Question

A thin circular metal disc of the radius $500mm$ is set rotating about a central axis normal to its plane. Upon raising in temperature gradually, the radius increases to $507.5mm$ . 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, $K=\frac{L^2}{2\left(\frac{M{R}^2}{2}\right)}=\frac{L^2}{M\left(R^2\right)}$
Taking log and differentiating, $\frac{\text{dK}}{\text{K}}=-\frac{2\text{dR}}{\text{R}}$
or, the percentage change in $\text{K}=\frac{\text{dK}\times 100}{\text{K}}=-\frac{2\text{dR}}{\text{R}}\times 100=-3\text{\%}$