The kinetic energy of a rolling ball can be calculated as the sum of its translational and rotational kinetic energies:
KE = 1/2 mv^2 + 1/2 Iω^2
where KE is the total kinetic energy, m is the mass of the ball, v is its velocity, I is the moment of inertia of the ball, and ω is its angular velocity.
To calculate the moment of inertia I of the ball, we can use the formula you provided:
I = (2/3)MR^2
where M is the mass of the ball and R is its radius.
Substituting the given values, we have:
M = 6.5 kg
R = 0.08 m
So, the moment of inertia of the ball is:
I = (2/3)(6.5 kg)(0.08 m)^2 ≈ 0.027 kg m^2
Now we can calculate the translational kinetic energy of the ball:
KE_trans = 1/2 mv^2 = 1/2 (6.5 kg)(5 m/s)^2 = 81.25 J
Next, we need to calculate the angular velocity ω of the ball. For a rolling ball without slipping, the tangential velocity v and the angular velocity ω are related by:
v = Rω
So, we have:
ω = v/R = 5 m/s / 0.08 m = 62.5 rad/s
Finally, we can calculate the rotational kinetic energy of the ball:
KE_rot = 1/2 Iω^2 = 1/2 (0.027 kg m^2)(62.5 rad/s)^2 ≈ 52 J
Therefore, the total kinetic energy of the rolling ball is:
KE = KE_trans + KE_rot = 81.25 J + 52 J ≈ 133.25 J
So, the kinetic energy of the rolling ball is approximately 133.25 J.