The 3.20 kg , 31.0-cm-diameter disk in the figure (Figure 1) is spinning at 310 rpm...

Question

Question

The 3.20 kg , 31.0-cm-diameter disk in the figure (Figure 1) is spinning at 310 rpm .

How much friction force must the brake apply to the rim to bring the disk to a halt in 2.60 s ? Brake

Answer 1

Answer #1

310 rpm = 32.46 rad/s

Torque = I*alpha

F*R = MR^2/2*alpha ( I = MR^2/2 for disc)

=> F = MR*alpha/2

now

W = W0 + alpha*t

0 = 32.46 + alpha*2.6

=> alpha = - 12.48 rad/s^2

=> F = 3.2*(0.31/4)*12.48 = 3.09 N

Answer 2

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