A solid sphere is rolling down an inclined plane. Thenthe ratio of its translational kinetic energy to its rotational kinetic energy is

A solid sphere is rolling down an inclined plane. Thenthe ratio of its

1.	A solid sphere is rolling down an inclined plane. Thenthe ratio of its translational kinetic energy to its rotational kinetic energy is
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5<br/>1.5<br/>1<br/>0.4</p>Solution :Let M and R be the mass and <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> of the solid sphere . <br/> As it is rolling down an inclinedplane , <br/> `therefore` Its <a href="https://interviewquestions.tuteehub.com/tag/translational-3229278" style="font-weight:bold;" target="_blank" title="Click to know more about TRANSLATIONAL">TRANSLATIONAL</a> kinetic energy is `K_(t) = (1)/(2) M v_(cm)^(2)` <br/> (where `v_(cm)` is the velocity of centre of mass) and its rotational kinetic energy is `K_(r) = (1)/(2) I omega^(2)` <br/> But `I = (2)/(5) <a href="https://interviewquestions.tuteehub.com/tag/mr-549185" style="font-weight:bold;" target="_blank" title="Click to know more about MR">MR</a>^(2)` (For solid sphere) <br/> and `omega = (v_(cm)) /(R) therefore K_(r) = (1)/(2) ((2)/(5) MR^(2)) ((v_(cm))/(R))^(2) = (1)/(5) M v_(cm)^(2)` <br/> Their corresponding ratio is `(K_(i))/(K_(r)) = ((1)/(2) M v_(cm)^(2))/((1)/(5) M v_(cm)^(2)) = (5)/(2) = 2.5`</body></html>