Solving Frames of Reference Homework: Static/Kinetic Friction

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JWSiow
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Homework Statement


A child sits 2m from the centre of a merry-go-round (a rotating wooden platform) which is rotating at 3.5revs/min in a clockwise direction (when viewed from above). She places a wooden block of mass m=0.25kg beside her on the platform.

a)Draw a diagram showing all the forces acting on the block, and describe each force.

b)If the block remains at rest relative to the platform, calculate its acceleration. Explain your reasoning.

c)If the coefficients of static and kinetic friction between the block and the platform are 0.2 and 0.18 respectively, do you expect the block to remain at rest or to slide relative to the platform?


Homework Equations


F(friction) = [tex]\mu[/tex]kFN
F(friction) = [tex]\mu[/tex]sFN


The Attempt at a Solution


a)I'm not sure if I had the write forces in my diagram, I had the weigh mg, normal force, centripetal force and friction force.

b)I'm not sure about this one either, but I think, since the frame of reference is the platform, and the block is not moving, acceleration = 0. ?

c)I didn't really know what to do for this, so I worked out the static and kinetic friction forces, and then compared it to the centripetal force. Since the centripetal force was smaller than the static friction force, the block would remain at rest. I don't know if this is correct, or if it's even a good explanation!
 
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JWSiow said:
a)I'm not sure if I had the write forces in my diagram, I had the weigh mg, normal force, centripetal force and friction force.

b)I'm not sure about this one either, but I think, since the frame of reference is the platform, and the block is not moving, acceleration = 0. ?

c)I didn't really know what to do for this, so I worked out the static and kinetic friction forces, and then compared it to the centripetal force. Since the centripetal force was smaller than the static friction force, the block would remain at rest. I don't know if this is correct, or if it's even a good explanation!
(a) friction force is the centripetal force.,so you have three forces ("centripetal force" is just the name of real force directed to the center. In this case - friction force)

(b) No. Platform is not inertial frame of reference. So in this frame of reference there is centrifugal force. You have two possibilities: use ground frame of refference and calculate centripetal acceleration (recommended :smile:) OR use platform as a frame of refference and calculate centrifugal acceleration.

(c) OK, but block would remain rest since centripetal force was smaller OR EQUAL to static friction force

regards
 
Ok, thanks. :)

So, for b), when calculating the centripetal acceleration, I'd use the velocity of the platform?
 
JWSiow said:
So, for b), when calculating the centripetal acceleration, I'd use the velocity of the platform?
Yes. And distance between mass and pivot. Find centripetal force first.

regards