In an amusement-park ride, riders stand with their backs against the wall of a spinning vertical cylinder. the floor falls away and the riders are held up by friction. the acceleration of gravity is 9.81 m/s 2 . if the radius of the cylinder is 3 m, find the minimum number of revolutions per minute necessary when the coefficient of static friction between a rider and the wall is 0.7. answer in units of rev/min

90.78 rev/min

Explanation:

In first place, we have to do the force balance to determine the minimum angular speed required to avoid slipping. The forces acting here are friction and the force due to circular movement, that is centripetal force. Then, we have:

μmg=mRω^2

ω=

Then, replacing the given values in the expression we have the following result:

ω=1.51 rev/s*60s=90.78 rev/min

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.