System of Particles and Rotational Motion Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A torque of \( 9 \, \text{Nm} \) is applied to a wheel with moment of inertia \( 3 \, \text{kg m}^2 \). What is its angular acceleration?

\( \tau = I \alpha \).

\( \tau = 9 \, \text{Nm} \), \( I = 3 \, \text{kg m}^2 \).

\( \alpha = \frac{\tau}{I} = \frac{9}{3} = 3 \, \text{rad/s}^2 \).

2 rad/s²
3 rad/s²
4 rad/s²
5 rad/s²
2

A uniform disk of mass \( 4 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) rotates at \( 8 \, \text{rad/s} \). What is its angular momentum about its axis?

\( L = I \omega \).

For a disk: \( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 4 \times (0.5)^2 = 0.5 \, \text{kg m}^2 \).

\( L = 0.5 \times 8 = 4 \, \text{kg m}^2/\text{s} \).

2 kg m²/s
3 kg m²/s
4 kg m²/s
5 kg m²/s
3

A uniform disk of mass \( 2 \, \text{kg} \) and radius \( 0.2 \, \text{m} \) rotates about its center with an angular speed of \( 10 \, \text{rad/s} \). What is its rotational kinetic energy?

Rotational KE: \( K = \frac{1}{2} I \omega^2 \).

Moment of inertia of a disk: \( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04 \, \text{kg m}^2 \).

\( K = \frac{1}{2} \times 0.04 \times (10)^2 = 0.02 \times 100 = 2 \, \text{J} \).

1 J
2 J
3 J
4 J
2

Why does a symmetric body rotating about its symmetry axis have angular momentum aligned with the axis?

Symmetry ensures perpendicular components of angular momentum cancel out, leaving \( \mathbf{L} = I \mathbf{\omega} \) aligned with the axis of rotation.

Due to high torque
Due to cancellation of perpendicular components
Due to zero angular velocity
Due to uneven mass distribution
2

A uniform rod of length \( 2 \, \text{m} \) and mass \( 4 \, \text{kg} \) lies along the x-axis with its left end at the origin. What is the position vector of its center of mass?

For a uniform rod, the center of mass is at its geometric center.

Length = \( 2 \, \text{m} \), so the center is at \( 1 \, \text{m} \) from the origin along the x-axis.

Position vector: \( \mathbf{R} = 1 \, \hat{\mathbf{i}} + 0 \, \hat{\mathbf{j}} + 0 \, \hat{\mathbf{k}} = 1 \, \hat{\mathbf{i}} \, \text{m} \).

\( 0.5 \, \hat{\mathbf{i}} \, \text{m} \)
\( 1 \, \hat{\mathbf{i}} \, \text{m} \)
\( 1.5 \, \hat{\mathbf{i}} \, \text{m} \)
\( 2 \, \hat{\mathbf{i}} \, \text{m} \)
2

In rotational motion, what is the relationship between linear velocity and angular velocity for a particle?

The linear velocity \( v \) of a particle in rotational motion is related to angular velocity \( \omega \) by \( v = \omega r \), where \( r \) is the perpendicular distance from the axis.

\( v = \omega / r \)
\( v = \omega r \)
\( v = \omega r^2 \)
\( v = \omega / r^2 \)
2

A solid sphere of mass \( 3 \, \text{kg} \) and radius \( 0.4 \, \text{m} \) has an angular momentum of \( 12 \, \text{kg m}^2/\text{s} \). What is its angular velocity?

\( I = \frac{2}{5} M R^2 = \frac{2}{5} \times 3 \times (0.4)^2 = 0.192 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{12}{0.192} \approx 62.5 \, \text{rad/s} \).

60 rad/s
62.5 rad/s
65 rad/s
67 rad/s
2

For a body rotating about a fixed axis, what is true about particles on the axis?

Particles on the axis of rotation have a perpendicular distance \( r = 0 \) from the axis, so their linear velocity \( v = \omega r = 0 \), meaning they remain stationary.

They move in circular paths
They remain stationary
They have the maximum linear velocity
They have zero angular velocity
2

Which of the following is true about the angular velocity of a rigid body rotating about a fixed axis?

For a rigid body rotating about a fixed axis, all particles have the same angular velocity, regardless of their distance from the axis, as it’s a property of the body’s rotation as a whole.

It varies with distance from the axis
It is the same for all particles in the body
It is zero for points farthest from the axis
It depends on the mass of each particle
2

A solid sphere of mass \( 3 \, \text{kg} \) and radius \( 0.1 \, \text{m} \) rolls without slipping down an incline. If its linear speed is \( 5 \, \text{m/s} \), what is its angular velocity?

For rolling without slipping: \( v = \omega r \).

\( v = 5 \, \text{m/s} \), \( r = 0.1 \, \text{m} \).

\( \omega = \frac{v}{r} = \frac{5}{0.1} = 50 \, \text{rad/s} \).

40 rad/s
50 rad/s
60 rad/s
70 rad/s
2

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