System of Particles and Rotational Motion Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Three particles of masses \( 2 \, \text{kg} \), \( 3 \, \text{kg} \), and \( 5 \, \text{kg} \) are placed at coordinates \( (0, 0) \), \( (4, 0) \), and \( (2, 3) \) respectively. What is the x-coordinate of their center of mass?

Using the formula for the x-coordinate of the center of mass: \( X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \).

Masses: \( m_1 = 2 \, \text{kg} \), \( m_2 = 3 \, \text{kg} \), \( m_3 = 5 \, \text{kg} \).

Coordinates: \( x_1 = 0 \), \( x_2 = 4 \), \( x_3 = 2 \).

\( X = \frac{(2 \times 0) + (3 \times 4) + (5 \times 2)}{2 + 3 + 5} = \frac{0 + 12 + 10}{10} = \frac{22}{10} = 2.2 \, \text{m} \).

1.8 m
2.0 m
2.2 m
2.4 m
3

Why is torque zero when the force is applied along the line passing through the pivot point?

Torque \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \) depends on the perpendicular distance (\( r \sin \theta \)). If the force acts along the line through the pivot, \( \theta = 0^\circ \) or \( 180^\circ \), so \( \sin \theta = 0 \), making torque zero.

Because the force is zero
Because the perpendicular distance is zero
Because the angular velocity is zero
Because the mass is evenly distributed
2

A cylinder of mass \( 10 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) rolls without slipping. If its angular momentum about its axis is \( 25 \, \text{kg m}^2/\text{s} \), what is its angular velocity?

Angular momentum: \( L = I \omega \).

For a solid cylinder: \( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 10 \times (0.5)^2 = 1.25 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{25}{1.25} = 20 \, \text{rad/s} \).

15 rad/s
20 rad/s
25 rad/s
30 rad/s
2

A force \( \mathbf{F} = 4 \, \hat{\mathbf{i}} - 6 \, \hat{\mathbf{j}} \, \text{N} \) acts at \( \mathbf{r} = 5 \, \hat{\mathbf{i}} + 2 \, \hat{\mathbf{j}} \, \text{m} \). What is the magnitude of the torque about the origin?

\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & 2 & 0 \\ 4 & -6 & 0 \end{vmatrix} = \hat{\mathbf{k}} (5 \times (-6) - 2 \times 4) = \hat{\mathbf{k}} (-30 - 8) = -38 \, \hat{\mathbf{k}} \, \text{Nm} \).

Magnitude = \( 38 \, \text{Nm} \).

36 Nm
37 Nm
38 Nm
39 Nm
3

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

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

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

-18 kg m²/s
-15 kg m²/s
-12 kg m²/s
-10 kg m²/s
2

A torque of \( 18 \, \text{Nm} \) acts on a disk with moment of inertia \( 6 \, \text{kg m}^2 \) starting from rest. What is its angular speed after \( 4 \, \text{s} \)?

\( \alpha = \frac{\tau}{I} = \frac{18}{6} = 3 \, \text{rad/s}^2 \).

\( \omega = \omega_0 + \alpha t = 0 + 3 \times 4 = 12 \, \text{rad/s} \).

10 rad/s
12 rad/s
14 rad/s
16 rad/s
2

A solid sphere of mass \( 4 \, \text{kg} \) and radius \( 0.3 \, \text{m} \) has an angular momentum of \( 7.2 \, \text{kg m}^2/\text{s} \). What is its angular velocity?

\( I = \frac{2}{5} M R^2 = \frac{2}{5} \times 4 \times (0.3)^2 = 0.144 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{7.2}{0.144} = 50 \, \text{rad/s} \).

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

A solid cylinder 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?

\( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 3 \times (0.4)^2 = 0.24 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{12}{0.24} = 50 \, \text{rad/s} \).

40 rad/s
45 rad/s
50 rad/s
55 rad/s
3

Vectors \( \mathbf{a} = 3 \, \hat{\mathbf{i}} - 5 \, \hat{\mathbf{j}} \) and \( \mathbf{b} = 2 \, \hat{\mathbf{i}} + 4 \, \hat{\mathbf{j}} \) are given. What is the magnitude of \( \mathbf{a} \times \mathbf{b} \)?

\( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -5 & 0 \\ 2 & 4 & 0 \end{vmatrix} = \hat{\mathbf{k}} (3 \times 4 - (-5) \times 2) = \hat{\mathbf{k}} (12 + 10) = 22 \, \hat{\mathbf{k}} \).

Magnitude = \( 22 \).

20
21
22
23
3

Two particles of masses \( 7 \, \text{kg} \) and \( 3 \, \text{kg} \) are at \( (0, 6) \) and \( (5, 0) \) respectively. What is the distance of their center of mass from the origin?

\( X = \frac{(7 \times 0) + (3 \times 5)}{7 + 3} = \frac{15}{10} = 1.5 \).

\( Y = \frac{(7 \times 6) + (3 \times 0)}{7 + 3} = \frac{42}{10} = 4.2 \).

Distance = \( \sqrt{(1.5)^2 + (4.2)^2} = \sqrt{2.25 + 17.64} = \sqrt{19.89} \approx 4.46 \, \text{m} \).

4.0 m
4.2 m
4.46 m
4.8 m
3

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