Question

A piece of metal is pressed against the rim of a 2.2- kg, 24- cm-diameter grinding wheel that is turning at 2600 rpm. The metal has a coefficient of friction of 0.85 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20.0 s?

Answer 1

given that

mass m = 2.2 kg

diameter d = 2r = 0.24 m

radius r = 0.24/2 = 0.12 m

angular speed w = 2600 rpm = 2600*2*3.14/60 = 272.13 rad/s

coefficient of friction = = 0.85

time t = 20 s

from the relation

net torque = T = I*alpha

here alpha = angular acceleration = wf-wi/t

alpha = -272.13/20 = -13.61 rad/s^2

torque = T = -f*r

-f*r = 1/2*mr^2*alpha

f = 1/2*2.2*0.12*13.61

f = 1.8 N

f = *N = 1.8

N = 1.8/0.85 = 2.12 N