Torque and revolution question

Question

Starting from rest, a 12-cm-diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia is 2.5*10^-5 kg m^2.

How much torque is applied to the disk?

How many revolutions does it make before reaching full speed?

Answer 1

2000rpm is actually a frequency, multiplying it by 2π makes it an angular velocity

r = 6cm = 0.06m

t = 3s

f = 2000rev/min = 33.3rev/s

I = 2.5E-5kgm²

Torque = Iα

α = ?

ω = 2πf

v = at

ωr = αrt

ω = αt

so

α = ω/t = 2πf/t

then

Torque = Iα = I(2πf/t) = 1.7*10^-3 kgm²/s²

s = ½at²

rθ = ½αrt²

θ = ½αt² = ½(2πf/t)t² = πft

number of turns = θ/2π = πft/2π = ½ft = 50 rev

,,.,.