Question

Calculate the rotational kinetic energy (in J) of a 14.0 kg motorcycle wheel if its angular velocity is 130 rad/s and its inner radius is 0.270 m and outer radius 0.300 m. (Assume the wheel can be modeled as an annular cylinder rotating about the cylinder axis.)

Answer 1

The rotational kinetic energy is 19271.07 Joules.

Step-by-step explanation:

Given that,

Mass of the motorcycle, m = 14 kg

Angular velocity of the wheel,
`\omega=130\ rad/s`

Inner radius is 0.270 m,
`r_1=0.27\ m`

Outer radius,
`r_2=0.3\ m`

The rotational kinetic energy of the wheel is given by :

`K=(1)/(2)I\omega^2`

I is moment of inertia of the wheel,
`I=m(r_1^2+r_2^2)`

So,

`K=(1)/(2)m(r_1^2+r_2^2)\omega^2\\\\K=(1)/(2)* 14* ((0.27)^2+(0.3)^2)* (130)^2\\\\K=19271.07\ J`

So, the rotational kinetic energy is 19271.07 Joules.