Turntable Rotating on Shaft Problem

AI Thread Summary
The discussion revolves around solving a problem related to the torque and friction force on a turntable. The user initially calculates angular acceleration and torque due to friction, assuming only the turntable's moment of inertia. Confusion arises regarding the correct application of the moment of inertia of both the turntable and the shaft. After clarification, it is determined that using the shaft's radius alone yields the correct friction force, resolving the discrepancy in the expected answer. The final conclusion confirms that the correct friction force is achieved by adjusting the radius used in the calculations.
BrainMan
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Homework Statement


upload_2016-11-13_19-23-39.png


Homework Equations


ωf = ωo = αt

∑τ = Iα

The Attempt at a Solution


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1) Find the acceleration

ωf = ωo + αt

fo) / t = α = -0.231 rad/sec

2) Find ∑τ

∑τ =Iα

I assumed that the only torque acting on the turntable was the torque due to friction

τfriction = (1/2)MR2(-0.231) = -.00312 N*m
 
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Looks OK so far. How do you get the friction force from the torque?
 
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BrainMan said:
τfriction = (1/2)MR2(-0.231) = -.00312 N*m
I didn't see a justification for ignoring the moment of inertia of the shaft (I'm assuming that the M and R above are for the disk).
 
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gneill said:
I didn't see a justification for ignoring the moment of inertia of the shaft (I'm assuming that the M and R above are for the disk).

So if I factor in the moment of inertia of the shaft I get the ∑τ = Itotalα

=> [(1/2)MR2 + (1/2)mr2 ](-0.231) = -0.00312 N*m

=> Ffriction = ∑τ / (R+r) = -0.020 N

The correct answer should be 0.52 N so I'm not sure how I'm so far off.
 
BrainMan said:
=> Ffriction = ∑τ / (R+r) = -0.020 N

The correct answer should be 0.52 N so I'm not sure how I'm so far off.
Why use R + r ? Note that the brake pads are applied to the shaft.
 
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TSny said:
Why use R + r ? Note that the brake pads are applied to the shaft.

Okay if I use r instead of R + r I get the right answer. Thanks!
 
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