Question

A thin circular metal disc of the radius $500 \, mm$ is set rotating about a central axis normal to its plane. Upon raising in 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)

Given $\frac{\text{dR}}{\text{R}}\times 100=\frac{\left(\text{507.5} - \text{500.0}\right)}{\text{500.0}}\times 100=\text{1.5\%}$
Now, $\text{K}=\frac{1}{2}\text{I}\left(\omega \right)^{2}=\frac{1}{2 \text{I}}\left(\text{I} \omega \right)^{2}=\frac{\left(\text{L}\right)^{2}}{2 \text{I}}$
or, or, $\quad K=\frac{L^{2}}{2\left(\frac{M R^{2}}{2}\right)}=\frac{L^{2}}{M\left(R^{2}\right)}$
Taking log and differentiating, $\frac{\text{dK}}{\text{K}} = - \frac{2 \text{dR}}{\text{R}}$
or, the percentage change in $\text{K} = \frac{\text{dK} \times 1 0 0}{\text{K}} = - \frac{2 \text{dR}}{\text{R}} \times 1 0 0 = - 3 \text{\%}$