A solid sphere of mass m is released from rest at the rim of a...

Let $R$ be the radius of the hemispherical cup and $r$ be the radius of the solid sphere.

As the sphere rolls down from the rim to the bottom, its center of mass descends by a height $h = R - r$.

By conservation of mechanical energy, the loss in potential energy is equal to the gain in translational and rotational kinetic energy:

$mg(R - r) = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

For a solid sphere, the moment of inertia about its center of mass is $I = \dfrac{2}{5}mr^2$. For pure rolling, the velocity of the center of mass is $v = \omega r$.

Substituting these values into the energy equation:

$mg(R - r) = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{v}{r}\right)^2$

$mg(R - r) = \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 = \dfrac{7}{10}mv^2$

From this, we can find the centripetal term:

$\dfrac{mv^2}{R - r} = \dfrac{10}{7}mg$

At the lowest point, the center of mass of the sphere moves in a circular path of radius $R - r$. The net force towards the center of the path provides the centripetal force:

$N - mg = \dfrac{mv^2}{R - r}$

Substituting the value of the centripetal term:

$N - mg = \dfrac{10}{7}mg$

$N = mg + \dfrac{10}{7}mg = \dfrac{17mg}{7}$