Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 7.29:

Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

(b) What is the force of friction after perfect rolling begins?

Answer:

A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.

Question 7.30:

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s^-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μ_k = 0.2.

Answer:

Disc

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω₀ =10 π rad s^–1

Coefficient of kinetic friction, μ_k = 0.2

Initial velocity of both the objects, u = 0

Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma

μ_kmg= ma

Where,

a = Acceleration produced in the objects

m = Mass

∴a = μ_kg … (i)

As per the first equation of motion, the final velocity of the objects can be obtained as:

v = u + at

= 0 + μ_kgt

= μ_kgt … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.

Torque, τ= –Iα

α = Angular acceleration

μxmgr = –Iα

Using the first equation of rotational motion to obtain the final angular speed:

Rolling starts when linear velocity, v = rω

Equating equations (ii) and (v), we get:

Since t_d > t_r, the disc will start rolling before the ring.

Question 7.31:

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µ_s = 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?

Answer:

Mass of the cylinder, m = 10 kg

Radius of the cylinder, r = 15 cm = 0.15 m

Co-efficient of kinetic friction, µ_k = 0.25

Angle of inclination, θ = 30°

Moment of inertia of a solid cylinder about its geometric axis, 

The various forces acting on the cylinder are shown in the following figure:

The acceleration of the cylinder is given as:

(a) Using Newton’s second law of motion, we can write net force as:

f_net = ma

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:

Question 7.32:

Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

(b) The instantaneous speed of the point of contact during rolling is zero.

(c) The instantaneous acceleration of the point of contact during rolling is zero.

(d) For perfect rolling motion, work done against friction is zero.

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Answer:

(a) False

Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

(b) True

Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False

When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True

When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True

The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

Question 7.33:

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

(a) Show pi = p’i + miV

Where pi is the momentum of the i^th particle (of mass mi) and p′ i = mi v′ i. Note v′ i is the velocity of the i^th particle relative to the centre of mass.

Also, prove using the definition of the centre of mass 

(b) Show K = K′ + ½MV²

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.

(c) Show L = L′ + R × MV

Where  is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember r’i = ri – R; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(d) Show 

Further, show that

where τ′_ext is the sum of all external torques acting on the system about the centre of mass.

(Hint: Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Answer:

(a)Take a system of i moving particles.

Mass of the i^th particle = mi

Velocity of the i^th particle = vi

Hence, momentum of the i^th particle, pi = mi vi

Velocity of the centre of mass = V

The velocity of the i^th particle with respect to the centre of mass of the system is given as:

v’i = vi – V … (1)

Multiplying mi throughout equation (1), we get:

mi v’i = mi vi – mi V

p’i = p_i – _­m_i V

Where,

pi’ = mivi’ = Momentum of the i^th particle with respect to the centre of mass of the system

∴pi = p’i ­+ mi V

We have the relation: p’i = mivi’

Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:

(b) We have the relation for velocity of the i^th particle as:

vi = v’i + V

… (2)

Taking the dot product of equation (2) with itself, we get:

Where,

K =   = Total kinetic energy of the system of particles

K’ =   = Total kinetic energy of the system of particles with respect to the centre of mass

 = Kinetic energy of the translation of the system as a whole

(c) Position vector of the i^th particle with respect to origin = ri

Position vector of the i^th particle with respect to the centre of mass = r’i

Position vector of the centre of mass with respect to the origin = R

It is given that:

r’i = ri – R

ri = r’i + R

We have from part (a),

pi = p’i ­+ mi V

Taking the cross product of this relation by ri, we get:

(d) We have the relation:

We have the relation: