A sphere of mass `M` rolls without slipping on rough surface with centre of mass has constant speed `v_0`. If mass of the sphere is `m` and its radius be R`, then the angular momentum of the sphere about the point of contact is.

Options

A. `(4)/(5) Mv_0R(- hat k)`
B. `(9)/(5) Mv_0 R(- hat k)`
C. `(8)/(5) Mv_0 R(- hat k)`
D. `(7)/(5) Mv_0 R (- hat k)`

Answer

D

Explanation

(d) Since `vec L_p = vec L_(cm) + vec r xx vec p_(cm)`
`vec L_p = v = (10 - t) ms^-1`
=`I_(cm) omega_0 (- hat k) + M v_0 R(- hat k)`
Since sphere is in pure rolling motion hence
`omega = v_0//R`
`rArr vec L_p = [(2)/(5) MR^2((v_0)/(R)) + M v_0 R(- hat k)]=(7)/(5) M v_0 R (- hat k)`.
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