Rotational Motion Study Guide for Competitive Exams

Rotational Motion

Angular Momentum Moment of Inertia Rolling Motion Torque

Q. In a system of two particles, if one particle has an angular momentum of L1 and the other has L2, what is the total angular momentum of the system?

A. L1 + L2
B. L1 - L2
C. L1 * L2
D. L1 / L2

Solution

Total angular momentum of the system is the vector sum of individual angular momenta: L_total = L1 + L2.

Correct Answer: A — L1 + L2

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Q. In a system of two rotating bodies, if one body has twice the moment of inertia of the other, how does their angular momentum compare if they rotate with the same angular velocity?

A. The same
B. Twice as much
C. Half as much
D. Four times as much

Solution

Angular momentum L = Iω; if one has twice the moment of inertia, it will have twice the angular momentum at the same angular velocity.

Correct Answer: B — Twice as much

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Q. In a system of two rotating disks, if disk A has twice the moment of inertia of disk B and they are both rotating with the same angular velocity, what can be said about their angular momentum?

A. LA = LB
B. LA = 2LB
C. LA = 4LB
D. LA = 1/2 LB

Solution

Angular momentum L = Iω; since IA = 2IB and ωA = ωB, LA = 2LB.

Correct Answer: B — LA = 2LB

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Q. In rolling motion, which type of energy is associated with the rotation of the object?

A. Translational kinetic energy
B. Rotational kinetic energy
C. Potential energy
D. Elastic potential energy

Solution

Rotational kinetic energy is associated with the rotation of the object in rolling motion.

Correct Answer: B — Rotational kinetic energy

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Q. The moment of inertia of a composite body can be calculated using which theorem?

A. Pythagorean theorem
B. Parallel axis theorem
C. Perpendicular axis theorem
D. Conservation of energy

Solution

The moment of inertia of a composite body can be calculated using the parallel axis theorem.

Correct Answer: B — Parallel axis theorem

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Q. The moment of inertia of a hollow cylinder about its central axis is given by which formula?

A. 1/2 MR^2
B. MR^2
C. 1/3 MR^2
D. 2/3 MR^2

Solution

The moment of inertia of a hollow cylinder about its central axis is I = MR^2.

Correct Answer: B — MR^2

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Q. The moment of inertia of a system of particles is calculated by summing which of the following?

A. Mass times distance from the axis
B. Mass times square of distance from the axis
C. Mass times angular velocity
D. Mass times linear velocity

Solution

The moment of inertia is calculated by summing the mass of each particle times the square of its distance from the axis of rotation: I = Σ(m_i * r_i²).

Correct Answer: B — Mass times square of distance from the axis

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Q. Two particles A and B of masses m1 and m2 are moving in a circular path with angular velocities ω1 and ω2 respectively. What is the total angular momentum of the system?

A. m1ω1 + m2ω2
B. m1ω1 - m2ω2
C. m1ω1m2ω2
D. m1ω1 + m2ω2/2

Solution

Total angular momentum L = m1ω1 + m2ω2 for particles moving in the same direction.

Correct Answer: A — m1ω1 + m2ω2

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Q. Two particles A and B of masses m1 and m2 are moving in a straight line with velocities v1 and v2 respectively. If they collide elastically, which of the following statements is true regarding their angular momentum about the center of mass?

A. It is conserved
B. It is not conserved
C. Depends on the masses
D. Depends on the velocities

Solution

Angular momentum about the center of mass is conserved in an elastic collision.

Correct Answer: A — It is conserved

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Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about a point O located at the center of mass?

A. (m1v1 + m2v2)
B. (m1v1 - m2v2)
C. m1v1 + m2v2
D. 0

Solution

Total angular momentum is the sum of individual angular momenta, which is m1v1 + m2v2.

Correct Answer: C — m1v1 + m2v2

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Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about the origin if they are at a distance r from the origin?

A. m1v1r + m2v2r
B. m1v1r - m2v2r
C. m1v1r + m2(-v2)r
D. 0

Solution

Total angular momentum L = m1v1r - m2v2r, but since they are in opposite directions, it simplifies to m1v1r + m2v2r.

Correct Answer: A — m1v1r + m2v2r

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Q. Two particles A and B of masses m1 and m2 are moving in opposite directions with velocities v1 and v2 respectively. What is the total angular momentum of the system about the origin?

A. m1v1 + m2v2
B. m1v1 - m2v2
C. m1v1 + m2(-v2)
D. m1v1 + m2v2

Solution

Total angular momentum L = m1v1 + m2(-v2) = m1v1 - m2v2.

Correct Answer: C — m1v1 + m2(-v2)

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A. (m1v1 + m2v2)r
B. (m1v1 - m2v2)r
C. 0
D. (m1v1 + m2v2)/2

Solution

Since they are moving in opposite directions, the total angular momentum about point O is zero.

Correct Answer: A — (m1v1 + m2v2)r

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Q. Two particles A and B of masses m1 and m2 are moving with velocities v1 and v2 respectively. If they collide elastically, which of the following statements is true regarding their angular momentum about the center of mass?

A. It is conserved
B. It is not conserved
C. Depends on the masses
D. Depends on the velocities

Solution

Angular momentum is conserved in an elastic collision about the center of mass.

Correct Answer: A — It is conserved

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Q. Two particles of masses m1 and m2 are moving in a circular path of radius r with angular velocities ω1 and ω2 respectively. What is the total angular momentum of the system?

