Question

A turntable rotates about a fixed vertical axis, making one revolution in $10\text{s}$. The moment of inertia of the turntable about the axis of rotation is $1200{\text{kg.m}}^2$. A man of mass $80\text{kg}$, initially standing at the centre of the turntable, runs outwards along the radius. What is the approximate angular velocity of the turntable when the man is at a distance of $2\text{m}$ from the centre?

• A. $1.25\text{rad/s}$
• B. $2.75\text{rad/s}$
• C. $0.50\text{rad/s}$
• D. $0.1\text{rad/s}$
  

Answer 1

The correct option is C $0.50\text{rad/s}$

Let $I_0$ be the initial moment of inertia of the system (man + table)
$I_0=I_{\text{man}}+I_{\text{table}}$
$=0+1200=1200{\text{kg.m}}^2$
($I_{\text{man}}=0$ as the man is at the axis of rotation)
$I=$final moment of inertia of the system
$I=I_{\text{man}}^{\prime }+I_{\text{table}}$
$\Rightarrow I=m{r}^2+1200$
$\therefore I=80(2{)}^2+1200=1520{\text{kg.m}}^2$

$\because$ weight of disc is passing through it's centre and weight of man is acting $\perp$ to the axis of rotation hence we can conserve angular momentum as ${\tau }_{\text{ext}}=0$ about axis of rotation,

By applying $P.C.A.M$:
$I_0{\omega }_0=I\omega ....\left(i\right)$
${\omega }_0=\frac{2\pi }{T_0}=\frac{\pi }{5}\text{rad/s}$
From Eq. $\left(i\right)$
$\Rightarrow \omega =\frac{I_0{\omega }_0}{I}=\frac{1200\times \pi }{1520\times 5}\simeq 0.50\text{rad/s}$