A ring of mass `M` and radius `R` sliding with a velocity `v_(0)` suddenly enters inthrough surface where the coefficient of friction is `mu`, as shown in figure.
.
Choose the correct statement (s).

Options

A. The linear distance moved by the centre of mass before the ring starts rolling is `(3v_(0)^(2))/(8 mu g)`
B. The net work done by friction force is `- (3)/(8) mv_(0)^(2)`
C. The loss is kinetic energy of the ring is `(mv_(0)^(2))/(4)`
D. The gain in rotational kinetic energy is `+ (mv_(0)^(2))/(8)`.

Answer

A, C, D

Explanation

(A) `v^(2) = v_(0)^(2) - 2 as` at pure rolling `v = (v_(0))/(2)`
`((V_(0))/(2))^(2) = v_(0)^(2) - 2 mu gs rArr s = (3 v_(0)^(2))/(8 mu g)`
(B) `W_("friction") = Delta K rArr W_("friction") = K_(f) - K_(i)`
`W_("friction") = (1)/(2) m((v_(0))/(2))^(2) (1+K^(2)/(R^(2))) - (1)/(2) mv_(0)^(2)`
`W_("friction") = (1)/(2) m (v_(0)^(2))/(4) (1 +1) -(1)/(2) mv_(0)^(2)`
`W_("friction") = (1)/(2) mv_(0)^(2) [(1)/(2) -1] rArr W_("friction") = -(1)/(4) mv_(0)^(2)`
( C) `W_("friction") = Delta K , Delta K = (1)/(4) mv_(0)^(2)`
(D) `KE_("rotational") = (1)/(2) I omega^(2)`
`KE_("rotational") = (1)/(2) mR^(2) xx ((v_(0) //2)/(R))^(2)`
`K_("rotational") = (mv_(0)^(2))/(8)`.