Torque and couple - MCQs

Let the weight of the rod be W. The boy carries W/4 and the man carries 3W/4. Taking moments about the boy's end, if the man is at distance x from the boy: Solving for x: \( x = \frac{2L}{3} \) Distance from the other end = \( L - \frac{2L}{3} = \frac{L}{3} \)

When two equal and opposite forces are applied at different points, they form a couple that produces pure rotation without translation.

Torque is the rate of change of angular momentum:

Torque is a vector quantity as it has both magnitude and direction (given by the right-hand rule). Work, power, and gravitational constant are scalar quantities.

Torque is the rate of change of angular momentum:

Torque \( \vec{\tau} = \vec{r} \times \vec{F} \) is perpendicular to both position vector \( \vec{r} \) and force \( \vec{F} \). Therefore:

A couple consists of two equal and opposite parallel forces that produce a net torque but no net force, resulting in pure rotation without translation.

According to the principle of conservation of angular momentum, if the net external torque on a system is zero, the angular momentum remains constant.

Torque \( \tau = I\alpha \), where I is moment of inertia and α is angular acceleration. If τ = 0, then α = 0, meaning there is no angular acceleration (the body rotates with constant angular velocity or is at rest).

This is the rotational analogue of Newton's second law (F = ma). For rotational motion, torque equals moment of inertia times angular acceleration: \( \tau = I\alpha \).

First, find angular acceleration: \( \alpha = \frac{\tau}{I} = \frac{30}{2} = 15 \text{ rad/s}^2 \)
Then, angle covered: \( \theta = \frac{1}{2} \alpha t^2 = \frac{1}{2} \times 15 \times 10^2 = 750 \text{ rad} \)

Moment of inertia: \( I = mr^2 = 10 \times (0.3)^2 = 0.9 \text{ kg-m}^2 \)
Angular acceleration: \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 10}{10} = -1 \text{ rad/s}^2 \)
Torque: \( \tau = I\alpha = 0.9 \times 1 = 0.9 \text{ N-m} \)

Initial angular velocity: \( \omega_i = 60 \text{ rpm} = 60 \times \frac{2\pi}{60} = 2\pi \text{ rad/s} \)
Angular acceleration: \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \text{ rad/s}^2 \)
Torque: \( \tau = I\alpha = 2 \times \frac{\pi}{30} = \frac{\pi}{15} \text{ N-m} \)

Power \( P = \tau \omega \)
Angular velocity: \( \omega = 1800 \text{ rpm} = 1800 \times \frac{2\pi}{60} = 60\pi \text{ rad/s} \)
Torque: \( \tau = \frac{P}{\omega} = \frac{100000}{60\pi} \approx 531 \text{ N-m} \)

Initial angular velocity: \( \omega_i = 20 \text{ rev/s} = 20 \times 2\pi = 40\pi \text{ rad/s} \)
Angular acceleration: \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 40\pi}{10} = -4\pi \text{ rad/s}^2 \)
Torque: \( \tau = I\alpha = 5 \times 10^{-3} \times 4\pi = 20\pi \times 10^{-3} = 2\pi \times 10^{-2} \text{ N-m} \)

Torque \( \vec{\tau} = \vec{r} \times \vec{F} \)
So, \( \vec{\tau} = -\hat{i} - 6\hat{j} + 5\hat{k} \)

Force due to weight: \( F = mg = 10 \times 9.8 = 98 \text{ N} \)
Radius of cylinder: \( r = \frac{0.2}{2} = 0.1 \text{ m} \)
Torque: \( \tau = F \times r = 98 \times 0.1 = 9.8 \text{ N-m} \)

Just as force causes linear acceleration (F = ma), torque causes angular acceleration (τ = Iα). Therefore, torque is the rotational analogue of force.

Angular acceleration: \( \alpha = \frac{\tau}{I} = \frac{1000}{200} = 5 \text{ rad/s}^2 \)
Assuming initial angular velocity is 0:
Angular velocity after 3 s: \( \omega = \alpha t = 5 \times 3 = 15 \text{ rad/s} \)

When a ring rotates about its center, centrifugal forces act outward, causing the ring to expand and its radius to increase.

A couple produces pure rotational motion without translation, as it consists of two equal and opposite parallel forces that create a net torque but no net force.

Position vector relative to point: \( \vec{r'} = \vec{r} - \vec{r_0} = (3\hat{i} + 2\hat{j}) - (\hat{i} + \hat{j}) = 2\hat{i} + \hat{j} \)
Torque: \( \vec{\tau} = \vec{r'} \times \vec{F} \)
Wait, let me recalculate: \( \vec{\tau} = \vec{r'} \times \vec{F} = (2\hat{i} + \hat{j}) \times (2\hat{i} + 3\hat{j}) \)
Correction: The correct answer should be \( 4\hat{k} \), not \( -2\hat{k} \)

Moment of inertia of Earth: \( I = \frac{2}{5}MR^2 \)
Initial angular velocity: \( \omega_i = \frac{2\pi}{T} = \frac{2\pi}{24 \times 3600} \text{ rad/s} \)
Time to stop: t = 1 day = 86400 s
Angular acceleration: \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - \omega_i}{86400} \)
Torque: \( \tau = I\alpha = F \times R \)
Solving for F gives \( F = \frac{\pi MR}{14} \)

For a particle in circular motion, centripetal force \( F = m\omega^2 r \)
For particles of same mass: \( \frac{F_1}{F_2} = \frac{m\omega^2 R_1}{m\omega^2 R_2} = \frac{R_1}{R_2} \)

The tangential component of force produces torque, which causes rotational motion. The radial component only changes the direction of motion but doesn't produce rotation.