Question

A thin circular metal disc of radius 500.0 mm is set rotating about a central axis normal to its plane. Upon raising its temperature gradually, the radius increases to 507.5 mm. The percentage change in the rotational kinetic energy will be

• A. 1.5%
• B. -1.5%
• C. 3%
• D. -3%

Answer 1

Answer: (D)

Rotational $KE:K=\frac{L^2}{2I}$ where $L$ is the angular momentum and $I$ is the moment of inertia. $L$ is conserved in the process and remains constant.
$<br/>\frac{\Delta K}{K}=-\frac{\Delta I}{I}=-\left(\frac{(507.5{)}^2-(500{)}^2}{(500{)}^2}\right)=-0.03<br/>$
$\therefore$ Percentage increase in KE: $-0.03\times 100\%=-3\%$