Question

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

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

Answer 1

The correct option is B

$\frac{1}{r}\sqrt{\frac{4gh}{3}}$

Step 1. Given data

Mass of the bob is $m$

Radius of the fly wheel is $r$

Step 2. Finding the Rotational kinetic energy, $E_R$

Let the bob of mass $m$ fall through a height $h$.

Rotational kinetic energy of flywheel,

$E_R=\frac{1}{2}I{\omega }^2$ [Where, $I$ is the moment of inertia and $\omega$ is the angular velocity.]

$E_R=\frac{1}{2}\left(\frac{m{r}^{{}^2}}{2}\right){\omega }^2$

$E_R=\frac{m{r}^2{\omega }^2}{4}$

Step 3. Finding the angular velocity

According to the Work energy theorem, the change in the potential energy of the bob is converted to the kinetic energy of the bob and flywheel.

Since, the string is tied around the flywheel, the linear velocity, $v$ of the bob

$v=\omega r$

Kinetic energy of the bob$=\frac{1}{2}m{v}^2$

$Potentialenergy=Kineticenergyofbob+kineticenergyofflywheel$

$mgh=\frac{1}{2}m{\omega }^2{r}^2+\frac{1}{4}m{r}^2{\omega }^2$

$\omega =\frac{1}{r}\sqrt{\frac{4gh}{3}}$

Hence, the correct option is $\left(B\right)$.