System of Particles and Rotational Motion Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does a ceiling fan continue to rotate after the power is turned off?

With no significant external torque (assuming negligible friction), angular momentum is conserved (\( \mathbf{L} = I \mathbf{\omega} \)), so the fan continues rotating until friction slows it down.

Due to increasing torque
Due to conservation of angular momentum
Due to zero moment of inertia
Due to external forces accelerating it
2

A uniform rod of mass \( 7 \, \text{kg} \) and length \( 1.5 \, \text{m} \) is pivoted at one end. What is its moment of inertia about the pivot?

For a rod pivoted at one end: \( I = \frac{1}{3} M L^2 \).

\( M = 7 \, \text{kg} \), \( L = 1.5 \, \text{m} \).

\( I = \frac{1}{3} \times 7 \times (1.5)^2 = \frac{7 \times 2.25}{3} = 5.25 \, \text{kg m}^2 \).

5.0 kg m²
5.25 kg m²
5.5 kg m²
5.75 kg m²
2

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

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 0 & 0 \\ 2 & 3 & 0 \end{vmatrix} = \hat{\mathbf{k}} (4 \times 3 - 0 \times 2) = 12 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Z-component = \( 12 \, \text{kg m}^2/\text{s} \).

10 kg m²/s
12 kg m²/s
14 kg m²/s
16 kg m²/s
2

A solid sphere of mass \( 2 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) has an angular momentum of \( 10 \, \text{kg m}^2/\text{s} \). What is its angular velocity?

\( I = \frac{2}{5} M R^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{10}{0.2} = 50 \, \text{rad/s} \).

45 rad/s
48 rad/s
50 rad/s
52 rad/s
3

A system consists of two particles: \( 1 \, \text{kg} \) at \( (2, 3) \) and \( 2 \, \text{kg} \) at \( (4, 1) \). What is the y-coordinate of the center of mass?

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

\( m_1 = 1 \, \text{kg} \), \( y_1 = 3 \); \( m_2 = 2 \, \text{kg} \), \( y_2 = 1 \).

\( Y = \frac{(1 \times 3) + (2 \times 1)}{1 + 2} = \frac{3 + 2}{3} = \frac{5}{3} \approx 1.67 \, \text{m} \).

1.33 m
1.67 m
2.0 m
2.33 m
2

A \( 5 \, \text{kg} \) particle moves with velocity \( \mathbf{v} = 4 \, \hat{\mathbf{i}} \, \text{m/s} \) at \( \mathbf{r} = -2 \, \hat{\mathbf{j}} \, \text{m} \). What is the magnitude of its angular momentum about the origin?

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 0 \\ 4 & 0 & 0 \end{vmatrix} = \hat{\mathbf{k}} (0 \times 0 - (-2) \times 4) = 8 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Magnitude = \( 8 \, \text{kg m}^2/\text{s} \).

7 kg m²/s
8 kg m²/s
9 kg m²/s
10 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 \( \mathbf{r} = 2 \, \hat{\mathbf{i}} \, \text{m} \). What is the z-component of its angular momentum?

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 0 & 0 \\ -3 & 4 & 0 \end{vmatrix} = \hat{\mathbf{k}} (2 \times 4 - 0 \times (-3)) = 8 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Z-component = \( 8 \, \text{kg m}^2/\text{s} \).

6 kg m²/s
8 kg m²/s
10 kg m²/s
12 kg m²/s
2

A force \( \mathbf{F} = 4 \, \hat{\mathbf{i}} - 3 \, \hat{\mathbf{j}} \, \text{N} \) acts at \( \mathbf{r} = 2 \, \hat{\mathbf{i}} + 1 \, \hat{\mathbf{j}} \, \text{m} \). What is the magnitude of the torque about the origin?

\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & 0 \\ 4 & -3 & 0 \end{vmatrix} = \hat{\mathbf{k}} (2 \times (-3) - 1 \times 4) = \hat{\mathbf{k}} (-6 - 4) = -10 \, \hat{\mathbf{k}} \, \text{Nm} \).

Magnitude = \( 10 \, \text{Nm} \).

8 Nm
9 Nm
10 Nm
11 Nm
3

A \( 5 \, \text{kg} \) particle moves with velocity \( \mathbf{v} = 4 \, \hat{\mathbf{i}} \, \text{m/s} \) at \( \mathbf{r} = -3 \, \hat{\mathbf{j}} \, \text{m} \). What is the magnitude of its angular momentum about the origin?

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -3 & 0 \\ 4 & 0 & 0 \end{vmatrix} = \hat{\mathbf{k}} (0 \times 0 - (-3) \times 4) = 12 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Magnitude = \( 12 \, \text{kg m}^2/\text{s} \).

10 kg m²/s
12 kg m²/s
14 kg m²/s
16 kg m²/s
2

What is the direction of torque relative to the force and position vector?

Torque \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \) is perpendicular to both the position vector \( \mathbf{r} \) and force \( \mathbf{F} \), following the right-hand rule.

Parallel to the force
Parallel to the position vector
Perpendicular to both force and position vector
In the direction of rotation
3

A \( 3 \, \text{kg} \) mass rotates in a circle of radius \( 0.6 \, \text{m} \) with a speed of \( 4 \, \text{m/s} \). What is its angular momentum about the center?

\( L = m v r \).

\( m = 3 \, \text{kg} \), \( v = 4 \, \text{m/s} \), \( r = 0.6 \, \text{m} \).

\( L = 3 \times 4 \times 0.6 = 7.2 \, \text{kg m}^2/\text{s} \).

6.5 kg m²/s
7.0 kg m²/s
7.2 kg m²/s
7.5 kg m²/s
3

What distinguishes rolling motion from pure rotational motion?

Rolling motion combines translation (motion of the center) and rotation about an axis, while pure rotational motion involves only rotation about a fixed axis without translation.

Rolling motion involves only translation
Rolling motion combines translation and rotation
Rolling motion has no rotational component
Rolling motion occurs without an axis
2

What is the significance of the right-hand rule in determining the direction of a vector product?

The right-hand rule provides a consistent method to determine the direction of the resulting vector (e.g., \( \mathbf{a} \times \mathbf{b} \)), ensuring it is perpendicular to the plane of the two vectors in a direction based on their relative orientation.

It determines the magnitude of the vector
It ensures the vector lies in the plane of the two vectors
It defines the direction perpendicular to the plane of the vectors
It calculates the angle between the vectors
3

A particle in a rigid body rotates about a fixed axis at a distance of \( 0.8 \, \text{m} \) with an angular velocity of \( 5 \, \text{rad/s} \). What is its linear speed?

Linear speed: \( v = \omega r \).

\( \omega = 5 \, \text{rad/s} \), \( r = 0.8 \, \text{m} \).

\( v = 5 \times 0.8 = 4 \, \text{m/s} \).

3 m/s
4 m/s
5 m/s
6 m/s
2

A disk with moment of inertia \( 3 \, \text{kg m}^2 \) rotates at \( 8 \, \text{rad/s} \). A torque of \( 9 \, \text{Nm} \) acts for \( 2 \, \text{s} \). What is its final angular velocity?

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

\( \Delta \omega = \alpha t = 3 \times 2 = 6 \, \text{rad/s} \).

\( \omega = \omega_0 + \Delta \omega = 8 + 6 = 14 \, \text{rad/s} \).

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