A 85.6 g drink is left on a turntable causing it to turn with an angular speed of 8.20 rad/s. If the drink has coefficients of friction of μs = 0.830 and μk = 0.520 with the turntable, what is the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide?

Question
Asked Oct 25, 2019

A 85.6 g drink is left on a turntable causing it to turn with an angular speed of 8.20 rad/s. If the drink has coefficients of friction of μs = 0.830 and μk = 0.520 with the turntable, what is the furthest distance that the drink could have been placed from the centre of the turntable for the drink not to slide?

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Step 1

Normal reaction acting on the drink is:

N
mg
drink
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N mg drink

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Step 2

When the drink is in static condition, the frictional force will be static frictional force and is equal to:

fasciak = 4,Ndrink
Sui'n
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fasciak = 4,Ndrink Sui'n

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Step 3

The drink is on a turn table which is in circular motion. Therefore, there will be a centripetal force which will act on the drink. Now for the drink no...

= facnk
centripetal
mor= u,mg
0.830(9.81 m/s2
(8.20 rad/s)
=0.12 m
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= facnk centripetal mor= u,mg 0.830(9.81 m/s2 (8.20 rad/s) =0.12 m

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