Question

A rope is wrapped around the rim of a large uniform solid disk of mass 155 kg and radius 2.50 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.55 s?

Answer 1

Mass of the solid disk = M = 155 kg

Radius of the solid disk = R = 2.5 m

Moment of inertia of the solid disk = I

I = MR²/2

I = (155)(2.5)²/2

I = 484.375 kg.m²

Force with which the rope is pulled = F = 195 N

Angular acceleration of the disk =

I = FR

(484.375) = (195)(2.5)

= 1.006 rad/s²

Initial angular speed of the disk = ₁ = 0 rad/s (At rest)

Angular speed of the disk after 2.55 sec = ₂

Time period = T = 2.55 sec

₂ = ₁ + T

₂ = 0 + (1.006)(2.55)

₂ = 2.565 rad/s

Converting from rad/s to rev/s, (1 revolution = 2 radians)

₂ = 0.408 rev/s

Angular speed of the disk at the end of 2.55 sec = 0.408 rev/s