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?

Answers: 3

Show answers

Answers

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

c. bronchi

i know this is true mom has it.

mark me branliest only got it 1 time!

Answer from: Quest

magnitude and direction

Another question on Physics

Physics, 22.06.2019 07:00

Abucket full of water weights 4 kg and water well is 10m deep. a girl draws water from the well. it takes the girl 3 minutes to draw a bucket full of water from the well. what is the power of the girl?

Answers: 1

Answer

Physics, 22.06.2019 08:30

Hey student studies gravity using objects that have the same mass which two objects have the greatest gravitational force acting between them a. 100kg 1.0m 100kg b. 100kg 2.0m 100kg c. 100kg 2.0m 100kg big d. 100kg big 3.0m 100kg big

Answers: 1

Answer

Physics, 22.06.2019 09:50

If an athlete runs once around a track, back to the starting line, her average velocity is zero. true or false?

Answers: 3

Answer

Physics, 22.06.2019 10:20

Assume that a person skiing high in the mountains at an altitude of h = 15100 ft takes in the same volume of air with each breath as she does while walking at sea level. determine the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level. assume that the air composition (i.e. % of air that is oxygen) is the same at sea level as it is at 15100 ft.

Answers: 2

Answer

You know the right answer?

Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the me...

Questions