Question:

Two discs of moments 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 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 ω1​=ω2​.

Answer:

(a) The angular speed of the two-disc system is the same as the angular speed of either of the individual discs, ω1​=ω2​.

(b) The kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs because some of the energy is lost due to friction between the two discs. This loss in energy is due to the heat generated by the friction between the two discs.

Question:

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) True. During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body because it is the force of friction that helps the body to roll.

(b) False. The instantaneous speed of the point of contact during rolling is not zero because the point of contact is constantly changing due to the rolling motion.

(c) False. The instantaneous acceleration of the point of contact during rolling is not zero because the point of contact is constantly changing due to the rolling motion.

(d) True. For perfect rolling motion, work done against friction is zero because there is no slipping of the body, thus no work is done against friction.

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

Question:

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (a) Show p=pi′​+mi​V where pi​ is the momentum of the ith particle (of mass mi​) and pi′​=mi​vi′​. Note vi′​ is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the center of mass ∑pi′​=0 (b) Show K=K′+1​/2MV^2 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 MV^2/2 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 L′=ri′​×pi′​ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember ri′​=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 dtdL′​=∑ri′​×dp′​/dt Further, show that dL′​/dt=τext′​ where τext′​ 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 Newtons Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Answer:

(a) p=pi′​+mi​V

Proof:

Let R be the position vector of the centre of mass of the system of particles.

Then,

∑mi​R=∑mi​(ri​−ri′​)

=∑mi​ri​−∑mi​ri′​

=∑pi​−∑mi​vi′​

⇒∑pi′​=0

Therefore, pi′​=mi​vi′​

⇒p=pi′​+mi​V

(b) K=K′+1​/2MV^2

Proof:

Let K′ be the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass.

Then,

K′=∑mi​v′2​i

=∑mi​(vi​−V)2

=∑mi​vi2​−2V∑mi​vi​+V2∑mi​

=∑mi​vi2​−2V∑pi​+V2∑mi​

=∑mi​vi2​−2V∑pi​+V2M

Where M is the total mass of the system of particles.

Also,

K=∑mi​v2​i

=∑mi​vi2​+V2∑mi​

=∑mi​vi2​+V2M

Therefore,

K=K′+1​/2MV^2

(c) L=L′+R×MV

Proof:

Let L′=ri′​×pi′​ be the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass.

Then,

L′=∑ri′​×pi′​

=∑(ri​−R)×mi​vi′​

=∑ri​×mi​vi′​−R×∑mi​vi′​

=∑ri​×mi​vi′​−R×MV

Also,

L=∑ri​×pi​

=∑ri​×mi​vi​

Therefore,

L=L′+R×MV

(d) dtdL′​=∑ri′​×dp′​/dt

Proof:

Let τext′​ be the sum of all external torques acting on the system about the centre of mass.

Then,

dL′​/dt=τext′​

=∑ri′​×Fext′​

=∑ri′​×dp′​/dt

Where Fext′​ is the net external force acting on the system.

Therefore,

dtdL′​=∑ri′​×dp′​/dt

Question:

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^2=2gh​/1+k^2​/R^2 using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Answer:

1. Calculate the gravitational force (Fg) acting on the body. This can be done by multiplying the mass of the body (m) by the acceleration due to gravity (g). Fg = mg

2. Calculate the normal force (Fn) acting on the body. This can be done by multiplying the mass of the body by the normal acceleration (a) of the body. Fn = ma

3. Calculate the torque (T) acting on the body. This can be done by multiplying the force (F) by the radius of gyration (k) of the body about its symmetry axis. T = Fk

4. Calculate the angular acceleration (α) of the body. This can be done by dividing the torque (T) by the moment of inertia (I) of the body. α = T/I

5. Calculate the linear acceleration (a) of the body. This can be done by multiplying the angular acceleration (α) by the radius (R) of the body. a = αR

6. Calculate the velocity (v) of the body. This can be done by multiplying the linear acceleration (a) by the time (t) it takes for the body to reach the bottom of the inclined plane. v = at

7. Calculate the final velocity (v) of the body. This can be done by solving the equation v^2=2gh/1+k^2/R^2 for v. v = √2gh/1+k^2/R^2

Question:

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m^2. (a) What is his new angular speed? (Neglect friction.) (b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Answer:

(a) The new angular speed can be calculated using the equation Iω = Iω’ where I is the moment of inertia, ω is the initial angular speed, and ω’ is the new angular speed.

