A solid sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about the centre v0 / 2 r. The translational velocity after the sphere starts pure rolling is (x v0/y) in forward direction. Find (x+y).

Question

A solid sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about the centre v0 / 2 r. The translational velocity after the sphere starts pure rolling is (x v0/y) in forward direction. Find (x+y). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/78d4fd10fa3be2659631a8ba122e7c76-.png" /> Angular momentum can be conserved about lowest point m v0 r+(2/5) m r2 (v0/2 r) =m v' r+(2/5) m r2 (v'/r) (10 m v0 r+2 M v0 r/10) =(7 m v' r/5) v'=(6 v0/7)

Q. A solid sphere of mass M and radius r slips on a rough horizontal plane. At some instant it has translational velocity v0​ and rotational velocity about the centre v0​/2r. The translational velocity after the sphere starts pure rolling is yxv0​​ in forward direction. Find (x+y).

Angular momentum can be conserved about lowest point mv0​r+52​mr22rv0​​ =mv′r+52​mr2rv′​ 1010mv0​r+2Mv0​r​ =57mv′r​v′=76v0​​

Angular momentum can be conserved about lowest point
$m v_{0} r + \frac{2}{5} m r^{2} \frac{v _{0}}{2 r}$
$= m v^{'} r + \frac{2}{5} m r^{2} \frac{v ^{'}}{r}$
$\frac{10 m v _{0} r + 2 M v _{0} r}{10}$
$= \frac{7 m v ^{'} r}{5} v^{'} = \frac{6 v _{0}}{7}$