Answer
The magnitude of the net force that the track exerts on the ball is 0.38 N
Work Step by Step
We can express the angular velocity in units of rad/s
$\omega = (60~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 2\pi~rad/s$
The horizontal force $F_x$ from the track on the ball provides the centripetal force to keep the ball moving in a circle.
$F_x = m\omega^2~r$
$F_x = (0.030~kg)(2\pi)^2(0.20~m)$
$F_x = 0.237~N$
The vertical force $F_y$ from the track is equal to the ball's weight.
$F_y = mg$
$F_y = (0.030~kg)(9.80~m/s^2)$
$F_y = 0.294~N$
We can find the magnitude of the net force that the track exerts on the ball.
$F = \sqrt{(F_x)^2+(F_y)^2}$
$F = \sqrt{(0.237~N)^2+(0.294~N)^2}$
$F = 0.38~N$
The magnitude of the net force that the track exerts on the ball is 0.38 N
