Physics, 07.05.2020 01:01 Queen9336

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?

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Answer from: twinkie7078

Complete Question

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.

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?

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

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 $v = 1.3 \ m/s$

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

$I_g = \frac{MR^2}{2}$

Substituting values

$I_g = \frac{211 * 1.5 ^2}{2}$

$I_g = 237.38 \ kg \cdot m^3$

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

I_b = mR^2

Substituting values

I_b = 48 * 1.5^2

$I_b = 108\ kg \cdot m^2$

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

I_T = I_g + I _b

So I_T = 237.38 + 108

$I_T = 345.38 \ kg \cdot m^2$

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

p = mvR

This can also be defined as

p = I_T * w_g

Where w_g is the angular speed of the merry-go-round

mvR = I_T * w_g

Making w_g the subject

$w_g = \frac{mvR}{I_T}$

Substituting values

$w_g = \frac{48 * 1.3 *1.5 }{345.38}$

$w_g = 0.271 \ rad/s$

From above

p_1 = p_2

Where p_1 is the angular momentum of the merry-go-round with the boy on it

And p_2 is the angular momentum of the merry-go-round without the boy on it

I_T * w_g = I_g * w_l

Where w_l is the speed of the merry-go-round without the boy on it

Making w_l the subject of the formula

$w_l = \frac{I_T * w_g}{I_g}$

$w_l = \frac{345.38 * 0.271 }{237.38}$

$w_l = 0.394 \ rad/s$

Answer from: Quest

true

explanation:

Answer from: Quest

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