System of Particles and Rotational Motion Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A torque of \( 10 \, \text{Nm} \) is applied to a wheel with moment of inertia \( 5 \, \text{kg m}^2 \). What is its angular acceleration?

\( \tau = I \alpha \).

\( \tau = 10 \, \text{Nm} \), \( I = 5 \, \text{kg m}^2 \).

\( \alpha = \frac{\tau}{I} = \frac{10}{5} = 2 \, \text{rad/s}^2 \).

1.5 rad/s²
2.0 rad/s²
2.5 rad/s²
3.0 rad/s²
2

A system of two particles, \( 3 \, \text{kg} \) at \( (1, 2) \) and \( 1 \, \text{kg} \) at \( (5, 6) \), moves as a rigid body. What is the distance of the center of mass from the origin?

\( X = \frac{(3 \times 1) + (1 \times 5)}{3 + 1} = \frac{3 + 5}{4} = 2 \).

\( Y = \frac{(3 \times 2) + (1 \times 6)}{3 + 1} = \frac{6 + 6}{4} = 3 \).

Distance = \( \sqrt{X^2 + Y^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6 \, \text{m} \).

3.2 m
3.6 m
4.0 m
4.5 m
2

A \( 5 \, \text{kg} \) object moves with a velocity of \( 4 \, \hat{\mathbf{i}} + 2 \, \hat{\mathbf{j}} \, \text{m/s} \). What is the magnitude of the velocity of its center of mass?

For a single object, the center of mass velocity equals the object’s velocity.

Magnitude = \( \sqrt{(4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47 \, \text{m/s} \).

4.0 m/s
4.47 m/s
4.8 m/s
5.0 m/s
2

A solid cylinder of mass \( 3 \, \text{kg} \) and radius \( 0.1 \, \text{m} \) has a kinetic energy of \( 4.5 \, \text{J} \). What is its angular speed?

\( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 3 \times (0.1)^2 = 0.015 \, \text{kg m}^2 \).

\( K = \frac{1}{2} I \omega^2 \Rightarrow 4.5 = \frac{1}{2} \times 0.015 \times \omega^2 \Rightarrow 9 = 0.015 \omega^2 \Rightarrow \omega^2 = 600 \Rightarrow \omega = \sqrt{600} \approx 24.5 \, \text{rad/s} \).

22 rad/s
24.5 rad/s
26 rad/s
28 rad/s
2

A torque of \( 12 \, \text{Nm} \) is applied to a disk with moment of inertia \( 4 \, \text{kg m}^2 \). What is its angular acceleration?

\( \tau = I \alpha \).

\( \tau = 12 \, \text{Nm} \), \( I = 4 \, \text{kg m}^2 \).

\( \alpha = \frac{\tau}{I} = \frac{12}{4} = 3 \, \text{rad/s}^2 \).

2 rad/s²
3 rad/s²
4 rad/s²
5 rad/s²
2

Three particles of masses \( 4 \, \text{kg} \), \( 8 \, \text{kg} \), and \( 12 \, \text{kg} \) are at \( (1, 0) \), \( (3, 2) \), and \( (0, 4) \) respectively. What is the y-coordinate of their center of mass?

Formula: \( Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \).

Masses: \( 4, 8, 12 \, \text{kg} \); y-coordinates: \( 0, 2, 4 \).

\( Y = \frac{(4 \times 0) + (8 \times 2) + (12 \times 4)}{4 + 8 + 12} = \frac{0 + 16 + 48}{24} = \frac{64}{24} = 2.67 \, \text{m} \).

2.5 m
2.67 m
2.8 m
3.0 m
2

A \( 4 \, \text{kg} \) particle moves with velocity \( \mathbf{v} = 3 \, \hat{\mathbf{j}} \, \text{m/s} \) at \( \mathbf{r} = 2 \, \hat{\mathbf{i}} \, \text{m} \). What is the magnitude of its angular momentum about the origin?

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 0 & 0 \\ 0 & 3 & 0 \end{vmatrix} = \hat{\mathbf{k}} (2 \times 3 - 0 \times 0) = 6 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Magnitude = \( 6 \, \text{kg m}^2/\text{s} \).

5 kg m²/s
6 kg m²/s
7 kg m²/s
8 kg m²/s
2

A \( 1 \, \text{kg} \) particle moves with velocity \( \mathbf{v} = 3 \, \hat{\mathbf{i}} + 4 \, \hat{\mathbf{j}} \, \text{m/s} \) at position \( \mathbf{r} = 2 \, \hat{\mathbf{j}} \, \text{m} \). What is the z-component of its angular momentum about the origin?

