = ½ I
ω02 = ½ x ½ m r22 = ½ x ½ m r2ω02
                                             = ¼ x 20 x 0.122 x 1252 J
                                             = 1125 J
     Initial total energy = 1125 J
(c) Final rotational energy = ½ I
2
                                            = ¼ x 20 x 0.122 x 1252 J
                                            = 1125 J
     Initial total energy = 1125 J
(c) Final rotational energy = ½ I
ω2 
     Angular acceleration
 
     Angular acceleration
α= - (μk mg r/ I) [ from (2) above]
                                      = - (
mg r/ I) [ from (2) above]
                                    = - (
μk mg r/ ½ m r2
                                      = - 2
mg r/ ½ m r2
                                    = - 2
μk g / r
     t =
g / r
t =
ω0 r / 3 μk g [ from eqn (A) above]
                          
g [ from eqn (A) above]
                          
ω = ω0+ α t [ eqn (4) ]
                              =
ω0 - (2μk g / r) / ( g / r) / ( ω0  r / 3 μk g)
                              =
g)
                              =
ω0+ ( 2/3) ω0 = 1/3 ω0
Final rotational kinetic energy = ½ I (ω0/3)2 = 1125 / 9 J
                                          = 125 J
Final translational energy = ½ m v2
                                 v =
ω r = ω0 / 3 r
Final translational energy = ½ m v2  
                                    = ½ x 20 x ( 125 / 3 x 0.12)2
                                    = 250 J
Final total energy = 125 J + 250 J = 375 J.
The final total energy is not the same as initial total energy. The difference is 750 J. the decrease in the energy is due to the work done against friction. This is dissipated as heat . 

Question-19

Read each statement below carefully and state with reasons if it is true or false:
(i)  During rolling, the force of friction acts in the same direction as the direction of motion of the centre of mass of the body.
(ii)  The instantaneous speed of the point of contact during rolling is zero.
(iii) The instantaneous acceleration of the point of contact during rolling is zero.
(iv) For perfect rolling motion, work done against friction is zero.
(v)  A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Solution:

(i) True. Friction always opposes the relative motion at the point of contact.
(ii) True. The points do not move relative to each other.
(iii) Not true. The point of contact on the rolling object has a centripetal acceleration.
(iv) True. Since the point of contact does not move there is no displacement of the friction force.
(v) True.

The condition of perfect rolling is that the force of friction at that point B should be zero. Hence after perfect rolling begins, the force of friction is zero.





Test Your Skills Now!
Take a Quiz now
Reviewer Name