A uniform disc of mass `m` is fitted (pivoted smoothly) with a rod of mass `m//2`. If the bottom of the rod os pulled with a velocity `v`, it moves without changing its and of orientation and the disc rolls without sliding. Take kinetic energy of the system (rod + disc) is.
.

Options

A. `(mv^2)/(4)`
B. `(mv^2)/(2)`
C. `(5 mv^2)/(16)`
D. `mv^2`

Answer

D

Explanation

(d) Rolling can be considered as pure rotation about point of contact
`K_("rolling") = (1)/(2)I_p omega^2 = (1)/(2) ((mR^2)/(2) + mR^2) omega^2 = (3)/(4) mR^2 omega^2` …(i)
The rod translates with the velocity `v` hence velocity of centre of disc will also be `v`
`v = omega R` ...(ii)
From (i) and (ii) , `K_("rolling") = (3)/(4) mv^2`
Kinetic energy of the rod `K_(rod) = (1)/(2) ((m)/(2)) v^2 = (mv^2)/(4)` ...(iii)
`K_(total) = k_("rolling") + k_(rod) = mv^2`.