A 9.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 12.0 m/s, determine (a) the translational kinetic energy of its center of gravity.

a translational=1/2 m v^2

b. rotational: I w^2=I (v^2/r)
where I= lookup. Is it a solid cylinder, or hollow?

how do you get the radius though?

To determine the translational kinetic energy of the center of gravity of a rolling cylinder, we need to know its mass and speed.

The formula for translational kinetic energy is given by:

KE = (1/2) * m * v^2

Where:
KE is the translational kinetic energy,
m is the mass of the object,
v is the speed of the object.

In this case, the mass of the cylinder is given as 9.0 kg, and the speed of the center of gravity is given as 12.0 m/s.

Substituting the given values into the formula, we get:

KE = (1/2) * 9.0 kg * (12.0 m/s)^2

Simplifying further:

KE = (1/2) * 9.0 kg * 144 m^2/s^2

Now, we can multiply the values inside the parentheses:

KE = 0.5 * 9.0 kg * 144 m^2/s^2

Multiplying further:

KE = 648 J

Therefore, the translational kinetic energy of the center of gravity of the rolling cylinder is 648 Joules (J).