10 points SeCP11 8.P039 A large grinding wheel in the shape of a solid cylinder

Question

10 points SeCP11 8.P039 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 0.330 m is causes the wheel to have an angular acceleration of 0.806 rad/s? (a) What is the moment of inertia of the wheel? kg·m,2 (b) What is the mass of the wheel? te) Ir the wheel starts from rest, what is its angular velocity after 4.10 s have elapsed,assuming the force is acting during rad/s Need Help?RaER 7 -10 points SerCP11 8.P.050 Four objects-a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell-each have a mass of 4.79 kg and a radius of 0.210 m My Not 5 6 8 9

Explanation / Answer

Torque may be expressed in two useful ways for this problem:

(1) = I

(2) = rF

Substituting (2) into (1):

rF = I

Solved for I:

(a)

I = rF /

= (0.330m)(290N) / 0.806rad/s²

= 118.73kgm²

(b)

Then, the moment of inertia for a solid uniform cylinder is:

I = 0.5mr²

m = 2I / r²

= 2(118.73kgm²) / (0.330m)²

= 2180.6kg

(c)

Its angular velocity after 4.10s is:

= + t

= 0 + (0.806rad/s²)(4.10s)

= 3.30rad/s

Hope this helps.

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