Question 1

A uniform rod of mass 2 kg and length 4 m is rotating about its center. If the moment of inertia about the center is 2.67 kg·m², what is the radius of gyration of the rod about its center?

Accepted Answer

Radius of gyration k = √(I/M) = √(2.67/2) = √1.335 ≈ 1.15 m.

Question 2

A system consists of three point masses: 2 kg at (0, 0), 3 kg at (2, 0), and 5 kg at (2, 2). The y-coordinate of the center of mass is:

Accepted Answer

y_cm = (m₁y₁ + m₂y₂ + m₃y₃)/(m₁ + m₂ + m₃) = (2×0 + 3×0 + 5×2)/(2+3+5) = 10/10 = 1.0... wait, recalculating: = 10/8 = 1.25.

Question 3

A torque of 12 N·m is applied to a wheel with moment of inertia 3 kg·m². If the wheel starts from rest, what is the angular acceleration?

Accepted Answer

Using τ = Iα, we get α = τ/I = 12/3 = 4 rad/s².

Question 4

Two particles of mass 3 kg and 5 kg are separated by 2 m. The center of mass of the system is at a distance of x m from the 3 kg mass. What is the value of x?

Accepted Answer

Using x_cm formula: x = (m₂ × d)/(m₁ + m₂) = (5 × 2)/(3 + 5) = 10/8 = 1.25 m from 3 kg mass.

Question 5

A solid disc of mass 4 kg and radius 0.5 m rotates with angular velocity 20 rad/s. What is its rotational kinetic energy? (Use I = ½MR² for disc)

Accepted Answer

KE_rot = ½Iω² = ½(½ × 4 × 0.25) × 400 = ½ × 0.5 × 400 = 200 J.

Question 6

Two particles of equal mass m are placed at the corners of a square of side a. What is the moment of inertia of the system about an axis passing through one corner and perpendicular to the plane of the square?

Accepted Answer

One mass is at the axis (I = 0), one is at distance a (I = ma²), and two others are at distance a√2 (I = 2m(a√2)² = 4ma²); but with only 2 particles given: I = 0 + ma² + 2ma² = 3ma².

Question 7

A solid sphere of mass 5 kg and radius 0.2 m rotates about a diameter with angular velocity ω = 4 rad/s. Calculate its rotational kinetic energy. (Use I = (2/5)MR²)

Accepted Answer

I = (2/5) × 5 × (0.2)² = 0.08 kg·m²; KE_rot = (1/2) × 0.08 × 16 = 6.4 J.

Question 8

Three point masses of 1 kg each are fixed at the vertices of an equilateral triangle of side 2 m. What is the moment of inertia of the system about an axis passing through one vertex and perpendicular to the plane?

Accepted Answer

One mass is at the axis (I = 0); two masses at distance 2 m each give I = 1×(2)² + 1×(2)² = 4 + 4 = 8 kg·m².

Question 9

Two blocks of mass 2 kg and 3 kg are connected by a light inextensible string over a pulley (treat as a point mass). If the system is released from rest, what is the magnitude of acceleration of the blocks? (Use g = 10 m/s², neglect pulley mass and friction)

Accepted Answer

Using a = (m₂ - m₁)g / (m₁ + m₂) = (3 - 2) × 10 / (2 + 3) = 10 / 5 = 2 m/s².

Question 10

A uniform disc of mass M = 2 kg and radius R = 0.4 m is rotating about its central axis. If its angular velocity increases from 0 to 10 rad/s in 5 seconds, calculate the angular acceleration.

Accepted Answer

Angular acceleration α = (ω₂ - ω₁) / t = (10 - 0) / 5 = 2 rad/s².

Question 11

A wheel of moment of inertia I = 0.5 kg·m² is rotating at 20 rad/s. A constant retarding torque of 2 N·m is applied. How long does it take to stop?

Accepted Answer

Using τ = Iα, α = τ/I = 2/0.5 = 4 rad/s²; using ω = ω₀ - αt, 0 = 20 - 4t gives t = 5 s.

Question 12

A uniform rod of length L = 0.6 m and mass M = 3 kg is pivoted at one end. What is its moment of inertia about the pivot point? (Use I = (1/3)ML²)

Accepted Answer

I = (1/3) × 3 × (0.6)² = (1/3) × 3 × 0.36 = 0.36 kg·m².

Question 13

A system of two particles of masses 2 kg and 3 kg are separated by 0.5 m. The centre of mass is located at a distance of 0.3 m from the 3 kg mass. Find the distance of centre of mass from the 2 kg mass.

Accepted Answer

Using the centre of mass formula: distance from 2 kg mass = (3 × 0.3) / (2 + 3) = 0.9 / 5 = 0.18 m ≈ 0.2 m.

Question 14

A rigid body rotates about a fixed axis with angular acceleration α = 2 rad/s². If the moment of inertia is I = 4 kg·m², what is the net torque acting on the body?

Accepted Answer

Using τ = Iα, net torque = 4 × 2 = 8 N·m.

Question 15

A uniform rod of mass 2 kg and length 4 m rotates about its center. If the moment of inertia about the center is 2.67 kg·m², calculate the radius of gyration of the rod about its center.

Accepted Answer

Using I = MK², we get K = √(I/M) = √(2.67/2) ≈ 1.15 m.

Question 16

Two particles of masses 3 kg and 5 kg are separated by 2 m. Their center of mass lies at a distance from the 3 kg mass of:

Accepted Answer

Center of mass position: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (3×0 + 5×2)/(3+5) = 1.25 m from 3 kg mass.

Question 17

A disc of mass 4 kg and radius 0.5 m rotates with angular velocity 10 rad/s. If another disc of mass 2 kg and radius 0.5 m is placed coaxially on it and friction causes them to rotate together, find the final common angular velocity.

Accepted Answer

Using conservation of angular momentum: I₁ω₁ = (I₁ + I₂)ω_f; (½×4×0.25×10) = (½×4×0.25 + ½×2×0.25)ω_f; 5 = 0.75ω_f; ω_f ≈ 6.67 rad/s.

Question 18

A system consists of three point masses: 1 kg at origin, 2 kg at (2, 0) m, and 3 kg at (0, 2) m. The y-coordinate of the center of mass is:

Accepted Answer

y_cm = (m₁y₁ + m₂y₂ + m₃y₃)/(m₁ + m₂ + m₃) = (1×0 + 2×0 + 3×2)/(1+2+3) = 6/6 = 1.0 m.

CDS System of Particles & Rotational Motion

System of Particles & Rotational Motion — Practice Questions