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.