A solid, uniform cylinder with mass 8.45kg and diameter 12.0cm is spinning with

Question

A solid, uniform cylinder with mass 8.45kg and diameter 12.0cm is spinning with angular velocity 225rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.332.

A)What must the applied normal force be to bring the cylinder to rest after it has turned through 5.15 revolutions?

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Explanation / Answer

Given:-

Mass,m = 8.45kg

Diameter,d =12.0cm = 0.12m

radius,r=0.06m

Initial angular velocity,wi= 225rpm = 23.55 rad/s                        [1rpm=2*pie/60 rad/s]

Coefficient of kinetic friction , ?k=0.332.

Final angular velocity,wf= 0 rad/s

?=5.15 revolutions = 32.34 rad                   [1revolution=2*pie rad]

Normal force, Fn =?

Use the kinematic equation , wf2= wi2+2??

we get ?= -8.58rad/s2

Moment of inertia of the solid, uniform cylinder, I=1/2mr2= 1/2*8.45*0.062 = 0.015kgm2

Torque, T= I*?= 0.131Nm

we have , T= ?k * Fn*r

Thereby, Fn= T/?kr = 0.131/ [0.332*0.06] = 6.58N

The normal force be to bring the cylinder to rest after it has turned through 5.15 revolutions is 6.58N

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