Question

As shown in the figure, a bob of mass $m$ is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius $r$ and mass $m.$ When released from rest the bob starts falling vertically. When it has covered a distance of $h,$ the angular speed of the wheel will be:

• A. $\frac{1}{r}\sqrt{\frac{4 gh}{3}}$
• B. $r\sqrt{\frac{3}{2 gh}}$
• C. $\frac{1}{r}\sqrt{\frac{2 gh}{3}}$
• D. $r\sqrt{\frac{3}{4 gh}}$

Answer 1

Answer: (A)

Using work-energy theorem
$mgh=\frac{1}{2}mv^{2}+\frac{1}{2}Iω^{2}$
$mgh=\frac{1}{2}mω^{2}r^{2}+\frac{1}{2}\frac{mr^{2}}{2}\omega ^{2}$ ( since $I=\frac{mr^{2}}{2}$ and $v=rω$ )
$mgh=\frac{3}{4}mω^{2}r^{2}$
Hence, $\omega =\sqrt{\frac{4 gh}{3 r^{2}}}=\frac{1}{r}\sqrt{\frac{4 gh}{3}}$