Question

A merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed w. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round of afterwards is(a) 2ω(b) ω(c) ω/2(d) 0

Answer 1

hence option a is correct

Answer 2

Answer:

(a) $2w$

Explanation:

Given,

• mass of the ring = M
• mass of the man = M
• Radius of the ring = R
• Angular speed of the ring and man = $w$

As we know that the moment of inertial of the ring about its center of rotation is $I_1\ =\ MR^2$

and the moment of inertial of the man about the center of rotation is $I_2\ =\ MR^2$

person jumps off  therefore the final angular speed of the man will be zero.

Let $w_f$ be the final angular speed of the ring.

Now From the conservation of angular momentum,

$\therefore (I_1\ +\ I_2)w\ =\ I_1w_f\ +\ 0\\\Rightarorw w_f\ =\ \dfrac{(I_1\ +\ I_2)w}{I_1}\\\Rightarrow w_f\ =\ \dfrac{(MR^2\ +\ MR^2)w}{MR^2}\\\Rightarrow w_f\ =\ \dfrac{2MR^2w}{MR^2}\\\Rightarrow w_f\ =\ 2w$

Hence the final angular speed of the ring after the jump of the person is $2w$.