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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Asked Oct 25, 2019
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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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Expert Answer

Step 1

Given information:

Mass of the drink (m) = 85.6 g = 0.0856 kg

Let the farthest distance the drink can be placed without siding = r

Angular velocity of the table (ω) = 8.20 rad/s

The coefficient of static friction (µs) = 0.830

The coefficient of kinetic friction (µk) = 0.520

Step 2

In this case the required centripetal force is provided by the static friction force for the d...

smg = mrw2
0.83(9.81)
0.121 m
w2
8.22
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smg = mrw2 0.83(9.81) 0.121 m w2 8.22

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Science

Physics

Angular Motion

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