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Toy car conservation of angular momentum

  • Thread starter bcjochim07
  • Start date
374
0
1. Homework Statement
A 200g toy car is placed on a narrow 60 cm diameter track with wheel grooves that keep the car going in a circle. The 1 kg. track is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car's switch is turned on, it soon reaches a steady speed of .75 m/s relative to the track. What then is the track's angular velocity in rpm?


2. Homework Equations
L=I*angular velocity
L=mrvsintheta
Moment of Inertia for a hoop: I=MR^2

3. The Attempt at a Solution

I found the angular momentum of the car and added it with the expression for the angular momentum of the track. The two combined should have a momentum of 0.
(.200kg)(.30m)(.75m/s) + (1kg)(.30m)^2*angular velocity = 0

angular velocity = -.5 rad/s = -4.77 rpm

According to my book the answer is 4 rpm and I was just wondering what I am doing wrong.
 

Answers and Replies

Doc Al
Mentor
44,827
1,083
(.200kg)(.30m)(.75m/s) + (1kg)(.30m)^2*angular velocity = 0
.75 m/s is the speed of the car with respect to the track, not with respect to the lab frame.
 
454
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The .75 m/s and the angular velocity you compute from that is relative to the track. The angular velocity of the track you want to compute is relative to the earth. You need to transform the angular velocity of the car to the non-rotating frame.
 
374
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Ok, I see. Can you help me with that transformation? Relative velocity is one of those things that makes a little uncomfortable. Would you just subtract the velocity of the car relative to the track from the answer that I got up above?
 
Doc Al
Mentor
44,827
1,083
Try this:
speed of car w.r.t ground = speed of car w.r.t track + speed of track w.r.t ground
 

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