A. (m1 + m2)r(ω1 + ω2)
B. m1rω1 + m2rω2
C. m1r^2ω1 + m2r^2ω2
D. m1ω1 + m2ω2

Solution

Total angular momentum L = m1rω1 + m2rω2.

Correct Answer: B — m1rω1 + m2rω2

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Q. Two particles of masses m1 and m2 are moving in a circular path with radii r1 and r2 respectively. If they have the same angular velocity, what is the ratio of their angular momenta?

A. m1r1/m2r2
B. m1/m2
C. r1/r2
D. m1r2/m2r1

Solution

Angular momentum L = mvr, thus L1/L2 = (m1r1)/(m2r2) when ω is constant.

Correct Answer: A — m1r1/m2r2

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Q. Two particles of masses m1 and m2 are moving in a straight line with velocities v1 and v2 respectively. If they collide elastically, what is the expression for the change in angular momentum about the center of mass?

A. m1v1 + m2v2
B. m1v1 - m2v2
C. 0
D. m1v1 + m2v2 - (m1v1' + m2v2')

Solution

In an elastic collision, the total angular momentum about the center of mass is conserved.

Correct Answer: D — m1v1 + m2v2 - (m1v1' + m2v2')

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Q. What happens to the moment of inertia of a rigid body if it is rotated about an axis that is not its principal axis?

A. It increases
B. It decreases
C. It remains the same
D. It becomes zero

Solution

The moment of inertia can change when rotating about an axis that is not a principal axis due to the distribution of mass relative to the new axis.

Correct Answer: A — It increases

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Q. What is the acceleration of a rolling object down an incline if the incline angle is θ?

A. g sin(θ)
B. g sin(θ)/2
C. g sin(θ)/3
D. g sin(θ)/4

Solution

The acceleration of a rolling object down an incline is given by g sin(θ)/2, considering both translational and rotational motion.

Correct Answer: B — g sin(θ)/2

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Q. What is the angular momentum of a rolling object about its center of mass?

A. mv
B. Iω
C. mv + Iω
D. 0

Solution

The angular momentum L of a rolling object about its center of mass is given by L = mv + Iω, where I is the moment of inertia and ω is the angular velocity.

Correct Answer: C — mv + Iω

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Q. What is the condition for an object to be in rotational equilibrium?

A. Net force is zero
B. Net torque is zero
C. Both net force and net torque are zero
D. None of the above

Solution

For rotational equilibrium, the net torque acting on the object must be zero.

Correct Answer: B — Net torque is zero

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Q. What is the condition for rolling without slipping?

A. v = Rω
B. v = 2Rω
C. v = 0
D. v = R^2ω

Solution

The condition for rolling without slipping is that the linear velocity v of the center of mass is equal to the product of the radius R and the angular velocity ω, i.e., v = Rω.

Correct Answer: A — v = Rω

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Q. What is the effect on the angular momentum of a system if an external torque is applied?

A. It increases
B. It decreases
C. It changes
D. It remains constant

Solution

An external torque changes the angular momentum of the system.

Correct Answer: C — It changes

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Q. What is the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to its plane?

A. 1/2 MR^2
B. MR^2
C. 1/4 MR^2
D. 2/3 MR^2

Solution

The moment of inertia of a disk about an axis through its center is I = 1/2 MR^2.

Correct Answer: A — 1/2 MR^2

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Q. What is the moment of inertia of a solid cylinder of mass M and radius R about its central axis?

A. 1/2 MR^2
B. 1/3 MR^2
C. MR^2
D. 2/5 MR^2

Solution

The moment of inertia of a solid cylinder about its central axis is given by I = 1/2 MR^2.

Correct Answer: A — 1/2 MR^2

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Q. What is the moment of inertia of a solid disk about its central axis?

A. (1/2)MR^2
B. (1/3)MR^2
C. (1/4)MR^2
D. MR^2

Solution

The moment of inertia of a solid disk about its central axis is (1/2)MR^2.

Correct Answer: A — (1/2)MR^2

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Q. What is the moment of inertia of a solid sphere about an axis through its center?

A. (2/5)mr^2
B. (1/2)mr^2
C. (1/3)mr^2
D. (5/2)mr^2

Solution

The moment of inertia of a solid sphere about an axis through its center is given by I = (2/5)mr^2.

Correct Answer: A — (2/5)mr^2

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Q. What is the moment of inertia of a solid sphere of mass M and radius R about an axis through its center?

A. 2/5 MR^2
B. 3/5 MR^2
C. 1/2 MR^2
D. MR^2

Solution

The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.

Correct Answer: A — 2/5 MR^2

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Q. What is the moment of inertia of a thin circular hoop of mass M and radius R about an axis through its center?

A. MR^2
B. 1/2 MR^2
C. 1/3 MR^2
D. 2/5 MR^2

Solution

The moment of inertia of a thin circular hoop about an axis through its center is I = MR^2.

Correct Answer: A — MR^2

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Q. What is the moment of inertia of a thin circular plate of mass M and radius R about an axis through its center and perpendicular to its plane?

A. 1/2 MR^2
B. MR^2
C. 1/4 MR^2
D. 1/3 MR^2

Solution

The moment of inertia of a thin circular plate about an axis through its center is I = 1/2 MR^2.

Correct Answer: A — 1/2 MR^2

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Rotational Motion

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