Question

A grinding wheel is a uniform cylinder with a radius of 8.70 cm and a mass of 0.600 kg.

(a) Calculate its moment of inertia about its center.
( kg·m2)

(b) Calculate the applied torque needed to accelerate it from rest to 1500 rpm in 3.00 s if it is known to slow down from 1500 rpm to rest in 59.0 s.
( m·N)

Answer 1

I = 1/2 mr^2

m = 0.6kg
r = 0.087 m, taking radius as 0.09 m

(a) I = .00243 kg·m2

ω = αt
1500 rpm = 157.08 rad/s

α = 157.08/3 = 52.36rad/s^2

theoretical T = Iα = 0.00243 kg·m2 x 52.36 rad/s^2 = 0.127m·N

Also....

Required additional torque due to friction, found from the additional data I overlooked:

"Slowing down" acceleration = α = 157.08/59 = 2.66rad/s^2
This same acceleration will be present when the wheel speeds back up to 1400 rpm, and must be added to the theoretical value obtained above.

T (friction) = 0.00243 kg·m2 x 2.66 rad/s^2 = 0.0065m·N

(b) T (total) = 0.127m·N + 0.0065m·N = 0.133m·N