Question

A horizontal disc rotating freely about a vertical axis through its centre makes $90$ revolutions per minute. $A$ small piece of wax of mass $m$ falls vertically on the disc and sticks to it at a distance $r$ from the axis. If the number of revolutions per minute reduce to $60$, then the moment of inertia of the disc is

• A. $mr ^{2}$
• B. $\frac{3}{2} m r^{2}$
• C. $2\, mr ^{2}$
• D. $3\, mr ^{2}$

Answer 1

Answer: (C)

$w_{1}=\frac{2 \pi 90}{60}=3 \pi\, rps$
$w_{2}=\frac{2 \pi 60}{60}=2 \pi\, rps$
Using angular momentum conservation
$I(1.5)=\left(I+m^{2}\right)(1)$
$\frac{I}{2}=m r^{2}$
$I=2\, mr ^{2}$