A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ?

Question

A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ? (A)  (2/7)  (B)  (3/7)  (C)  (4/7)  (D)  (5/7)

Tardigrade Admin · Accepted Answer

When a body rolls down an inclined plane, it makes rotational as well as translational motions. Thus, it is associated with rotational and translational kinetic energies. Hence, total kinetic energy of sphere

K = K_{rot.} + K_{trans}

= \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}

= \frac{1}{2} \frac{2}{5} m r^{2} (\frac{v ^{2}}{r ^{2}}) + \frac{1}{2} m v^{2}

= \frac{1}{5} m v^{2} + \frac{1}{2} m v^{2} = \frac{7}{10} m v^{2}

Also,

K_{rot.} = \frac{1}{2} I ω^{2} = \frac{1}{5} m v^{2}

Hence,

\frac{K _{rot.}}{K} = \frac{\frac{1}{5} m v ^{2}}{\frac{7}{10} m v ^{2}} = \frac{2}{7}

Q. A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ?

$\frac{2}{7}$

$\frac{3}{7}$

$\frac{4}{7}$

$\frac{5}{7}$

Q. A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ?

72​

73​

74​

75​

$\frac{2}{7}$

$\frac{3}{7}$

$\frac{4}{7}$

$\frac{5}{7}$