Question

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

• A. 2:07
• B. 3:01
• C. 1:02
• D. 1:01

Answer 1

Answer: (B)

3:01

Answer 2

${{E}_{k}}=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}$ ?(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.