A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 290 N applied to its edge causes the wheel to have an angular acceleration of 0.906 rad/s2.
(a) What is the moment of inertia of the wheel?
kg·m2
(b) What is the mass of the wheel?
kg
(c) If the wheel starts from rest, what is its angular velocity after 5.10 s have elapsed, assuming the force is acting during that time?
rad/s
4 Answers
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(a) The moment inerttia of the wheel :
τ = I * α ----> τ = r * F
0.33 * 290 = I * 0.906
I = 105.63 kg.m^2
(b) I = 0.5 * m * r^2
m = 105.63 /0.5 * 0.33^2 = 1939.93 kg
(c) ωf = ωo + α * t
= 0 + 0.906 * 5.10 = 4.62 rad/s
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10
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Solid Cylinder
Source(s): https://shrinke.im/bamJB -
Angular acceleration is torque divided by the moment of inertia:
α = τ / I
So I = τ / α
I = ( 290 N )( 0.330 m ) / 0.906 sec^-2
I = 105.63 kg·m² ◄---
For a uniformly dense cylinder:
I = ½MR²
2I = MR²
M = 2I / R²
M = 2( 105.63 kg·m² ) / ( 0.330 m )²
M = 1940 kg ◄---
∆ω = αt
∆ω = ( 0.906 sec^-2 )( 5.10 sec )
∆ω = 4.62 rad/sec
Since it started from rest,
ω = ∆ω = 4.62 rad/sec◄---