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

Let the horizontal position on the loop track correspond to $\theta = 0$. At this point, the ball is at a height $R$ from the lowest point.

By the principle of conservation of mechanical energy from the release point to the horizontal position:

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

For a solid spherical ball, the moment of inertia is $I = \dfrac{2}{5}mr^2$. For rolling without slipping, $v = \omega r$. Substituting these into the energy equation:

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

$mg(H - R) = \dfrac{7}{10}mv^2 \Rightarrow v^2 = \dfrac{10}{7}g(H - R)$

The normal force $N$ at the horizontal position provides the necessary centripetal force:

$N = \dfrac{mv^2}{R} = \dfrac{10}{7}mg \left(\dfrac{H - R}{R}\right)<br><br>Given$m = 70 $\text{ g} =$0.07 $\text{ kg},$g = 9.8 $\text{ m/s}^2$, $H = 60$\text{ cm} =$0.6$\text{ m}, and $R = 10$\text{ cm} =$0.1$\text{ m}:

$N =$\dfrac{10}{7} \times$0.07$\times$9.8$\times \left(\dfrac{0.6 - 0.1}{0.1}\right) =$0.1$\times$9.8$\times$5 = 4.9$\text{ N}

To find the frictional force $f$, we analyze the tangential motion. Let $f$ act in the upward direction. The downward tangential force is gravity $mg$. The equation of linear motion is:

$mg - f = ma_t$

The torque about the center of mass is provided by the frictional force, which causes angular deceleration:

$\tau = f r = I$\alpha = \left(\dfrac{2}{5}mr^2\right)\left(\dfrac{a_t}{r}\right) \Rightarrowf = \dfrac{2}{5}ma_t \Rightarrow ma_t = \dfrac{5}{2}f$<br><br>Substituting$ma_t$into the linear motion equation:<br><br>$mg - f = \dfrac{5}{2}f \Rightarrow \dfrac{7}{2}f = mg \Rightarrow f = \dfrac{2}{7}mg$<br><br>Calculating the magnitude of the frictional force:<br><br>$f = \dfrac{2}{7} \times0.07 \times9.8 = 0.02 \times9.8 = 0.196 \text{ N}$<br><br>Since the calculated value of$f$is positive, the assumed upward direction is correct. The friction acts upward to provide the torque required to slow down the ball's rotation as it ascends.<br><br>Answer:$4.9 \text{ N},0.196 \text{ N}upward