A coin is placed 10.5 cm from the axis of a rotating turntable (anyone remember

Question

A coin is placed 10.5 cm from the axis of a rotating turntable (anyone remember record players?) of variable speed. When the speed of the record is slowly increased, the coin remains fixed on the turntable until a rate of 38 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Hint: You'll need to think about how to convert rpm to m/sec ... the method I've shown in class works great here! It is extremely helpful to realize that the coin travels a distance 2 ? R each time during each revolution. That is to say, 1 revolution equals a distance of 2 ? R.

Explanation / Answer

w = 38 rpm = 38*2*pi/60 = 3.98 rad/s

v = r*w = 0.105*3.98 = 0.4176 m/s

m*v^2/r = N*mue

m*v^2/r = m*g*mue

==> mue = v^2/(g*r) = 0.17

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