Rotation 2 Question 17
17. The net reaction of the disc on the block is
(a) $m \omega^{2} R \sin \omega \hat{\mathbf{j}}-m g \hat{\mathbf{k}}$
(b) $\frac{1}{2} m \omega^{2} R\left(e^{\omega t}-e^{-\omega t}\right) \hat{\mathbf{j}}+m g \hat{\mathbf{k}}$
(c) $\frac{1}{2} m \omega^{2} R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{\mathbf{j}}+m g \hat{\mathbf{k}}$
(d) $-m \omega^{2} R \cos \omega \hat{\mathbf{j}}-m g \hat{\mathbf{k}}$
Integer Answer Type Questions
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Answer:
Correct Answer: 17. (b)
Solution:
$$ \mathbf{F} _{\text {rot }}=\mathbf{F} _{\text {in }}+2 m\left(v _{\text {rot }} \hat{\mathbf{i}}\right) \times \omega \hat{\mathbf{k}}+m(\omega \hat{\mathbf{k}} \times r \hat{\mathbf{i}}) \times \omega \hat{\mathbf{k}} $$
$$ m r \boldsymbol{\omega}^{2} \hat{\mathbf{i}}=\mathbf{F} _{\text {in }}+2 m v _{\text {rot }} \boldsymbol{\omega}(-\hat{\mathbf{j}})+m \boldsymbol{\omega}^{2} r \hat{\mathbf{i}} $$
$$ \begin{aligned} \mathbf{F} _{\text {in }} & =2 m v _r \omega \hat{\mathbf{j}} \\ r & =\frac{R}{4}\left[e^{\omega t}+e^{-\omega t}\right] \\ \frac{d r}{d t} & =v _r=\frac{R}{4}\left[\omega e^{\omega t}-\omega e^{-\omega t}\right] \\ \mathbf{F} _{\text {in }} & =2 m \frac{R \omega}{4}\left[e^{\omega t}-e^{-\omega t}\right] \omega \hat{\mathbf{j}} \\ \mathbf{F} _{\text {in }} & =\frac{m R \omega^{2}}{2}\left[e^{\omega t}-e^{-\omega t}\right] \hat{\mathbf{j}} \end{aligned} $$
Also, reaction is due to disc surface then
$$ \mathbf{F} _{\text {reaction }}=\frac{m R \omega^{2}}{2}\left[e^{\omega t}-e^{-\omega t}\right] \hat{\mathbf{j}}+m g \hat{\mathbf{k}} $$
