A 5.0gcoin is placed 15cmfrom the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs=0.80and μk=0.50.The turntable very slowly speeds up to 60rpm.Does the coin slide off?

Short Answer

Expert verified

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

Step by step solution

01

Given information

mass of the coin m=0.005kg

radius of the turntable r=0.15m

Static friction coefficient for coin and turntable μs=0.80

kinetic friction coefficient for coin and turntable μk=0.50

speedf=60rpm

acceleration due to gravityg=9.8m/s2

02

Explanation

The angular speed of the turntable

ω=2πf=2π(60)160=6.3rad/s

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force Fs=ma=μsFn=μsmgFs=0.8×0.005×9.8=0.392N

similarly,

Fs=ma0.0392=0.005xaa=7.84m/s2

The maximum velocity of the turntable is given by,

Vmax=axr=7.84x0.15Vmax=1.08m/s

The maximum angular velocity of the turntable is given by

ωmax=Vmaxr=1.080.15ωmax=7.2rad/s

now that we have the maximum angular acceleration of the turntable, we can calculate its maximum speed in rpm

Fmax=7.2×9.549=68.7rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

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