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Question

Physics Question on System of Particles & Rotational Motion

If a spherical ball rolls on a table without slipping the friction of its total energy associated with rotational energy is:

A

44625

B

44599

C

44597

D

44627

Answer

44599

Explanation

Solution

Rotational energy of sphere
ER=12Iω2E_{R}=\frac{1}{2} I \omega^{2}
For sphere, moment of inertia
I=25mR2I =\frac{2}{5} m R^{2}
ER=12(25mR2)(vR)2\therefore E_{R} =\frac{1}{2}\left(\frac{2}{5} m R^{2}\right)\left(\frac{v}{R}\right)^{2}
=15mv2=\frac{1}{5} m v^{2}
Translational kinetic energy Er=12mv2E_{r}=\frac{1}{2} m v^{2}
\therefore Total energy =15mv2+12mv2=\frac{1}{5} m v^{2}+\frac{1}{2} m v^{2}
=710mv2=\frac{7}{10} m v^{2}
\therefore Required fraction
=15mv2710mv2=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}
=27=\frac{2}{7}