Question

If a 0.15-kg baseball with a radius of 3.7 cm is thrown with a linear speed of 41 m/s and an angular speed of 45 rad/s , how much of its kinetic energy is translational energy? assume the ball is a uniform, solid sphere.

Answer 1

Total Kinetic energy of the ball is given by

`KE = (1)/(2)Iw^2 +(1)/(2) mv^2`

`KE = (1)/(2)((2)/(5)mR^2)w^2 +(1)/(2) mv^2`

`KE = (1)/(2)((2)/(5)*0.15*0.037^2)*45^2 +(1)/(2) *0.15*41^2`

`KE = 0.083 + 126`

`KE = 126.083 J`

Total translational KE of the ball is given by

`KE_T = (1)/(2)mv^2`

`KE_T = 126 J`

Fractional part of translational KE is given by

`fraction = (KE_T)/(KE)`

`fraction = (126)/(126.083)`

`fraction = 0.99`

So translational KE is 99% of total KE of the ball