help with physics problem please!?

Question

An amusement park ride consists of a rotating vertical cylinder with rough canvas walls. The floor is initial about halfway up the cylinder wall. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down. The rider has a mass of 50 kg, the radius of the cylinder is 5m, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficent of static friction between the rider and the wall is .6

a. calculate the centripetal force on the rider when the cylinder is rotating and state what provides that force

b. Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force

c. at the same rotational speed, would a rider of twice the mass slide down the wall? explain

Answer 1

An object in motion continues in motion, in a straight line, unless acted on by some outside force. Newton's 1st.

When an object is traveling in a circle, the outside force that diverts it, from a straitght line to a circular path, is called centripetal force. The centripetal force continually bends the object's path so it moves in a circle and the force points toward the center of the circle. "Toward the center" is what centripetal means (from Latin or Greek or whatever).

The magnitude of centripetal force, Fc, needs to be (no more no less if the result is to be a circular path)
Fc = m*v^2/r.

So plug in the data of the problem and you've got the centripetal force for answer a. What provides it? The rider feels it on his/her back. ... Where the wall is.

b. If the rider is in equilibrium in terms of vertical motion, it has no net force in the vertical direction. The rider has weight
W = m*g
Therefore something must be counter-acting the weight. Perhaps friction? You got it.
Fnet = 0 = W - Ff
Solve for Ff.

C. If another rider has 2X the mass, his weight is 2X the other rider's weight.

Friction is calculated
Ff = mu*N
where mu is the coefficient of friction and N is the normal force against the surface. Look back at question a. N is a force against the wall. From Newton's 3rd, the wall exerts a force on the rider and that force is called centripetal force,
m*v^2/r.

So the normal force is also a function of m so if mass is 2X, the normal force is 2X. So the friction force upwards is 2X higher, and the weight is a force downwards, which is also 2X that for the other rider. Again, the net force ends up being zero. So the larger rider stays put too.