Rotation 2 Question 16
16. The distance $r$ of the block at time $t$ is
(a) $\frac{R}{2} \cos 2 \omega t$
(b) $\frac{R}{2} \cos \omega t$
(c) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$
(d) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$
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Answer:
Correct Answer: 16. (c)
Solution:
- Force on block along slot $=m \omega^{2} r=m a=m \frac{v d v}{d r}$
$$ \begin{aligned} & \int _0^{y} v d v=\int _{R / 2}^{r} \omega^{2} r d r \Rightarrow \frac{v^{2}}{2}=\frac{\omega^{2}}{2} r^{2}-\frac{R^{2}}{4} \\ & \Rightarrow \quad v=\omega \sqrt{r^{2}-\frac{R^{2}}{4}}=\frac{d r}{d t} \Rightarrow \int _{R / 4} \frac{d r}{\sqrt{r^{2}-\frac{R^{2}}{4}}}=\int _0^{t} \omega d t \\ & \ln \frac{r+\sqrt{r^{2}-\frac{R^{2}}{4}}}{\frac{R}{2}}-\ln \frac{R / 2+\sqrt{\frac{R^{2}}{4}-\frac{R^{2}}{4}}}{\frac{R}{4}}=\omega t \\ & \Rightarrow \quad r+\sqrt{r^{2}-\frac{R^{2}}{4}}=\frac{R}{2} e^{\omega t} \\ & \Rightarrow \quad r^{2}-\frac{R^{2}}{4}=\frac{R^{2}}{4} e^{2 \omega t}+r^{2}-2 r \frac{R}{2} e^{\omega t} \\ & \Rightarrow \quad r=\frac{\frac{R^{2}}{4} e^{2 \omega t}+\frac{R^{2}}{4}}{R e^{\omega t}}=\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right) \end{aligned} $$
