Turntable Rotating on Shaft Problem

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Homework Help Overview

The discussion revolves around a problem involving a turntable rotating on a shaft, focusing on the dynamics of angular motion and torque. Participants are examining the relationships between angular acceleration, torque, and friction forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of angular acceleration and the resulting torque due to friction. Questions arise regarding the treatment of the moment of inertia of the shaft and its impact on the overall torque calculation. There is also a discussion about the appropriate method for determining the friction force from torque.

Discussion Status

The conversation is active, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the calculations, and there is an exploration of different interpretations of the problem setup, particularly concerning the moment of inertia and the application of forces.

Contextual Notes

Participants are navigating assumptions about the system, including which components' moments of inertia should be considered and the implications of using different radii in their calculations. There is a noted discrepancy between calculated and expected values, prompting further inquiry into the assumptions made.

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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