Merry-Go-Round Velocity

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In summary: So the problem was that you forgot to include the speed of the child in the initial angular momentum calculation.In summary, the problem involves a heavy child jumping onto a merry-go-round with no friction in the bearing. Two methods are used to solve for the angular velocity of the merry-go-round and child. The correct solution is found using method #1, where the initial angular momentum of the system is calculated by including the speed of the child.
  • #1
blue5t1053
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Problem:
In a plyground there is a small merry-go-round of radius 3.91 m and rotational inertia 5.68e+03 kg m^2. A child of mass 199 kg runs at a speed of 3.09 m/sec tangent to the rim of the merry-go-round when it is at rest and then jumps on. Assume no friction in the bearing of the merry-go-round. What is the angular velocity of the merry-go-round and child?

Besides the child be a heavy example, I wasn't able to come up with the right answer of 0.276 rad/sec. What might I be doing wrong?

My Work:
I searched the archives and found two ways of solving the problem.
[tex]method \ 1 \ \Rightarrow m*v*r = (I_{merry-go-round} + I_{child}) \omega \ ; solving \ for \ \omega[/tex]

[tex]method \ 2 \ \Rightarrow I_{merry-go-round} * \omega \ + \ m*v*r = (I_{merry-go-round} + I_{child}) \omega^{prime} \ ; \ solving \ for \ \omega^{prime}[/tex]

What I came up with:

[tex]method \ 1 \ \Rightarrow (199 kg \ + (\frac{5680 kg*m^{2}}{3.91m^{2}}))*(3.09 \frac{m}{sec})*(3.91m) = ((5680 kg*m^{2}) + (199 kg * ((3.91m)^{2}))) \omega \ ; solving \ for \ \omega[/tex]

[tex]\omega \ = \ 2.87[/tex]

[tex]method \ 2 \ \Rightarrow ((5680 kg*m^{2}) * (\frac{3.09 \frac{m}{sec}}{3.91 m} * 2 \pi)) + (199 kg \ + (\frac{5680 kg*m^{2}}{(3.91m)^{2}})*(3.09 \frac{m}{sec})*(3.91m) = (5680 kg*m^{2} + (199 kg * (3.91m^{2})) \omega^{prime} \ ;[/tex]
[tex]solving \ for \ \omega^{prime}[/tex]

[tex]\omega^{prime} \ = \ 5.52[/tex]
 
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  • #2
This is a conservation of angular momentum problem, so method #1 makes sense. (You could use method #2 by setting [itex]\omega = 0[/itex]--since it starts at rest, but then you're back to method #1.) But you didn't plug in the numbers properly. Hint: What's "m" stand for?
 
  • #3
Am I to assume that 'm' is total mass of the system (merry-go-round + child) or merry-go-round for interia of the merry-go-round?
 
  • #4
That would be incorrect. What's the total angular momentum of the system initially?
 
  • #5
[tex]L = r *p[/tex]

[tex]L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child[/tex]
 
  • #6
blue5t1053 said:
[tex]L = r *p[/tex]
This is correct.

[tex]L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child[/tex]
But this isn't quite right. What's 371.53 kg the mass of? What does "p" stand for?

Hint: What's the only thing moving initially?
 
  • #7
I thought 'p' was for momentum of the merry-go-round. The only object moving initially is the child. If I apply the child attached to the merry-go-round initially with the starting speed, if I understand this correctly, would be

[tex]L = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}[/tex]
 
Last edited:
  • #8
You forgot the speed of the child. (Yes, "p" stands for momentum, not just mass.)
 
  • #9
I had a lapse of thought there. So I shouldn't include the mass of the merry-go-round?
 
  • #10
Until the kid jumps on it, the merry-go-round isn't moving and thus has no angular momentum. Only the kid has angular momentum (with respect to the center of the merry-go-round).
 
  • #11
So if I revisit method one...

[tex]L_{child} = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}[/tex]

[tex]L_{merry-go-round} = (3.91 m) * (371.531 kg) * (0 \frac{m}{sec}) = 0 \frac{kg*m^{2}}{sec}[/tex]

[tex]I_{child} = (199 kg) * (3.91 m)^{2} = 3042.33 kg*m^{2}[/tex]

[tex]method \ 1 \ \Rightarrow 2404.3 \frac{kg*m^{2}}{sec} = (5680 kg*m^{2} + 3042.33kg*m^{2} ) \omega \ ; solving \ for \ \omega \ = 0.2756 \frac{rad}{sec} = \ 0.276 \frac{rad}{sec}[/tex]

Thanks!
 
  • #12
Looks good!
 

What is Merry-Go-Round Velocity?

Merry-Go-Round Velocity refers to the speed at which a merry-go-round or carousel rotates. It is measured in units of distance per time, such as meters per second or feet per minute.

How is Merry-Go-Round Velocity calculated?

The velocity of a merry-go-round can be calculated by dividing the distance traveled (circumference of the circle) by the time taken to travel that distance. This can be represented by the formula v = 2πr/t, where v is the velocity, r is the radius of the circle, and t is the time taken to make one full rotation.

Does the Merry-Go-Round Velocity change?

Yes, the Merry-Go-Round Velocity can change depending on factors such as the size and weight of the riders, the speed at which the operator pushes the merry-go-round, and any external forces acting on the carousel (such as wind).

What is the average Merry-Go-Round Velocity?

The average Merry-Go-Round Velocity can vary, but it is typically around 3-4 meters per second (6-9 miles per hour). However, this can vary greatly depending on the size and design of the merry-go-round.

How is Merry-Go-Round Velocity relevant to physics?

Merry-Go-Round Velocity is relevant to physics because it involves the concepts of circular motion, velocity, and acceleration. It can also be used to demonstrate the principles of centripetal force and inertia.

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