In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section.

(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R = 2.16 m. Assume R>>r, and the mass of the ball is 3.2 kg.)
(b) If the brass ball is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
*My work so far*
for finding I for a sphere I believe that I need to use I = (2/5)MR^2 which is I = 5.97 kg*m^2. Can I use V = sqrt(gR) to find the V...if so, V = 4.6 m/s. I am not sure what to do now. Can I use -Fs = m(-V^2/R)???

To stay on track at the top of the loop, MV^2/R must equal or exceed Mg there. Use conservation of enewrgy to relate the velocity there to the distance of the release point above the bottom of the loop.
The total kinetic energy when the velocity is V is
KE = (1/2) M V^2 + (1/2)(2/5)Mr^2*(V/r)^2 = (3/5) M V^2
If H is the initial height of the ball above the bottom of the loop, then at the top of the loop, 2R from the bottom,
MgH = (3/5) M V^2 + 2 MgR
V^2 = (5/3) gH -(10/3)gR > g
Solve for the minimum required H

To find the minimum height above the bottom of the track at which the ball must be released to ensure it does not leave the track at the top of the loop, we can use conservation of energy.

Let's consider the initial potential energy of the ball at height H above the bottom of the track. This potential energy is converted into kinetic energy as the ball reaches the top of the loop. At the top of the loop, the entire potential energy is converted into kinetic energy.

The total kinetic energy when the velocity is V is given by:
KE = (1/2)MV^2 + (1/2)(2/5)MR^2(V/r)^2 = (3/5)MV^2 (1)

Where M is the mass of the ball, V is its velocity, R is the radius of the loop, and r is the radius of the ball.

At the top of the loop, 2R from the bottom, the kinetic energy is equal to the potential energy:
MgH = (3/5)MV^2 + 2MgR (2)

Where g is the acceleration due to gravity.

To find the minimum required height H, we need to solve equation (2) for H. First, rearrange equation (2):
(3/5)MV^2 = MgH - 2MgR

Substituting V^2 from equation (1):
(5/3)gH - (10/3)gR > g

Solving for H:
H > (13/5)R

Therefore, the minimum height H above the bottom of the track at which the ball must be released is greater than (13/5) times the radius of the loop R.

For part (b) of the question, we need to find the magnitude of the horizontal component of the force acting on the ball at point Q when it is released from a height 6R above the bottom of the track.

To find the magnitude of the horizontal force, we can use Newton's second law, which states that the sum of all horizontal forces acting on an object is equal to its mass times its acceleration in the horizontal direction.

At point Q, the only horizontal force acting on the ball is the centripetal force required to keep it moving in a circular path. The centripetal force is given by:

Fc = M(V^2/R)

Where Fc is centripetal force, M is the mass of the ball, V is its velocity, and R is the radius of the loop.

Substituting V from the equation V = √(gR) (which you correctly calculated), we have:

Fc = M((gR)/R) = Mg

Therefore, the magnitude of the horizontal component of the force acting on the ball at point Q is equal to its weight, which is Mg.