A 2.5kg solid homogenous ball is rolling without slipping at 3.0 meters per second to the right on a level surface. The diameter of the ball is 80 centimeters. What is the total kinetic energy of the ball? Can someone give me the equation for this?

To calculate the total kinetic energy of the ball, we need to consider both the linear kinetic energy and the rotational kinetic energy.

The linear kinetic energy (KE_linear) is given by the equation:

KE_linear = (1/2) * mass * velocity^2

In this case, the mass of the ball is given as 2.5 kg, and the velocity is given as 3.0 m/s.

So, plugging in the numbers into the equation:

KE_linear = (1/2) * 2.5 kg * (3.0 m/s)^2
= 11.25 Joules

Now, let's calculate the rotational kinetic energy (KE_rotational).

The moment of inertia (I) for a solid homogeneous sphere rotating about its diameter is given by the equation:

I = (2/5) * mass * radius^2

The radius of the ball can be determined by dividing the diameter by 2:

radius = 80 cm / 2
= 40 cm = 0.4 m

Plugging in the numbers:

I = (2/5) * 2.5 kg * (0.4 m)^2
= 0.16 kg·m^2

The rotational kinetic energy (KE_rotational) is given by the equation:

KE_rotational = (1/2) * I * omega^2

Here, we need to calculate the angular velocity (omega). In the case of a ball rolling without slipping, the linear velocity (v) and the angular velocity (omega) are related by the equation:

v = r * omega

Solving for omega:

omega = v / r
= 3.0 m/s / 0.4 m
= 7.5 rad/s

Plugging in the numbers:

KE_rotational = (1/2) * 0.16 kg·m^2 * (7.5 rad/s)^2
= 45 Joules

Finally, to get the total kinetic energy (KE_total), we need to add both the linear kinetic energy and the rotational kinetic energy:

KE_total = KE_linear + KE_rotational
= 11.25 Joules + 45 Joules
= 56.25 Joules

Therefore, the total kinetic energy of the ball is 56.25 Joules.