System of Particles and Rotational Motion Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 2 \, \text{kg} \) mass rotates in a circle of radius \( 0.5 \, \text{m} \) with a speed of \( 8 \, \text{m/s} \). What is its angular momentum about the center?

\( L = m v r \).

\( m = 2 \, \text{kg} \), \( v = 8 \, \text{m/s} \), \( r = 0.5 \, \text{m} \).

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

7.5 kg m²/s
8.0 kg m²/s
8.5 kg m²/s
9.0 kg m²/s
2

Why does a body with zero net torque maintain its rotational state?

Zero net torque (\( \mathbf{\tau} = 0 \)) means no change in angular momentum (\( \frac{d\mathbf{L}}{dt} = 0 \)), so the rotational state (angular velocity) remains constant.

Due to zero linear velocity
Due to no change in angular momentum
Due to increased mass
Due to changing axis
2

A force \( \mathbf{F} = 5 \, \hat{\mathbf{i}} + 4 \, \hat{\mathbf{j}} \, \text{N} \) acts at \( \mathbf{r} = -3 \, \hat{\mathbf{i}} - 2 \, \hat{\mathbf{j}} \, \text{m} \). What is the magnitude of the torque about the origin?

\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -3 & -2 & 0 \\ 5 & 4 & 0 \end{vmatrix} = \hat{\mathbf{k}} ((-3) \times 4 - (-2) \times 5) = \hat{\mathbf{k}} (-12 + 10) = -2 \, \hat{\mathbf{k}} \, \text{Nm} \).

Magnitude = \( 2 \, \text{Nm} \).

1 Nm
2 Nm
3 Nm
4 Nm
2

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

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & 0 & 0 \\ 0 & -2 & 0 \end{vmatrix} = \hat{\mathbf{k}} (5 \times (-2) - 0 \times 0) = -10 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

Magnitude = \( 10 \, \text{kg m}^2/\text{s} \).

8 kg m²/s
9 kg m²/s
10 kg m²/s
11 kg m²/s
3

A \( 3 \, \text{kg} \) mass rotates in a circle of radius \( 0.7 \, \text{m} \) with a speed of \( 4.2 \, \text{m/s} \). What is its angular momentum about the center?

\( L = m v r \).

\( m = 3 \, \text{kg} \), \( v = 4.2 \, \text{m/s} \), \( r = 0.7 \, \text{m} \).

\( L = 3 \times 4.2 \times 0.7 = 8.82 \, \text{kg m}^2/\text{s} \).

8.5 kg m²/s
8.82 kg m²/s
9.0 kg m²/s
9.5 kg m²/s
2

A \( 2 \, \text{kg} \) particle moves with velocity \( \mathbf{v} = -5 \, \hat{\mathbf{i}} - 3 \, \hat{\mathbf{j}} \, \text{m/s} \) at \( \mathbf{r} = 4 \, \hat{\mathbf{i}} \, \text{m} \). What is the z-component of its angular momentum?

\( \mathbf{L} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 0 & 0 \\ -5 & -3 & 0 \end{vmatrix} = \hat{\mathbf{k}} (4 \times (-3) - 0 \times (-5)) = -12 \, \hat{\mathbf{k}} \, \text{kg m}^2/\text{s} \).

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

-14 kg m²/s
-12 kg m²/s
-10 kg m²/s
-8 kg m²/s
2

A solid cylinder of mass \( 4 \, \text{kg} \) and radius \( 0.2 \, \text{m} \) rolls without slipping with a linear speed of \( 2 \, \text{m/s} \). What is its angular velocity?

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

\( v = 2 \, \text{m/s} \), \( r = 0.2 \, \text{m} \).

\( \omega = \frac{v}{r} = \frac{2}{0.2} = 10 \, \text{rad/s} \).

8 rad/s
10 rad/s
12 rad/s
14 rad/s
2

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

\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 0 \\ -2 & 3 & 0 \end{vmatrix} = \hat{\mathbf{k}} (1 \times 3 - 2 \times (-2)) = \hat{\mathbf{k}} (3 + 4) = 7 \, \hat{\mathbf{k}} \, \text{Nm} \).

5 Nm
6 Nm
7 Nm
8 Nm
3

Two particles of masses \( 6 \, \text{kg} \) and \( 4 \, \text{kg} \) are at \( (2, 3) \) and \( (8, 7) \) respectively. What is the distance of their center of mass from the origin?

\( X = \frac{(6 \times 2) + (4 \times 8)}{6 + 4} = \frac{12 + 32}{10} = 4.4 \).

\( Y = \frac{(6 \times 3) + (4 \times 7)}{6 + 4} = \frac{18 + 28}{10} = 4.6 \).

Distance = \( \sqrt{(4.4)^2 + (4.6)^2} = \sqrt{19.36 + 21.16} = \sqrt{40.52} \approx 6.36 \, \text{m} \).

6.0 m
6.36 m
6.5 m
6.8 m
2

A torque of \( 20 \, \text{Nm} \) is applied to a wheel with moment of inertia \( 4 \, \text{kg m}^2 \). What is its angular acceleration?

\( \tau = I \alpha \).

\( \tau = 20 \, \text{Nm} \), \( I = 4 \, \text{kg m}^2 \).

\( \alpha = \frac{\tau}{I} = \frac{20}{4} = 5 \, \text{rad/s}^2 \).

4 rad/s²
5 rad/s²
6 rad/s²
7 rad/s²
2

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