System of Particles and Rotational Motion Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 5 & 0 \\ -3 & -4 & 0 \end{vmatrix} = \hat{\mathbf{k}} (2 \times (-4) - 5 \times (-3)) = \hat{\mathbf{k}} (-8 + 15) = 7 \, \hat{\mathbf{k}} \, \text{Nm} \).

Magnitude = \( 7 \, \text{Nm} \).

6 Nm
7 Nm
8 Nm
9 Nm
2

A wheel of radius \( 0.4 \, \text{m} \) and moment of inertia \( 1.6 \, \text{kg m}^2 \) is accelerated by a torque of \( 8 \, \text{Nm} \). What is its angular acceleration?

Torque: \( \tau = I \alpha \).

\( \tau = 8 \, \text{Nm} \), \( I = 1.6 \, \text{kg m}^2 \).

\( \alpha = \frac{\tau}{I} = \frac{8}{1.6} = 5 \, \text{rad/s}^2 \).

4 rad/s²
5 rad/s²
6 rad/s²
7 rad/s²
2

Two particles of masses \( 2 \, \text{kg} \) and \( 8 \, \text{kg} \) are at \( (1, 5) \) and \( (4, 1) \) respectively. What is the distance of their center of mass from the origin?

\( X = \frac{(2 \times 1) + (8 \times 4)}{2 + 8} = \frac{2 + 32}{10} = 3.4 \).

\( Y = \frac{(2 \times 5) + (8 \times 1)}{2 + 8} = \frac{10 + 8}{10} = 1.8 \).

Distance = \( \sqrt{(3.4)^2 + (1.8)^2} = \sqrt{11.56 + 3.24} = \sqrt{14.8} \approx 3.85 \, \text{m} \).

3.5 m
3.85 m
4.0 m
4.2 m
2

Two particles of masses \( 4 \, \text{kg} \) and \( 6 \, \text{kg} \) are at \( (2, 0) \) and \( (0, 3) \) respectively. What is the distance of their center of mass from \( (0, 0) \)?

\( X = \frac{(4 \times 2) + (6 \times 0)}{4 + 6} = \frac{8}{10} = 0.8 \).

\( Y = \frac{(4 \times 0) + (6 \times 3)}{4 + 6} = \frac{18}{10} = 1.8 \).

Distance = \( \sqrt{(0.8)^2 + (1.8)^2} = \sqrt{0.64 + 3.24} = \sqrt{3.88} \approx 1.97 \, \text{m} \).

1.5 m
1.97 m
2.5 m
3.0 m
2

A disk with moment of inertia \( 1.5 \, \text{kg m}^2 \) rotates at \( 4 \, \text{rad/s} \). A torque of \( 6 \, \text{Nm} \) acts for \( 3 \, \text{s} \). What is its final angular velocity?

\( \alpha = \frac{\tau}{I} = \frac{6}{1.5} = 4 \, \text{rad/s}^2 \).

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

\( \omega = \omega_0 + \Delta \omega = 4 + 12 = 16 \, \text{rad/s} \).

12 rad/s
14 rad/s
16 rad/s
18 rad/s
3

Which statement is true about the axis of rotation in pure rotational motion?

In pure rotational motion, the axis of rotation is fixed, meaning it does not move or change direction, constraining the body to rotate about it.

It moves with the body
It is always fixed
It changes direction during rotation
It is perpendicular to the body
2

A torque of \( 21 \, \text{Nm} \) acts on a wheel with moment of inertia \( 7 \, \text{kg m}^2 \) starting from rest. What is its angular speed after \( 4 \, \text{s} \)?

\( \alpha = \frac{\tau}{I} = \frac{21}{7} = 3 \, \text{rad/s}^2 \).

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

10 rad/s
11 rad/s
12 rad/s
13 rad/s
3

A wheel with moment of inertia \( 2.5 \, \text{kg m}^2 \) rotates at \( 6 \, \text{rad/s} \). A torque of \( 10 \, \text{Nm} \) acts for \( 2 \, \text{s} \). What is its final angular velocity?

\( \alpha = \frac{\tau}{I} = \frac{10}{2.5} = 4 \, \text{rad/s}^2 \).

\( \Delta \omega = \alpha t = 4 \times 2 = 8 \, \text{rad/s} \).

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

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

What is the SI unit of moment of inertia?

Moment of inertia \( I = \sum m_i r_i^2 \) has units of mass times distance squared, so the SI unit is \( \text{kg m}^2 \).

N m
kg m²
kg m/s²
J s
2

A \( 6 \, \text{kg} \) object moves with a velocity of \( 3 \, \hat{\mathbf{i}} - 2 \, \hat{\mathbf{j}} \, \text{m/s} \). 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 = \( \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.6 \, \text{m/s} \).

3.2 m/s
3.6 m/s
4.0 m/s
4.4 m/s
2

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