System of Particles and Rotational Motion Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does a body with a larger radius of gyration have greater resistance to rotation?

Radius of gyration relates to moment of inertia (\( I = M k^2 \)), where a larger \( k \) means mass is distributed farther from the axis, increasing \( I \) and thus resistance to angular acceleration.

Because it has less mass
Because mass is farther from the axis
Because torque is reduced
Because angular velocity increases
2

In rolling motion of a rigid body without slipping, what is the relationship between linear and angular velocity?

For rolling without slipping, the linear velocity of the center of mass (\( v \)) equals the product of angular velocity (\( \omega \)) and radius (\( r \)): \( v = \omega r \).

\( v = \omega / r \)
\( v = \omega r \)
\( v = \omega r^2 \)
\( v = \omega / r^2 \)
2

Why is the angular momentum of a symmetric rigid body about a fixed axis often aligned with the axis?

For a symmetric rigid body rotating about a fixed axis (a symmetry axis), the perpendicular components of angular momentum cancel out due to symmetry, leaving \( \mathbf{L} = I \mathbf{\omega} \) aligned with the axis.

Due to high angular velocity
Due to symmetry about the axis
Due to zero torque
Due to uniform mass distribution
2

What is the physical quantity that measures the rotational inertia of a body?

Moment of inertia (\( I \)) measures rotational inertia, reflecting how mass is distributed relative to the axis and the body’s resistance to angular acceleration.

Torque
Angular momentum
Moment of inertia
Angular velocity
3

A thin ring of mass \( 4 \, \text{kg} \) and radius \( 0.3 \, \text{m} \) rotates about its center at \( 8 \, \text{rad/s} \). What is its rotational kinetic energy?

\( I = M R^2 = 4 \times (0.3)^2 = 0.36 \, \text{kg m}^2 \).

\( K = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.36 \times (8)^2 = 0.18 \times 64 = 11.52 \, \text{J} \).

11.0 J
11.52 J
12.0 J
12.5 J
2

What is the primary difference between angular momentum and torque?

Angular momentum (\( \mathbf{L} = I \mathbf{\omega} \)) is a measure of rotational motion, while torque (\( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \)) is the cause of rotational change.

Angular momentum causes rotation, torque measures it
Torque measures rotational motion, angular momentum causes it
Angular momentum measures rotational motion, torque causes it
Both measure rotational motion
3

A hollow cylinder of mass \( 5 \, \text{kg} \) and radius \( 0.4 \, \text{m} \) rolls without slipping with an angular velocity of \( 8 \, \text{rad/s} \). What is its linear speed?

\( v = \omega r \).

\( \omega = 8 \, \text{rad/s} \), \( r = 0.4 \, \text{m} \).

\( v = 8 \times 0.4 = 3.2 \, \text{m/s} \).

2.8 m/s
3.2 m/s
3.6 m/s
4.0 m/s
2

A thin ring of mass \( 5 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) has an angular momentum of \( 12.5 \, \text{kg m}^2/\text{s} \). What is its angular velocity?

\( I = M R^2 = 5 \times (0.5)^2 = 1.25 \, \text{kg m}^2 \).

\( \omega = \frac{L}{I} = \frac{12.5}{1.25} = 10 \, \text{rad/s} \).

8 rad/s
9 rad/s
10 rad/s
11 rad/s
3

What happens to the linear velocity of a particle as its distance from the axis of rotation increases?

Linear velocity \( v = \omega r \) increases with distance \( r \) from the axis because angular velocity \( \omega \) is constant for a rigid body, and \( v \) is directly proportional to \( r \).

It decreases
It remains constant
It increases
It becomes zero
3

Vectors \( \mathbf{a} = 5 \, \hat{\mathbf{i}} - 4 \, \hat{\mathbf{j}} \) and \( \mathbf{b} = 3 \, \hat{\mathbf{i}} + 6 \, \hat{\mathbf{j}} \) are given. What is the magnitude of \( \mathbf{a} \times \mathbf{b} \)?

\( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 5 & -4 & 0 \\ 3 & 6 & 0 \end{vmatrix} = \hat{\mathbf{k}} (5 \times 6 - (-4) \times 3) = \hat{\mathbf{k}} (30 + 12) = 42 \, \hat{\mathbf{k}} \).

Magnitude = \( 42 \).

40
41
42
43
3

A particle in a rigid body rotates at a distance of \( 0.25 \, \text{m} \) from a fixed axis with an angular velocity of \( 12 \, \text{rad/s} \). What is its linear speed?

\( v = \omega r \).

\( \omega = 12 \, \text{rad/s} \), \( r = 0.25 \, \text{m} \).

\( v = 12 \times 0.25 = 3 \, \text{m/s} \).

2.5 m/s
3.0 m/s
3.5 m/s
4.0 m/s
2

A \( 3 \, \text{kg} \) object moves with a velocity of \( -5 \, \hat{\mathbf{i}} - 2 \, \hat{\mathbf{j}} \, \text{m/s} \). 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 = \( \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \, \text{m/s} \).

5.0 m/s
5.39 m/s
5.6 m/s
6.0 m/s
2

Why does a larger moment of inertia result in smaller angular acceleration for a given torque?

From \( \mathbf{\tau} = I \mathbf{\alpha} \), angular acceleration \( \alpha \) is inversely proportional to moment of inertia \( I \). A larger \( I \) means more resistance to rotation, reducing \( \alpha \) for the same torque.

Because torque increases
Because it resists angular acceleration more
Because angular velocity increases
Because mass decreases
2

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

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

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 cylinder of mass \( 6 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) rolls without slipping with an angular velocity of \( 4 \, \text{rad/s} \). What is its linear speed?

\( v = \omega r \).

\( \omega = 4 \, \text{rad/s} \), \( r = 0.5 \, \text{m} \).

\( v = 4 \times 0.5 = 2 \, \text{m/s} \).

1.5 m/s
2.0 m/s
2.5 m/s
3.0 m/s
2

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