Rotation 3 Question 9
15. The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to
$(2008,4 M)$
(a) $\sqrt{\frac{k}{M}}$
(b) $\sqrt{\frac{2 k}{M}}$
(c) $\sqrt{\frac{2 k}{3 M}}$
(d) $\sqrt{\frac{4 k}{3 M}}$
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Answer:
Correct Answer: 15. (d)
Solution:
- $F _{\text {net }}=-\frac{4 k x}{3} x$
$$ \begin{array}{rlrl} & \therefore & a & =\frac{F _{\text {net }}}{M}=-\frac{4 k}{3 M} x=-\omega^{2} x \\ \therefore & \omega & =\sqrt{\frac{4 k}{3 M}} \end{array} $$
