Question

A disc of mass $10\,\text{kg$ and radius $0.1\,\text{m}$ is rotating at $120\,\text{r.p.m.}$. A retarding torque brings it to rest in $10\,\text{s}$. If the same torque is due to force applied tangentially on the rim of the disc, then the magnitude of force is}

• A. $0.2\pi\,\text{N}$
• B. $0.4\,\text{N}$
• C. $0.8\pi\,\text{N}$
• D. $0.1\pi\,\text{N}$

Hint:

Torque is the product of force and radius. Use it to relate rotational quantities.

Answer 1

The Correct Option is A

Step 1: Moment of inertia of the disc.
Moment of inertia of a disc about its axis:
\[ I = \frac{1}{2} m r^2 = \frac{1}{2} \times 10 \times (0.1)^2 = 0.05\,\text{kg m}^2 \]
Step 2: Angular acceleration.
The disc comes to rest from an initial angular speed of $120\,\text{r.p.m.} = 120 \times \frac{2\pi}{60} = 4\pi\,\text{rad/s}$.
The angular deceleration is:
\[ \alpha = \frac{\Delta \omega}{t} = \frac{4\pi}{10} = 0.4\pi\,\text{rad/s}^2 \]
Step 3: Torque and force relation.
Torque $\tau = I\alpha = 0.05 \times 0.4\pi = 0.02\pi\,\text{N m}$
Since torque is also given by $F r$, we have:
\[ F \times 0.1 = 0.02\pi \] \[ F = 0.2\pi\,\text{N} \]
Step 4: Conclusion.
The magnitude of force is $0.2\pi\,\text{N}$.