20 kg , 30 cm diameter disk is spinning at 300 rpm. How much friction force must be applied to rim to bring the disk to a halt in 3.0s?

Answer 1

The angular deceleration rate must be
alpha = (300*2 pi/60 rad/s)/3.0 s
= (10/3) pi = 10.47 rad/s^2

The required torque L is given by
L = I*alpha,
where I is the moment of inertia of the disc, (1/2) M R^2

Therefore

L = (1/2) M R^2 *10.47 rad/s^2
The units will be Newton-meters

Once you have determined L, use
F = L/R for the required friction force, in Newtons.

R is, of course, the radius, 0.15 m.

Answer 2

Thanks

Answer 3

Thanks for the help!

Answer 4

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