Question

Ratio of kinetic energy and rotational energy in the motion of a disc is:

• A. 2 : 7
• B. 3 : 1
• C. 1 : 2
• D. 1 : 1

Answer 1

Answer: (B)

$E_k=\frac{1}{2}m{\upsilon }^2+\frac{1}{2}I{\omega }^2$ ?(i) $\therefore$ Rotational energy $E_{Rotational}=\frac{1}{2}I{\omega }^2$ ?(ii) Also we know that $I=m{K}^2$ so $K^2=\frac{R^2}{2}$ and $v=R\omega$ Putting the value of $I$ in equation (i) and (ii), we get $\frac{E_k}{E_{rotational}}=\frac{\frac{1}{2}m{\upsilon }^2+\frac{1}{2}\frac{m{R}^2}{2}\times \frac{{\upsilon }^2}{R^2}}{\frac{1}{2}\times \frac{m{R}^2}{2}\times \frac{{\upsilon }^2}{R^2}}$ $=\frac{\frac{3}{4}m{\upsilon }^2}{\frac{1}{4}m{\upsilon }^2}=\frac{3}{1}$ hence ratio 3 : 1.