Question

A rotating table completes one rotation is $10sec$, and its moment of inertia is $100kg-m^2$. A person of $50kg.$ mass stands at the centre of the rotating table. If the person moves $2m$ from the centre, the angular velocity of the rotating table in ($rad/sec$). will be :

• A. $\frac{2\pi }{30}$
• B. $\frac{20\pi }{30}$
• C. $\frac{2\pi }{3}$
• D. $2\pi$

Answer 1

The correct option is A $\frac{2\pi }{30}$

when person moves 2m then new moment of inertia will be sum of MOI of the man and that of the table so $I^{\prime }=I+m{r}^2=100+50\times 4=300Kg{m}^2$

so according to conservation of angular momentum the new angular velocity will become$1/3$ of the initial because the moment of inertia get $3$ times so now it will take $10\times 3=30s$ to complete one rotation so angular velocity $w=2\pi /30rad/s$