A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is fre

Question

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 270 N applied to its edge causes the wheel to have an angular acceleration of 1.130 rad/s2.

a) What is the moment of inertia of the wheel?

b) What is the mass of the wheel?

c) If the wheel starts from rest, what is its angular velocity after 5.00 s have elapsed, assuming the force is acting during that time?

Ok I got a) & c) , a = 78.8496 & c = 5.65, but for some reason I cannot get b! And first I tried using the equation:

but then it said my answer was more than 10% off from the correct answer, so then I tried using F = m*a, and it still says I am more than 10% off from the correct answer.

WHAT DO I DO!?!

Explanation / Answer

Torque may be expressed in two useful ways for this problem: (1) t = Ia (2) t = rF Substituting (2) into (1): rF = Ia Solved for I: I = rF / a = (0.330m)(270N) / 1.130 rad/s² = 78.79kg·m² Then, the moment of inertia for a solid uniform cylinder is: I = 0.5mr² m = 2I / r² = 2(78.79kg·m²) / (0.330m)² = 1447kg Its angular velocity after 5.50s is: ? = ?0 + at = 0 + (1.13rad/s²)(5s) = 5.65rad/s

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