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?
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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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
----------------------------
{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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