Starting from rest, a 12--diameter compact disk takes 2.5 to reach its operating angular velocity of 2100 .
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Starting from rest, a 12--diameter compact disk takes 2.5 to reach its operating angular velocity of 2100 .

[From: ] [author: ] [Date: 12-11-12] [Hit: ]
Assuming that the operating angular velocity is 2100 rev/minute = 35 revolution /second.Angular acceleration isα =(ω2- ω1)/t = 2π (N2 – N1) / t = 2π N2/ t since N1 = 0α= 2π*35/2.{We know F = ma. Replace force by torque, m by I,τ= 2.......
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?

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No unit is given for angular velocity.

Assuming that the operating angular velocity is 2100 rev/minute = 35 revolution /second.
Angular acceleration is α = (ω2- ω1)/t = 2π (N2 – N1) / t = 2π N2/ t since N1 = 0
α = 2π*35/2.5 = 28 π rad /s²
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{We know F = ma. Replace force by torque, m by I, and acceleration by angular acceleration and you get the formula for torque as τ = Iα)

τ = 2.5 *10^-5*28 π = 2.2 e-3 Nm
Since it starts from rest ,the total angle turned = θ = 0.5 αt²

Total number of turns
= (θ / 2π )= 0.5 αt²/(2π)
= 0.5 t² (2π N2/ t) 2π)
= N2*t/2
=35*2.5/2
=43.75 Revolutions.
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