Problem 10.52

10.52. a)

First thing that we need to realize is that the center of mass of the sphere is a distance h+R above the ground. Therefore the initial energy of the system is given by
U_i=mg(h+r)
K_i=0
Now, if we are looking for the minimum height that is necessary for the ball to make it around the loop, then we need to look at the energy of the ball at the top of the loop, when the normal force is in the negative $\hat{y}$ direction. Here too, we have to take into account the non-zero radius of the sphere. The energy of the sphere at that point is given by
U_f = mg(2R-r)
$K_{f} = \frac{1}{2} m v_{f}^{2} + \frac{1}{2}I \omega^{2}$
we know that the moment of interia of a sphere is given by
$I = \frac{2}{5} mr^{2}$,
so plugging that into the equation for K_f, we find
$K_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}\frac{2}{5}mr^{2}\frac{v_{f}^{2}}{r^{2}}$
$K_{f} = \frac{7}{10}mv_{f}^{2}$.
So, equating the initial and final energies we find
$mg(h+r) = mg(2R-r) + \frac{7}{10}mv_{f}^{2}$
or simplifying yields
$g(h+2r) = 2Rg + \frac{7}{10}v_{f}^{2}$
Now, we need another equation for the final velocity. The condition for the minimum height is satisifed if the sphere is in perfect free fall at the top of the loop, or in other words, when the Normal force at the top of the loop is zero.
$\sum F_{y} = N - mg ~=~ -m\frac{v_{f}^{2}}{R-r}$
$mg=m\frac{v_{f}^{2}}{r}$
solving for the velocity squared, we find
v_f² = g(R-r).
Substituting this back in to the equation obtained from the conservation of energy, we find
$h_{min} = 2(R-r) + \frac{7}{10}(R-r)$
or $h_{min} = \frac{27}{10}(R-r)$

In part b, we are given the height and now we are looking for the components of the force at point P. From conservation of energy at P, we find
h = 3R
$mg(3R+r) = mgR + \frac{7}{10}mv_{P}^{2}$
solving for the velocity gives us
$v_{P}^{2} = \frac{10}{7}(2R+r)g$
Now, if we sum the force in the $\hat{y}$ direction, we find that the only force acting on the sphere in that direction is the force of gravity.
$\sum F_{y} = -mg$In the $\hat{x}$ direction, the only force acting is the Normal force, which we want to find in terms of the two radii and the mass of the sphere.
$\sum F_{x} = -N = m(\frac{-v_{P}^{2}}{(R-r)})$
$N = \frac{mv_{P}^{2}}{R-r}$
substituting in the value for v_P² that we found from conservation of energy, we find
$N = \frac{(10mg)(2R+r)}{7(R-r)}$