Question

A circular wooden loop of mass m and radius R rest on a flat horizontal friction less surface. A bullet, also of mass m, and moving with a velocity V, strikes the loop and gets embedded in it. The thickness of the loop is much smaller than R. The angular velocity with which the system angular velocity with which the system rotates just after the bullet strikes the loop is______.

Answer 1

Answer:

w = vR/3

Explanation:

The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.

From the principle of conservation of linear momentum , we have

m*v = 2*m* Vcm

Where v = velocity of bullet, Vcm = velocity of wood

Hence, we have  

Vcm = v2

Also, from the conservation of angular momentum about the centre of mass.

M*V*(R/2) = Ic*w - equation (I)

where Ic = moment of inertia and w = angular velocity

Ic for a ring is given by $mr^2 + m(r/2)^2$

Ic of a bullet is given by $m(r/2)^2$

Hence, the moment of inertia of the system  is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives

Ic(system) = $3*m*R^2/2$

Substituting back into equation (I), we have

$m*v*R^2=3*m*R^2*w/2$

Hence, we obtain w =vR/3

w=v3R