A uniform disk with mass 40.4 kg and radius 0.300 m has an axle through its cent

Question

A uniform disk with mass 40.4 kg and radius 0.300 m has an axle through its center and can rotate without friction. Starting from rest, a constant force 28.0 N is applied tangentially at the rim of the disk (visualize a hand pushing a bicycle wheel to get it spinning, but imagine that the force is applied constantly as the wheel speeds up, causeing it to accelerate its rotation).

Compute the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.320 revolution.    

Explanation / Answer

torque = r*f = I*alpha

alpha = rf/I = rf/1/2mr^2

alpha = 2f/mr = 2*28/40.4*0.3

alpha = 4.62 rad/s^2

from the equation

wf^2 = wi^1+2*alpha*theta

wf = sqrt(2*alpha*theta)

wf = sqrt(2*4.62*0.32*2*3.14) = 4.31 rad/s

v = r*wf = 1.29 m/s

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