Question

A grinding wheel is in the form of a uniform solid disk of radius 6.99 cm and mass 1.93 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.594 N · m that the motor exerts on the wheel.

(a) How long does the wheel take to reach its final operating speed of 1 100 rev/min?

______________s

(b) Through how many revolutions does it turn while accelerating?

___________rev

Answer 1

Given
r = 6.99 cm = 0.0699m
m = 1.930 kg

ω0 = 0 rad/s

τ = 0.594 Nm

a) ω = 1100 rev/min

moment of inertia is

I = 0.5 m r²

I = (0.5)(1.93kg)(0.0699 m)²

I = 0.0047 kg m²

and

τ= I α

so

α= τ/I

α = 0.594 Nm/0.0047 kg m²

α = 126.4 rad/s²

find time from

ω=ω0 + αt

(1100rev/min)(2πrad/rev)(1min/60s) = 0 + (126.4rad/s²)t

t = 115.13 rad/s / 126.4 rad/s²

t = 0.91 s

b) use

θ = θ0 + ω0 t + (0.5)αt²

θ = 0 + (0rad/s)(0.91s) + (0.5)(126.4 rad/s²)(0.91

s)²

θ = (52.33rad )(1 rev/2πrad)

θ = 8.33 rev