Rotation Ques 100
- When $\operatorname{disc} B$ is brought in contact with $\operatorname{disc} A$, they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is
(a) $\frac{2 I \omega}{3 t}$
(b) $\frac{9 I \omega}{2 t}$
(c) $\frac{9 I \omega}{4 t}$
(d) $\frac{3 I \omega}{2 t}$
(2007, 4M)
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Answer:
Correct Answer: 100.(a)
Solution:
- Let $\omega^{\prime}$ be the common velocity. Then from conservation of angular momentum, we have
$$ \begin{aligned} (I+2 I) \omega^{\prime} & =I(2 \omega)+2 I(\omega) \\ \omega^{\prime} & =\frac{4}{3} \omega \end{aligned} $$
From the equation,
angular impulse $=$ change in angular momentum, for any of the disc, we have
$$ \begin{aligned} \tau \cdot t & =I(2 \omega)-I \frac{4}{3} \omega=\frac{2 I \omega}{3} \\ \therefore \quad \tau & =\frac{2 I \omega}{3 t} \end{aligned} $$
BETA
