Angular momentum of merry go round

In summary, the question asks for the final angular velocity of a playground merry-go-round after a runner of mass 36 kg and velocity 2.9 m/s jumps onto its rim. The merry-go-round has a moment of inertia of 404 kgm^2 and a radius of 2m. The answer, according to the back of the book, is 0.381 rad/s. However, the asker is unable to obtain this answer, even after using the formula mvr = (Imgr + Ir)w and solving for w. The key to solving this problem is to understand that angular momentum is conserved, so the moment of inertia of the runner must be added to the moment of inertia of the merry
  • #1
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This question was asked a while ago by another user, but I do not understand how he got the answer:

"A runner of mass m=36 kg and running at 2.9 m/s runs and jumps on the rim of a playground merry-go-round which has a Moment Of Inertia of 404 kgm^2 and a radius of 2 m. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three decimal places?

"According to the back of the book, the answer is 0.381 rad/s; however, I can never come up with that answer."

He says he used the formula mvr = (Imgr + Ir)w where Imgr is the inertia of the merry-go-round and Ir is the inertia of runner, then solved for w and was able to get the correct answer. When I plug in the numbers, I don't get the right answer at all. Is he using a different "I" value for the runner, if so how does he calculate it? Please help!
 
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  • #2
angular momentum is given by [itex]L=I \omega = mvr[/itex]

Angular momentum is always conserved. So that the angular momentum as the runner jumps on the merry go round should be the same as the angular momentum as merry go round and the runner.

So, the moment of inertia of the merry-go-round alone is [itex]404 kgm^2[/itex], what is the moment of inertia of the runner at a distance of 2m from the centre of rotation of the merry-go-round? Then what would be the total moment of inertia of the entire system?
 
  • #3


I cannot verify the exact method used by the other user without further information. However, I can provide an explanation of the concept of angular momentum and how it applies in this scenario.

Angular momentum is a measure of rotational motion, and it is defined as the product of an object's moment of inertia (I) and its angular velocity (w). In simpler terms, it is the measure of how much an object is rotating and how fast it is rotating.

In this scenario, the runner is adding their own angular momentum to the merry-go-round when they jump onto it. The formula mvr = (Imgr + Ir)w can be used to calculate the final angular velocity (w) of the merry-go-round after the runner jumps on. Let's break down this formula:

- m is the mass of the runner
- v is the velocity of the runner
- r is the radius of the merry-go-round
- Imgr is the moment of inertia of the merry-go-round
- Ir is the moment of inertia of the runner

The moment of inertia is a measure of an object's resistance to rotational motion and is dependent on the distribution of mass and the object's shape. For a simple shape like a ring, the moment of inertia can be calculated using the formula I = mr^2, where m is the mass and r is the radius. However, for a more complex shape like a human body, the moment of inertia would need to be calculated using a more complex formula.

In this scenario, the moment of inertia of the merry-go-round is given (404 kgm^2), but we do not have the moment of inertia of the runner. It is possible that the other user used a simplified formula for the runner's moment of inertia, or they may have used a more complex formula that was not mentioned. Without knowing the exact method used, it is difficult to determine the discrepancy in the final answer.

In conclusion, the formula mvr = (Imgr + Ir)w is a valid way to calculate the final angular velocity of the merry-go-round after the runner jumps on, but the exact method used by the other user may not be clear. It is important to note that the moment of inertia can vary depending on the shape and distribution of mass, so it is crucial to use the correct values in calculations.
 

What is angular momentum and how is it related to a merry go round?

Angular momentum is a measure of the rotational motion of an object. In the case of a merry go round, it refers to the spinning motion of the ride.

What is the equation for calculating the angular momentum of a merry go round?

The equation for calculating angular momentum is L = Iω, where L is angular momentum, I is the moment of inertia, and ω is the angular velocity.

How does the speed of the merry go round affect its angular momentum?

The speed of the merry go round, or its angular velocity, has a direct relationship with its angular momentum. As the speed increases, the angular momentum also increases.

Why does the angular momentum of a merry go round remain constant?

According to the law of conservation of angular momentum, the total angular momentum of a closed system, such as a merry go round, remains constant unless an external torque is applied.

How does changing the shape of the merry go round affect its angular momentum?

The moment of inertia, which is a factor in the calculation of angular momentum, is affected by the shape of the object. Changing the shape of the merry go round will change its moment of inertia and therefore change its angular momentum.

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