Another friction problem (rotational this time)

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The discussion revolves around a physics problem involving a coin on a spinning record. The user calculates the linear speed of the coin to be 0.59 m/s and determines that the speed at which the coin would slide off, given a static friction coefficient of 0.5, is 0.84 m/s. Adding another coin on top does not change the sliding speed since the masses cancel out in the calculations. The user also describes their confusion regarding the diagram for linear velocity and instantaneous acceleration, noting that they correctly represented linear velocity tangent to the circular path and instantaneous acceleration directed inward. Overall, the calculations and concepts presented are mostly accurate, with a minor error in using gravity as 10 m/s² instead of 9.81 m/s².
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Homework Statement


Once again I just need someone to check my work, I think I have it correct. If a coin of mass .005kg is placed on a spinning record .14m from the center and it takes the coin 1.5 sec to complete a revolution then a) what is the linear speed of the coin. b)If the coefficient of static friction is .5 then at what speed would the coin slide off c) If you were to add another coin directly atop the first how would the affect your results in part b) and d) draw and label the linear velocity and instantaneous acceleration in a diagram.


Homework Equations


F=(mv^2)/r
Ff=umg u=coefficient of friction
Circumference/ Time = linear velocity

The Attempt at a Solution


part a
.88/1.5 = .59m/s

part b
(mv^2)/r=umg
(v^2)/r=ug
v=sqrt(rug)
v=.84m/s

part c
It wouldn't change because the masses canceled out

part d
This one I was a bit confused about. I drew the linear velocity tangent to the coins circular trajectory. I drew the instantaneous acceleration perpendicular to that going out, through the coin, from the center of the record.

Note: In the calculations for part b gravity was 10 instead of 9.81
 
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The centripetal acceleration of the object is inward isn't it?

Otherwise your numbers look OK.
 
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