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

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.