A solid sphere of mass m and radius r rolls d | vedclass

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(B) The total kinetic energy $(KE_{total})$ of a rolling object is the sum of its translational kinetic energy $(KE_{trans})$ and rotational kinetic e

Vedclass · Accepted Answer

$2/7$ rotational,$5/7$ translational

Vedclass · Answer

(B) The total kinetic energy $(KE_{total})$ of a rolling object is the sum of its translational kinetic energy $(KE_{trans})$ and rotational kinetic energy $(KE_{rot})$.$KE_{trans} = \frac{1}{2} mv^2$$KE_{rot} = \frac{1}{2} I\omega^2$For a solid sphere,the moment of inertia $I = \frac{2}{5} mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.Substituting these into the rotational kinetic energy formula:$KE_{rot} = \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v}{r})^2 = \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v^2}{r^2}) = \frac{1}{5} mv^2$.Now,the total kinetic energy is:$KE_{total} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{5+2}{10} mv^2 = \frac{7}{10} mv^2$.The fraction of rotational kinetic energy is $\frac{KE_{rot}}{KE_{total}} = \frac{1/5 mv^2}{7/10 mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.The fraction of translational kinetic energy is $\frac{KE_{trans}}{KE_{total}} = \frac{1/2 mv^2}{7/10 mv^2} = \frac{1}{2} 	imes \frac{10}{7} = \frac{5}{7}$.Thus,the rotational kinetic energy is $2/7$ of the total,and the translational kinetic energy is $5/7$ of the total.

$A$ solid sphere of mass $m$ and radius $r$ rolls down an inclined plane. What is the ratio of its rotational kinetic energy to its translational kinetic energy?

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