Need help on a circular motion question please

  • Thread starter chamonix
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Homework Statement



An air puck of mass .025kg is tied to a string and allowed to revolve in a circle of radius 1.0 m on a frictionless horizontal surface. The other end of the string passes through a hole in the center of the surface, and a mass of 1.0 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves on the surface.
a. what is the magnitude of the force that maintains circular motion acting on the puck?
b. What is the linear speed of the puck?

Homework Equations



for a. I have Fc=mVt^2/r

The Attempt at a Solution


well, I kinda got stuck on the part where there are two masses. I didnt know whether to add them together or to just use one of them. And I didnt know where else to start. Any guidance in the right direction would be most appreciated. Thanks.
 

Answers and Replies

  • #2
G01
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HINT: You know there has to be a centripetal force correct? WHAT is causing this force?
 
  • #3
Pythagorean
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well, I kinda got stuck on the part where there are two masses. I didnt know whether to add them together or to just use one of them. And I didnt know where else to start. Any guidance in the right direction would be most appreciated. Thanks.
Generally, the sum of all forces on the system should add up to zero if it has no transverse velocity. The system does have rotational velocity, but it doesn't sound like it's going anywhere as a whole system (it's just staying centered around that hole).

Since the transverse velocity is zero, the sum of the forces add up to zero, you can say:

F1 + F2 = 0

rearranged, we have

F1 = -F2

so if you just think about the string for a minute, and stand on it, so that you're rotating around with it, and ignore the ground or the hole, then you're just standing on a tightly stretched string that isn't going one way or the other, and on each side of the string is a force pulling it taut and those forces must be equal.

looking back at your system from the original point of view, what are the two forces?
 
  • #4
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Oh, ok, so then the two forces (not sure) are the outward force and the gravity, right?
 
  • #5
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Yes, the two forces are the centripetal and the gravitational, and they will be equal and opposite if the radius is constant.
 
  • #6
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Yes, the two forces are the centripetal and the gravitational, and they will be equal and opposite if the radius is constant.
huh?
The puck on the table has three forces on it: gravity, normal, and the tension. The tension is the centripetal force here. Okay I gave away too much now.
There is no "outward" force. The puck has inertia.
*edit* The sum of the forces is zero if the acceleration is zero. The velocity can change in magnitude or direction but either will constitute an acceleration.
 
Last edited:
  • #7
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So for a. I just use the equation F=ma.
then I get F=.025*9.81 and get F=.24525 N
and then for b. I use the Fc=mv^2/r
and get v=3.13209 m/s
Would this be correct?
 
  • #8
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Your reasoning is correct, the numbers look plausible. Check them one more time and if you get the same answer, go for it.
 
  • #9
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Thanks. All the help is greatly appreciated.
 

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