A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is 0.3406, how far from the center of the turntable can the coin be placed without having it slip off?

to get how far from the center of the turntable can the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r = m x g x μ

v^2 / r = g x μ .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

we have to convert the angular speed from rpm to rads/seconds

angular speed (rads/seconds) = (2π / 60 ) x rpm

angular speed (rads/seconds) = ((2 x π) / 60 ) x 33.3 = 3.49 rads/ seconds

now

velocity = 3.49 x r = 3.49r

now substituting the value of velocity into equation 1

v^2 / r = g x μ

(3.49r)^2 / r = 9.8 x 0.3406

12.18 x r = 3.34

r = 3.65 m

batolisis batolisis · Answer 2 · 2022-04-01T12:33:16+02:00

The distance where the coin can be placed without slipping off is : 0.2748 m

Given data :

Velocity of turntable = 33.3 rpm

coefficient of static friction = 0.3406

Determine how far from the center of the turntable the coin must be placed

angular velocity ( w ) = 33.3 * [tex]\frac{2\pi }{60} rad/s[/tex]

Given that

Mw²r = umg

( 33.3 * [tex]\frac{2\pi }{60}[/tex] )² * r = 0.3406 * 9.8

solve for r

r = ( 0.3406 * 9.8 ) / ( 3.4854 )²

= 0.2748 m

Hence we can conclude that The distance where the coin can be placed without slipping off is : 0.2748 m

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