Question:
If a spherical ball rolls on a table without slipping the friction of its total energy associated with rotational energy is:
If a spherical ball rolls on a table without slipping the friction of its total energy associated with rotational energy is:
Updated On: Jun 7, 2022
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The Correct Option is B
Solution and Explanation
Rotational energy of sphere
$E_{R}=\frac{1}{2} I \omega^{2}$
For sphere, moment of inertia
$I =\frac{2}{5} m R^{2}$
$\therefore E_{R} =\frac{1}{2}\left(\frac{2}{5} m R^{2}\right)\left(\frac{v}{R}\right)^{2}$
$=\frac{1}{5} m v^{2}$
Translational kinetic energy $E_{r}=\frac{1}{2} m v^{2}$
$\therefore$ Total energy $=\frac{1}{5} m v^{2}+\frac{1}{2} m v^{2}$
$=\frac{7}{10} m v^{2}$
$\therefore$ Required fraction
$=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}$
$=\frac{2}{7}$
$E_{R}=\frac{1}{2} I \omega^{2}$
For sphere, moment of inertia
$I =\frac{2}{5} m R^{2}$
$\therefore E_{R} =\frac{1}{2}\left(\frac{2}{5} m R^{2}\right)\left(\frac{v}{R}\right)^{2}$
$=\frac{1}{5} m v^{2}$
Translational kinetic energy $E_{r}=\frac{1}{2} m v^{2}$
$\therefore$ Total energy $=\frac{1}{5} m v^{2}+\frac{1}{2} m v^{2}$
$=\frac{7}{10} m v^{2}$
$\therefore$ Required fraction
$=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}$
$=\frac{2}{7}$
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Concepts Used:
System of Particles and Rotational Motion
- The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
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