System of Particles and Rotational Motion Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A hollow cylinder of mass \( 5 \, \text{kg} \) and radius \( 0.2 \, \text{m} \) rotates about its axis at \( 10 \, \text{rad/s} \). What is its rotational kinetic energy?

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

\( K = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.2 \times (10)^2 = 0.1 \times 100 = 10 \, \text{J} \).

9 J
10 J
11 J
12 J
2

A solid sphere of mass \( 3 \, \text{kg} \) and radius \( 0.2 \, \text{m} \) rotates at \( 15 \, \text{rad/s} \). What is its angular momentum about its axis?

\( I = \frac{2}{5} M R^2 = \frac{2}{5} \times 3 \times (0.2)^2 = 0.048 \, \text{kg m}^2 \).

\( L = I \omega = 0.048 \times 15 = 0.72 \, \text{kg m}^2/\text{s} \).

0.6 kg m²/s
0.72 kg m²/s
0.8 kg m²/s
1.0 kg m²/s
2

A uniform triangular lamina has vertices at \( (0, 0) \), \( (6, 0) \), and \( (0, 8) \). What are the coordinates of its center of mass?

For a uniform triangular lamina, the center of mass is at the centroid, given by the average of the vertices.

\( X = \frac{0 + 6 + 0}{3} = 2 \), \( Y = \frac{0 + 0 + 8}{3} = \frac{8}{3} \approx 2.67 \, \text{m} \).

Coordinates: \( (2, \frac{8}{3}) \).

\( (1, 2) \)
\( (2, \frac{8}{3}) \)
\( (3, 4) \)
\( (2, 3) \)
2

What is the effect of a force applied at the axis of rotation on a rigid body?

A force applied at the axis (\( r = 0 \)) produces no torque (\( \tau = r F \sin \theta = 0 \)), so it cannot cause rotation about that axis.

It causes maximum rotation
It causes no rotation
It increases angular velocity directly
It changes the axis of rotation
2

A thin ring of mass \( 3 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) rotates at \( 8 \, \text{rad/s} \). What is its angular momentum about its axis?

\( I = M R^2 = 3 \times (0.5)^2 = 0.75 \, \text{kg m}^2 \).

\( L = I \omega = 0.75 \times 8 = 6 \, \text{kg m}^2/\text{s} \).

5.5 kg m²/s
6.0 kg m²/s
6.5 kg m²/s
7.0 kg m²/s
2

A wheel with moment of inertia \( 4 \, \text{kg m}^2 \) rotates at \( 5 \, \text{rad/s} \). A torque of \( 8 \, \text{Nm} \) acts for \( 3 \, \text{s} \). What is its final angular velocity?

\( \alpha = \frac{\tau}{I} = \frac{8}{4} = 2 \, \text{rad/s}^2 \).

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

\( \omega = \omega_0 + \Delta \omega = 5 + 6 = 11 \, \text{rad/s} \).

9 rad/s
10 rad/s
11 rad/s
12 rad/s
3

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

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -3 & 0 & 0 \\ 0 & 5 & 0 \end{vmatrix} = \hat{\mathbf{k}} ((-3) \times 5 - 0 \times 0) = -15 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Magnitude = \( 15 \, \text{kg m}^2/\text{s} \).

13 kg m²/s
14 kg m²/s
15 kg m²/s
16 kg m²/s
3

Vectors \( \mathbf{a} = 7 \, \hat{\mathbf{i}} + 2 \, \hat{\mathbf{j}} \) and \( \mathbf{b} = -3 \, \hat{\mathbf{i}} + 4 \, \hat{\mathbf{j}} \) are given. What is the magnitude of \( \mathbf{a} \times \mathbf{b} \)?

\( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 7 & 2 & 0 \\ -3 & 4 & 0 \end{vmatrix} = \hat{\mathbf{k}} (7 \times 4 - 2 \times (-3)) = \hat{\mathbf{k}} (28 + 6) = 34 \, \hat{\mathbf{k}} \).

Magnitude = \( 34 \).

32
33
34
35
3

In rotational dynamics, what role does torque play?

Torque causes angular acceleration, as per \( \mathbf{\tau} = I \mathbf{\alpha} \), analogous to how force causes linear acceleration in Newton’s second law.

It measures the speed of rotation
It causes angular acceleration
It determines linear velocity
It balances the body’s mass
2

A thin ring of mass \( 1 \, \text{kg} \) and radius \( 0.2 \, \text{m} \) rotates about its center. If its kinetic energy is \( 4 \, \text{J} \), what is its angular speed?

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

For a ring: \( I = M R^2 = 1 \times (0.2)^2 = 0.04 \, \text{kg m}^2 \).

\( 4 = \frac{1}{2} \times 0.04 \times \omega^2 \Rightarrow 8 = 0.04 \omega^2 \Rightarrow \omega^2 = 200 \Rightarrow \omega = \sqrt{200} \approx 14.14 \, \text{rad/s} \).

10 rad/s
12 rad/s
14 rad/s
16 rad/s
3

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