Answer:
v = 156.78 m/s
Explanation:
To find the tangential velocity of a point on a rim of the sphere, you use the following formula for the kinetic energy of the sphere:
[tex]K=\frac{1}{2}I\omega ^2\\[/tex] (1)
K: kinetic energy = 238 J
I: moment of inertia of the sphere
w: angular velocity of the sphere
Furthermore, you take into account the following equation for the moment of inertia of a sphere (with an axis crossing a diameter), and also the following equation for the angular speed:
[tex]I=\frac{2}{5}MR^2[/tex] (2)
M: mass of the sphere = 23.0 kg
R: radius of the sphere = 0.330 m
[tex]\omega=\frac{v}{R}[/tex] (3)
v: tangential velocity of a point on the rim of the sphere
You replace the equation (2) and (3) into the equation (1):
[tex]K=\frac{1}{2}*\frac{2}{5}MR^2*\frac{v}{R}=\frac{1}{5}MRv[/tex]
You solve the previous equation for v, and then replace the values of the parameters:
[tex]v=\frac{5K}{MR}=\frac{5(238J)}{(23.0kg)(0.330m)}=156.78\frac{m}{s}[/tex]
The tangential velocity of a point on the rim is 156.78 m/s
