A wheel with moment of inertia 2.46kgm22.46kgm2 and radius R=0.78R=0.78m rotates

Question

A wheel with moment of inertia 2.46kgm22.46kgm2 and radius R=0.78R=0.78m rotates at 8282 rpm (revolutions per minute). Then a brake (in blue) applies a perpendicular force of F=14NF=14N to the wheel at a radius of r=0.12r=0.12m. The coefficient of kinetic friction between the brake and the wheel is 0.280.28.

---How long does it take for the wheel to come to come to rest?

Hint: Calculate the torque applied by the brake. Then get the angular deceleration from this and the moment of inertia. Then obtain the time it takes the angular velocity to become zero.

Explanation / Answer

friction force, f = u F

f = 0.12 x 14 = 1.68 N m  

Applying Net torque = I alpha

(1.68) (0.12) = (2.46)(alpha)

alpha = 0.082 rad/s^2

w0 = 8282 rpm = 8282 x 2pi rad / 60s = 867.3 rad/s

wf = wi + alpha t

0 = 867.3 - 0.082 t

t = 10576.7 sec OR 176.3 min Or 2.94 hrs

{ if w = 82 rpm then w = 8.59 rad/s and t = 104.7 sec }

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