Explanation:
It is given that,
Mass of the baseball, m = 144 g = 0.144 kg
Distance from axis of rotation, r = 82 cm = 0.82 m
Speed of the baseball, v = 82 mph = 36.65 m/s
(a) Let a is the ball's centripetal acceleration just before it is released. Its formula is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
[tex]a=\dfrac{(36.65)^2}{0.82}[/tex]
[tex]a=1638.07\ m/s^2[/tex]
[tex]a=1.6\times 10^3\ m/s^2[/tex]
(b) Let F is the magnitude of the net force that is acting on the ball just before it is released. It is given by :
F = ma
[tex]F=0.144\times 1638.07[/tex]
F = 235.88 N
[tex]F=2.3\times 10^2\ N[/tex]
Hence, this is the required solution.
