Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the ground with zero velocity with respect to the ground. What is the angular speed in radians/second of the merry-go-round after the boy jumps off?

A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.5 meters, and a mass M=211 kg. A small boy of mass m = 48 kg runs tangentially to the merry-go-round at a speed of v= 1.3 m/s, and jumps on.

Answer:

The speed of the merry-go-round after the boy jumps off is [tex]w_l = 0.394 \ rad/s[/tex]

Explanation:

From the question we are told that

The radius of the merry-go-round is M = 211 kg

The radius is R = 1.5 meters

The mass is of the boy is m = 48 kg

The speed of the boy is [tex]v = 1.3 \ m/s[/tex]

Generally the moment of inertia of the merry-go-round is mathematically represented as

[tex]I_g = \frac{MR^2}{2}[/tex]

Substituting values

[tex]I_g = \frac{211 * 1.5 ^2}{2}[/tex]

[tex]I_g = 237.38 \ kg \cdot m^3[/tex]

The moment inertia of the boy on the merry-go-round is

[tex]I_b = mR^2[/tex]

Substituting values

[tex]I_b = 48 * 1.5^2[/tex]

[tex]I_b = 108\ kg \cdot m^2[/tex]

The moment of inertia of both merry-go-round and the boy is

[tex]I_T = I_g + I _b[/tex]

So [tex]I_T = 237.38 + 108[/tex]

[tex]I_T = 345.38 \ kg \cdot m^2[/tex]

Since angular momentum is conserved in that the angular momentum of the merry-go-round with the boy on it is equal to the is angular momentum of the merry-go-round without the boy on it and it is mathematically evaluated as

[tex]p = mvR[/tex]

This can also be defined as

[tex]p = I_T * w_g[/tex]

Where [tex]w_g[/tex] is the angular speed of the merry-go-round

So

[tex]mvR = I_T * w_g[/tex]

Making [tex]w_g[/tex] the subject

[tex]w_g = \frac{mvR}{I_T}[/tex]

Substituting values

[tex]w_g = \frac{48 * 1.3 *1.5 }{345.38}[/tex]

[tex]w_g = 0.271 \ rad/s[/tex]

From above

[tex]p_1 = p_2[/tex]

Where [tex]p_1[/tex] is the angular momentum of the merry-go-round with the boy on it

And [tex]p_2[/tex] is the angular momentum of the merry-go-round without the boy on it

So

[tex]I_T * w_g = I_g * w_l[/tex]

Where [tex]w_l[/tex] is the speed of the merry-go-round without the boy on it

Making [tex]w_l[/tex] the subject of the formula

[tex]w_l = \frac{I_T * w_g}{I_g}[/tex]

[tex]w_l = \frac{345.38 * 0.271 }{237.38}[/tex]

[tex]w_l = 0.394 \ rad/s[/tex]