Question

A bowling ball has a mass of 5 kg, a moment of inertia of 0.0648 kg·m2, and a radius of 0.18m. It rolls along the plane without slipping a linear speed of 4.8 m/s.What is the total kinetic energy of the rolling ball?

Answer 1

Answer:

$K.E=80.6388J$

Explanation:

Given data

Mass m=5 kg

Moment of Inertia I=0.0648 kg.m²

Radius r=0.18 m

Linear Speed v=4.8 m/s

To find

Total Kinetic Energy K.E

Solution

The total kinetic energy K.E can be found as below

$K.E=(1/2)Iw^{2}+(1/2)mv^{2}$

First we need to find the angular speed ω

So

$w=(v/r)\\w=(4.8/0.18)\\w=26.66rad/s$

Substitute the angular speed and given values to find total kinetic energy

So

$K.E=(1/2)Iw^{2}+(1/2)mv^{2}\\K.E=(1/2)(0.0648kg.m^{2} )(26.666rad/s)^{2}+(1/2)(5kg)(4.8m/s)^{2}\\K.E=80.6388J$