Angular momentum: \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), where \( \mathbf{p} = m \mathbf{v} = 1 \times (3 \, \hat{\mathbf{i}} + 4 \, \hat{\mathbf{j}}) \).

\( \mathbf{L} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 2 & 0 \\ 3 & 4 & 0 \end{vmatrix} = \hat{\mathbf{k}} (0 \times 4 - 2 \times 3) = -6 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Z-component = \( -6 \, \text{kg m}^2/\text{s} \).

-8 kg m²/s
-6 kg m²/s
6 kg m²/s
8 kg m²/s
2

A thin ring of mass \( 4 \, \text{kg} \) and radius \( 0.6 \, \text{m} \) has an angular momentum of \( 12 \, \text{kg m}^2/\text{s} \). What is its angular velocity?

\( I = M R^2 = 4 \times (0.6)^2 = 1.44 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{12}{1.44} \approx 8.33 \, \text{rad/s} \).

8.0 rad/s
8.33 rad/s
8.5 rad/s
9.0 rad/s
2

A thin ring of mass \( 6 \, \text{kg} \) and radius \( 0.3 \, \text{m} \) rotates at \( 5 \, \text{rad/s} \). What is its angular momentum?

\( I = M R^2 = 6 \times (0.3)^2 = 0.54 \, \text{kg m}^2 \).

\( L = I \omega = 0.54 \times 5 = 2.7 \, \text{kg m}^2/\text{s} \).

2.5 kg m²/s
2.7 kg m²/s
3.0 kg m²/s
3.2 kg m²/s
2

A uniform circular ring of radius \( 2 \, \text{m} \) and mass \( 5 \, \text{kg} \) has its center at \( (3, 4) \). What is the position of its center of mass?

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

Given center at \( (3, 4) \), CM position = \( (3, 4) \).

\( (2, 3) \)
\( (3, 3) \)
\( (3, 4) \)
\( (4, 4) \)
3

A solid cylinder of mass \( 4 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) has a kinetic energy of \( 50 \, \text{J} \). What is its angular speed?

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

\( K = \frac{1}{2} I \omega^2 \Rightarrow 50 = \frac{1}{2} \times 0.5 \times \omega^2 \Rightarrow 100 = 0.5 \omega^2 \Rightarrow \omega^2 = 200 \Rightarrow \omega = \sqrt{200} \approx 14.14 \, \text{rad/s} \).

12 rad/s
14 rad/s
16 rad/s
18 rad/s
2

A hollow cylinder of mass \( 4 \, \text{kg} \) and radius \( 0.6 \, \text{m} \) rolls without slipping with a linear speed of \( 3 \, \text{m/s} \). What is its angular velocity?

\( v = \omega r \).

\( v = 3 \, \text{m/s} \), \( r = 0.6 \, \text{m} \).

\( \omega = \frac{v}{r} = \frac{3}{0.6} = 5 \, \text{rad/s} \).

4 rad/s
5 rad/s
6 rad/s
7 rad/s
2

A \( 2 \, \text{kg} \) mass rotates in a circle of radius \( 0.4 \, \text{m} \) with a speed of \( 6 \, \text{m/s} \). What is its angular momentum about the center?

\( L = m v r \).

\( m = 2 \, \text{kg} \), \( v = 6 \, \text{m/s} \), \( r = 0.4 \, \text{m} \).

\( L = 2 \times 6 \times 0.4 = 4.8 \, \text{kg m}^2/\text{s} \).

4.0 kg m²/s
4.5 kg m²/s
4.8 kg m²/s
5.0 kg m²/s
3

A uniform rod of mass \( 4 \, \text{kg} \) and length \( 2.5 \, \text{m} \) is pivoted at its center. What is its moment of inertia about the pivot?

For a rod pivoted at its center: \( I = \frac{1}{12} M L^2 \).

\( M = 4 \, \text{kg} \), \( L = 2.5 \, \text{m} \).

\( I = \frac{1}{12} \times 4 \times (2.5)^2 = \frac{4 \times 6.25}{12} = \frac{25}{12} \approx 2.08 \, \text{kg m}^2 \).

1.8 kg m²
2.08 kg m²
2.5 kg m²
3.0 kg m²
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0