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

Answer:

the rotational kinetic energy of the wheel is 1730J

Explanation:

Hello!

The concept of rotational kinetic energy refers to the amount of energy a body has when it is rotating, so it is crucial to know the moment of inertia and its speed of rotation, now we follow the next steps to solve

1. We find the moment of inertia of a hollow cylinder by the following equation

$I=0.5M(Re^2-Ri^2)$

where

I= moment of inertia

m=mass=13kg

Re=external radius=0.34m

Ri=internal radius=0.29m

solving

$I=0.5M(Re^2-Ri^2)\\I=0.5(13)((0.34^2-0.29^2)=0.2047 kg . m^2$

N ow we use the equation to find the kinetic energy of a rotating body

$E=0.5 I W^2$

where

E= rotational kinetic energy

I= moment of inertia =0.2047kg . m^2

W=angular velocity=130 rad/s

Solving

$E=0.5(0.2047)(130)^2=1730J$

the rotational kinetic energy of the wheel is 1730J