Engineering
Physics
Moment of Inertia
Static and Kinetic Friction
Circular Motion in Vertical Plane
Question

Disk A has mass of 4 kg and a radius r = 75 mm. It is at rest when placed in contact with the belt, which moves at a constant speed v = 18 m/s. μk = 0.25 between the disk and the belt. If the normal reaction by the belt on the disk is equal to the weight of the disc, then the number of revolutions executed by the disk before it reach as a constant angular velocity is 27βπ . Find β.

 

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Solution
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Torque about A

R(μ mg)=mR22 . α

2μgR=α

2 × 0.25 × 10R=α

α=(5R)

at constant angular speed

ω=(vR)

(vR)2= 2α(2πn)

n=(v24απR2)=18 × 184 × 5Rπ

n=(18 × 1820Rπ)=18 × 1820 × 75 × 10−3×π

n=18 × 18 × 10320 × 75 × π= (6 × 18 × 10320 × π × 25)

n=(6 × 18 × 4π × 2)

n=36 π 6π=216π