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Review: A uniform disk of mass 10.0 kg and radius 0.250 m spins at 300 rev/min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at an average distance 0.220 m from the axis. The coefficient of friction between pad and disk is 0.500. A piston in a cylinder of diameter 5.0 cm presses the brake pad against the disk. Find the pressure required for the brake fluid in the cylinder.

Short Answer

Expert verified

The pressure required for the brake fluid in the cylinder is P=758 Pa .

Step by step solution

01

Angular acceleration:

Angular acceleration is defined as the rate of change of angular velocity over time. It is usually expressed in radians per second per second.

02

The angular acceleration is:

α=ΔωΔt

Where, ais the angular acceleration, ωis the change in angular frequency, tis the change in time, and

The disk (mass M=10.0kg , radius R=0.250 m) has a moment of inertia is , l=12MR2

The disk slows from ωi=300rev/mintoωf=0intimeintervalt=60.0s

Its angular acceleration is,

a=ωt=ωf-ωit=-ωit

Frictional torque from the brake pad slows the wheel. Friction has moment arm d=0.220 m

The relation between friction and angular acceleration is:

f=μknn=fμk=MR2ωi2μkdΔt

The normal force and coefficient of friction (μk=0.500) between the brake pad and the disk determine the amount of friction. We can write an expression for the normal force:

P=nπr2=MR2ωi2μkdπr2Δt=(10.0kg)(0.250m)2300rev×2πrad×1minmin×1rev×60.0s2(0.500)(0.220m)(60.0s)π(0.0250m)2=758Pa

Hence, the pressure required for the brake fluid in the cylinder is P=758 Pa

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