Question 499: Centre of Mass and Rotational Motion

Question

A disc of mass 25 kg and radius 0.2 m is rotating at 240 r.p.m. A retarding torque brings it to rest in 20 second. If the torque is due to a force applied tangentially on the rim of the disc then the magnitude of the force is

Accepted Answer

Given:
Mass of disc, $m = 25\,	ext{kg}$
Radius, $R = 0.2\,	ext{m}$
Initial angular speed, $n = 240\,	ext{r.p.m}$ (revolutions per minute)
Time to stop, $t = 20\,	ext{sec}$
Final angular speed, $\omega_2 = 0$
Tangential force $F$ (to be found)
Step 1: Convert r.p.m. to angular velocity in rad/s
$$ \omega_1 = 2\pi n/60 $$
Substituting values:
$$ \omega_1 = 2\pi 	imes 240/60 = 2\pi 	imes 4 = 8\pi \ 	ext{rad/s} $$
Step 2: Find angular deceleration ($\alpha$)
Using equation:
$$ \omega_2 = \omega_1 + \alpha t $$
We have $\omega_2 = 0$, so:
$$ 0 = 8\pi + \alpha 	imes 20 \implies \alpha = -\frac{8\pi}{20} = -\frac{2\pi}{5} \ 	ext{rad/s}^2 $$
Step 3: Moment of inertia of a disc about its axis
$$ I = \frac{1}{2} m R^2 $$
Substitute:
$$ I = \frac{1}{2} 	imes 25 	imes (0.2)^2 = \frac{1}{2} 	imes 25 	imes 0.04 = 12.5 	imes 0.04 = 0.5\ 	ext{kg m}^2 $$
Step 4: Calculate retarding torque ($	au$)
$$ 	au = I \cdot \alpha $$
$$ 	au = 0.5 	imes \frac{2\pi}{5} = 0.5 	imes \frac{2\pi}{5} = \frac{\pi}{5} \ 	ext{N m} $$
(Negative sign is for direction; magnitude only.)
Step 5: Relate torque to force
Torque produced by force applied at rim:
$$ 	au = F 	imes R \implies F = \frac{	au}{R} $$
$$ F = \frac{\frac{\pi}{5}}{0.2} = \frac{\pi}{5} 	imes \frac{1}{0.2} = \frac{\pi}{5} 	imes 5 = \pi\ 	ext{N} $$
Final Answer:
Option C
$ \boxed{\pi \,	ext{N}} $

Answer

$\frac{\pi}{2} \mathrm{~N}$

Answer

$2 \pi \mathrm{~N}$

Answer

$\pi \mathrm{N}$

Answer

$\quad 4 \pi \mathrm{~N}$

Centre of Mass and Rotational Motion

A disc of mass 25 kg and radius 0.2 m is rotating at 240 r.p.m. A retarding torque brings it to rest in 20 second. If the torque is due to a force applied tangentially on the rim of the disc then the magnitude of the force is

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