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Merry go round problem

  • Thread starter phy21050
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  • #1
phy21050
Could someone walk me through the steps of the following problem step by step? Thanks. A merry go round with a moment of inertia of 1000kgm^2 is coasting at 2.20 rad/s. When an 80kg man steps onto the rim , a distance of 2 m from the axis of rotation the angular velocity decreases to ? rad/s
 

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  • #2
Tom Mattson
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Hi,

I would like to see how you would start this problem before telling you any detailed steps. Here is what you should be thinking about.

1. What shape can the merry-go-round be approximated as?
2. What can the man be approximated as?
3. What law of physics can be used to deterimine the initial and final angular velocities?

For #3, think about what quantity you learned in class lately that involve angular velocities. Is that quantity conserved? If so, can you use it here?
 
  • #3
phy21050
thats partly the reason I am writing because the textbook sux and lecture for the most part has been not helpful so being that I have a difficult time as it is with Physics that only makes it more difficult. As for the answers to your questions...
1. the merry go round is like a circle
2. I have no idea what the man is, a mass?
3. and for the 3rd I am not sure
 
  • #4
Tom Mattson
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Originally posted by phy21050
1. the merry go round is like a circle
Yes. Now, you know the rotational inertia of the merry-go-round, and you know the expression for the rotational inertia of a circle (from your book).

2. I have no idea what the man is, a mass?
Yes, he's a mass, but we are also interested in the configuration of the system (that's how you find the rotational inertia). So, try to find the rotational inertia of the (man+merry-go-round). I would treat the man as a point mass for this purpose.

3. and for the 3rd I am not sure
This should be readily apparent from the same chapter in which the problem appears. Try to look for it.
 
  • #5
phy21050
could you help me with the problem Tom? I don't really have the slightest idea as where to begin. The question is on a take home assignment and there is not really a similar question in the book.
 
  • #6
Tom Mattson
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I am trying to help you with the problem, but to be honest it sounds like you want me to do it for you. Could you try to read the chapter and come up with *something*? This problem is really not that difficult.

Here's a hint: The conservation law I alluded to earlier is the conservation of angular momentum. This is one of the most important principles in physics, and I am sure that your teacher must have given special attention to it.

Give it a try, show me how you start the problem, and I will help you from there.
 
  • #7
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We can solve this problem by using the conservation of angular momentum.

The initial angular momentum at the instant the man is about to jump on is:

L(total=L(man)-L(mgr)=Rp(man)-I(mgr)*omega(initial)


Now try to find the angular momentum after he jumps on.

L(total)=I(total)*omega(final)=
(I(mgr)+m(man)*R^2)*omega(final)

You should be able to work through this problem now.
Just remember what you are trying to solve for.
You are trying to solve for the final angular momentum.

*Hint* In order to get the decrease(in angular speed)what has to be equal?

peace out M2k
 

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