A small spherical ball of mass m rolls down a loop track. The ball...

Let the lowest point of the track be the reference level for potential energy. The height of point $A$ from the lowest point is $h = R + R\sin\theta$.

By conservation of mechanical energy, the loss in potential energy of the ball is equal to its gain in kinetic energy. For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energies:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

Substituting $I = \dfrac{2}{5}mr^2$ and $\omega = \dfrac{v}{r}$:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{v}{r}\right)^2 = \dfrac{7}{10}mv^2$

Equating the loss in potential energy to the kinetic energy:

$mg(H - h) = \dfrac{7}{10}mv^2$

$mg(H - R - R\sin\theta) = \dfrac{7}{10}mv^2$

$v^2 = \dfrac{10}{7}g(H - R - R\sin\theta)$

The radial acceleration $a_r$ of the centre of the ball is given by $\dfrac{v^2}{R}$ (assuming the radius of the ball $r \ll R$):

a_r = \dfrac{v^2}{R} = \dfrac{10}{7}g \left(\dfrac{H}{R} -1 - \sin\theta\right)<br><br>To find the tangential accelerationa_t$, we can differentiate the velocity equation with respect to time$t$. Differentiating$v^2$gives:<br><br>$2v $\dfrac{dv}{dt} = \dfrac{10}{7}$g \left(-R\cos\theta \dfrac{d\theta}{dt}\right)$<br><br>Since the tangential velocity is$v = R\dfrac{d\theta}{dt}$and tangential acceleration is$a_t = \dfrac{dv}{dt}$, we can substitute these into the equation:<br><br>$2v a_t = -\dfrac{10}{7}g\cos\theta \cdot v$<br><br>$a_t = -\dfrac{5}{7}g\cos\theta$<br><br>Thus, the radial and tangential accelerations are$\dfrac{10}{7}g \left(\dfrac{H}{R} -$1 - \sin$\theta\right)$and -$\dfrac{5}{7}$g\cos$\theta respectively.

Answer: Radial: \dfrac{10}{7}$g \left(\dfrac{H}{R} - 1 - \sin\theta\right)$, Tangential: $-\dfrac{5}{7}g\cos\theta$