A small button placed on a horizontal rotating platform withdiameter 0.310 m wil

Question

A small button placed on a horizontal rotating platform withdiameter 0.310 m will revolve with theplatform when it is brought up to a speed of 35.0 rev/min, provided the button is no more than0.128 m from the axis. (a) What is the coefficient of static frictionbetween the button and the platform?

(b) How far from the axis can the button be placed, withoutslipping, if the platform rotates at 50.0 rev/min?
(a) What is the coefficient of static frictionbetween the button and the platform?

(b) How far from the axis can the button be placed, withoutslipping, if the platform rotates at 50.0 rev/min?

Explanation / Answer

While the button doesn't move, the frictionalforce exactly balances the centripetal acceleration from therotating platform.

Friction : umg, u=coefficient of static friction
Centripetal acceleration : mw^2r, where w is the rotational speed(in rad/sec), r is the radius of placement.
Remember to match your units.

umg=m(w^2)r
u=(w^2)r/g

1. w=35 rpm=35*2*pi/60=3.665 rad/sec
u=(w^2)r/g=(3.665)^2*0.128/9.8=0.175
The coefficient of static friction is 0.175.
.
.
.
2.u=(w^2)r/g
r=ug/(w^2)

w=50*2*pi/60=5.236 rad/sec
r=ug/(w^2)
r=(0.175)(9.8)/(5.236)^2=0.0626 m
The button can be placed as far as 0.0626m without slipping.

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