Question

Calculate the rotational kinetic energy (in J) of a 13.0 kg motorcycle wheel if its angular velocity is 130 rad/s and its inner radius is 0.290 m and outer radius 0.340 m. (Assume the wheel can be modeled as an annular cylinder rotating about the cylinder axis.)

Answer 1

The rotational kinetic energy of the motorcycle wheel is calculated using the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia for an annular cylinder and ω is the angular velocity. Substituting the given values results in a rotational kinetic energy of 41510.125 Joules.

Step-by-step explanation:

To calculate the rotational kinetic energy of the motorcycle wheel, we can use the formula for the kinetic energy of a rotating object, which is:

KErot = (1/2)Iω2

where I is the moment of inertia and ω is the angular velocity. For an annular cylinder (a hollow cylinder), the moment of inertia is given by:

I = (1/2)M(R12 + R22)

Substituting the given values into the formula:

I = (1/2)(13.0 kg)(0.290 m2 + 0.340 m2) = 2.45625 kg·m2

Now, using the given angular velocity (130 rad/s), we calculate the rotational kinetic energy:

KErot = (1/2)(2.45625 kg·m2)(130 rad/s)2 = 2.45625 kg·m2×16900 rad2/s2

KErot = 41510.125 J

Therefore, the rotational kinetic energy of the motorcycle wheel is 41510.125 Joules.