Question

If a torque of 1.25 Nm acts on a circular ring for 4 s, then its angular momentum changes by ($kgm^2s^{-1}$):

• A. 25
• B. 50
• C. 15
• D. 5
• E. 10

Hint:

Angular impulse (Torque $\times$ time) is equal to the change in angular momentum, similar to linear impulse ($F \times t = \Delta P$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem relates torque, time, and the change in angular momentum. This is the rotational analogue of the impulse-momentum theorem in linear motion.
Step 2: Key Formula or Approach:
Torque (\( \tau \)) is defined as the rate of change of angular momentum (\( L \)). \[ \tau = \frac{dL}{dt} \] For a constant torque applied over a time interval \( \Delta t \), this can be written as: \[ \tau = \frac{\Delta L}{\Delta t} \] The change in angular momentum (\( \Delta L \)) is therefore: \[ \Delta L = \tau \times \Delta t \] This quantity, \( \tau \times \Delta t \), is known as angular impulse. Step 3: Detailed Explanation:
We are given: - Torque, \( \tau = 1.25 \text{ Nm} \) - Time duration, \( \Delta t = 4 \text{ s} \) We need to find the change in angular momentum, \( \Delta L \). Using the formula: \[ \Delta L = 1.25 \times 4 \] \[ \Delta L = 5.0 \] The units of angular momentum are kg m\(^2\) s\(^{-1}\).
Step 4: Final Answer:
The angular momentum changes by 5 kgm\(^2\) s\(^{-1}\).