Iω = Iω’ 7.6 kg m^2 * (30 revolutions/minute) = 7.6 kg m^2 * ω'

ω’ = 30 revolutions/minute

(b) Kinetic energy is not conserved in the process because the man is doing work on the weights, thus changing their kinetic energy. The change in kinetic energy comes from the work done by the man.

Question:

(a) Prove the theorem of perpendicular axes. (Hint : Square of the distance of a point (x, y) in the xy plane from an axis through the origin and perpendicular to the plane is x^2+y^2). (b) Prove the theorem of parallel axes.(Hint : If the centre of mass is chosen to be the origin ∑mi​ri​=0 ).

Answer:

(a) Proof of Theorem of Perpendicular Axes:

Let P be a point in the xy plane with coordinates (x, y).

We will prove that the square of the distance of P from the x-axis is equal to the square of the distance of P from the y-axis.

Let d1 be the distance of P from the x-axis and d2 be the distance of P from the y-axis.

We need to prove that d1^2 = d2^2

Let O be the origin in the xy plane.

Let P1 be the point on the x-axis such that OP1 = d1

Let P2 be the point on the y-axis such that OP2 = d2

By the Pythagorean theorem, we have

OP^2 = OP1^2 + OP2^2

Substituting the coordinates of P, we have

(x^2 + y^2) = (d1^2) + (d2^2)

Therefore, d1^2 = d2^2

Hence, the theorem of perpendicular axes is proved.

(b) Proof of Theorem of Parallel Axes:

Let P be a point in the xy plane with coordinates (x, y).

Let O be the origin in the xy plane.

Let P1 be the point on the x-axis such that OP1 = x

Let P2 be the point on the y-axis such that OP2 = y

We will prove that the square of the distance of P from the x-axis is equal to the square of the distance of P from the y-axis.

Let d1 be the distance of P from the x-axis and d2 be the distance of P from the y-axis.

We need to prove that d1^2 = d2^2

By the theorem of perpendicular axes, we have

(x^2 + y^2) = (d1^2) + (d2^2)

We also know that the centre of mass of a system of particles is at the origin, i.e. ∑mi​ri​=0

Substituting the coordinates of P1 and P2, we have

x + y = 0

Substituting this in the equation (x^2 + y^2) = (d1^2) + (d2^2), we have

x^2 + y^2 = (d1^2) + (d2^2)

x^2 + (x + y)^2 = (d1^2) + (d2^2)

x^2 + x^2 + 2xy + y^2 = (d1^2) + (d2^2)

2x^2 + 2xy = (d1^2) + (d2^2)

2x^2 = (d1^2) + (d2^2)

Therefore, d1^2 = d2^2

Hence, the theorem of parallel axes is proved.

Question:

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

Answer:

1. Determine the type of motion for both the disc and the ring. Answer: The disc will have a rotational motion and the ring will have a rolling motion.

2. Calculate the torque required to start the motion of both the disc and the ring. Answer: The torque required to start the motion of the disc is 100πNm and the torque required to start the motion of the ring is 20πNm.

3. Calculate the frictional force acting on both the disc and the ring. Answer: The frictional force acting on the disc is 20N and the frictional force acting on the ring is 4N.

4. Determine which of the two will start to roll earlier. Answer: The ring will start to roll earlier as it requires less torque and less frictional force to start its motion.

Question:

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

Answer:

1. Convert the mass of the bullet from grams to kilograms: 10 g = 0.01 kg

2. Calculate the momentum of the bullet: momentum = mass x velocity = 0.01 kg x 500 m/s = 5 kg m/s

3. Calculate the moment of inertia of the door: moment of inertia = mass x radius2 = 12 kg x 0.5m2 = 6 kg m2

4. Calculate the angular speed of the door: angular speed = momentum/moment of inertia = 5 kg m/s/6 kg m2 = 0.83 rad/s

Question:

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 θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly ?

Answer:

(a) The force of friction acting on the cylinder can be calculated using the equation Ffriction = μs * mg, where μs is the coefficient of static friction, m is the mass of the cylinder, and g is the acceleration due to gravity. Substituting the given values, we get:

Ffriction = 0.25 * 10 * 9.8 = 24.5 N

(b) The work done against friction during rolling can be calculated using the equation W = Ffriction * s, where Ffriction is the force of friction acting on the cylinder, and s is the distance traveled. Since the distance traveled is not given, we cannot calculate the work done against friction.

(c) The value of θ at which the cylinder begins to skid, and not roll perfectly, can be calculated using the equation μs = tanθ, where μs is the coefficient of static friction and θ is the angle of inclination. Substituting the given value of μs, we get:

tanθ = 0.25

Therefore, θ = tan-1(0.25) = 14.0°