# Crazy dude jumping on a Merry-Go-Round

1. Mar 19, 2007

### PhyzicsOfHockey

1. The problem statement, all variables and given/known data

A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.686 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

(a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round?
increase
stay the same
decrease

(b) Calculate the initial and final kinetic energies for this system.
Ki = kJ
Kf = kJ

2. Relevant equations
KErot= 1/2*Iw^2
I=1/2mr^2

3. The attempt at a solution
a) I would think it should increase but im probably wrong, lol.

b) The merry-go-round initially is just spinning so I trued finding the KE using the 1/2*I*w^2 formula so I got 1/4*m*r^2*w^2and I came up with 4.98 kJ and thats the wrong answer... is there something I did wrong?

I don't have any idea how to find the KEf. Can I assume that its the rotational KE of the merry-goround plus the KE of the guy jumping on it?

2. Mar 19, 2007

### robb_

Lets start with the initial KE. You have a rotating disc included in your system; is there anything else that has KE?

3. Mar 19, 2007

### Staff: Mentor

Try comparing the rim's tangential speed (where the person jumps on) with the speed of the individual, and keep in mind, when the person jumps on, they must have the 'same' tangential speed.

If the person is traveling faster, the person slows down, but the merry-go-round must increase its rev/s.

4. Mar 19, 2007

### PhyzicsOfHockey

No the disk is the only thing that has KE? Maybe there is something I am just not understanding about this disk.

5. Mar 19, 2007

### robb_

Sorry, maybe I do not understand what your system is?
The person is running.

6. Mar 19, 2007

### PhyzicsOfHockey

the edge of the merry go round is moving at 11.336 m/s much fast then the man. So the merry-go-round slows down. I did conservation of momentum.
Initally...
Lmgr= I*w= 2310.58
Lp=I*(v/r)=623.326
Ltot= 2933.9

using this I found the angular speed of the merry-go-round fonal
2933.9/(Imgr + Ip)=w=3.09832 rad/sec

But I can not see how this helps me find the initial or final conditions.

7. Mar 19, 2007

### PhyzicsOfHockey

So I must include the guy running although hes not on the merry-go-round yet?

8. Mar 19, 2007

### PhyzicsOfHockey

So I did KEi= .5*Imgr*Wmgr^2 + .5*Ip*Wp^2 and got 5.4524 kJ
then i did KEf= .5*(M1+M2)*r^2*.5*Wf^2 and got 3.559kJ

Am I even close?

9. Mar 20, 2007

### PhyzicsOfHockey

I guess Im not because thats still the wrong answer

10. Mar 20, 2007

### Mentz114

Robb has given the answer.

The initial KE = KE ( disc) + KE (man)

Final KE = initial KE = KE(disc)

Remember that when the man steps on the rim, he increases the MoI of the roundabout by m*r^2.

Last edited: Mar 20, 2007
11. Mar 20, 2007

### PhyzicsOfHockey

True Mentz144 but the Final KE does not equal the Initial KE. I finally figured it out. my KEi was right but I added masses when finding the new MoI not m*r^2 to the MoI of the disk.

12. Mar 20, 2007

### Mentz114

Phyz, The question asks is energy being lost or gained by the >>system<< during the process. Taking the man and turntable together as the system, I venture that the energy is the same before and after.

13. Mar 20, 2007

### PhyzicsOfHockey

Energy is being lost because energy is required to make the mans speed the same speed as the merry-go-round. But thank you for your input.

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