How Does a Boy Affect the Angular Velocity of a Merry-Go-Round When He Jumps On?

  • Thread starter Thread starter masamune
  • Start date Start date
AI Thread Summary
When a boy of mass 50 kg jumps onto a stationary merry-go-round with a mass of 150 kg and a radius of 2 m, the angular velocity of the system can be calculated using the conservation of angular momentum. The initial linear momentum of the boy, given by mv, must equal the angular momentum of the combined system after the jump. The moment of inertia (I) of the system should include both the merry-go-round and the boy, calculated as I = (1/2)MR^2 for the disk plus MR^2 for the boy. The correct formula for angular velocity (ω) after the boy jumps on is derived from mvr = Iω, ensuring both components are accounted for. Properly combining the moment of inertia for both the boy and the merry-go-round is crucial for an accurate calculation of the final angular velocity.
masamune
Messages
15
Reaction score
0
A boy of mass m = 50 kg running with speed v = 4 m/s jumps onto the outer edge of a merry-go-round of mass M = 150 kg and radius R = 2 m, as shown in the picture above. The merry-go-round is initially at rest, and can rotate about a frictionless pivot at its center. You may assume that the inital velocity of the boy is tangent to the edge of the merry-go round.

Treat the boy as a point particle and the merry-go-round as a uniform solid disk. What is the angular velocity of the merry-go-round after the boy has jumped onto it?

I don't know if I can do this, but I set the linear momentum of the boy equal to the angular momentum of the merry-go-round with the boy.
Basically, mv = Iw
For my moment of inertia, I used the sum of both masses and plugged my given information into ((M+m)R^2)/2. This was how I calculated moment of inertia. Then I plugged the boy's mass and his initial speed divided by my moment of inertia and tried to get omega (w). I got 0.5 exactly, but it's not correct. Any help would be appreciated.
 

Attachments

  • merry.gif
    merry.gif
    1.2 KB · Views: 1,400
Physics news on Phys.org
linear momentum and angular momentum are not the same thing.

What you want to do is set the boys rotational inertia to the rotational intertia of the entire system.

you should get something like

mvr = Iw

I is the I of the system, I am sure you can figure that out..
 
Is my method of calculating the I of the system correct? I took the sum of the boy and the merry-go-round, multiplied by the square of the radius and all that divided by 2.This should give me the the I of the ystem right?
 
masamune said:
Is my method of calculating the I of the system correct? I took the sum of the boy and the merry-go-round, multiplied by the square of the radius and all that divided by 2.This should give me the the I of the ystem right?


you need to add the I of the boy and the I of the disk

I for a uniform disk rotating about the center of mass is \frac{1}{2}MR^2

I for a point mass is MR^2

add them together you get (\frac{1}{2}M_{merry-go-round} + M_{boy})R^2
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top