A solid, uniform cylinder with mass 8.15 kg and diameter 15.0 cm is spinning with angular velocity 240 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.330.
What must the applied normal force be to bring the cylinder to rest after it has turned through 5.20 revolutions?

Normal force will be equal to 8.945 N

Explanation:

We have given mass of the cylinder m = 8.15 kg

Diameter d = 15 cm

So radius

Initial angular velocity

As the cylinder finally comes to rest so final angular velocity

Before coming to rest cylinder covers a distance of

From third equation of motion

Coefficient of kinetic friction

Moment of inertia of the solid cylinder

We know that

So normal force will be equal to

v 154.9= 0.9 length length s little over 154.9cm

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s