Question

Calculate the kinetic energy for a rolling ball.
Translational kinetic energy + Rotational kinetic energy:

\[K _{trans}=\dfrac{1}{2}mv ^{2}\]
and
\[K _{rot}=\dfrac{1}{2}I\omega^{2}\]

Given infomation:
\[I=\dfrac{2}{3}mR ^{2}\]
and
\[m=0,5kg\qquad v=1,2m/s\]

My problem is, that i have no idea how to calculate the rotational kinetic energy since
\[\omega\] and R aren't given.

Answer 1

\[K _{rot}=\dfrac{1}{2}(\dfrac{2}{3}mR^{2})\omega ^{2}\]
Which gives me \[K _{rot}=\dfrac{R^{2}w^{2}}{6}\]

Answer 2

The ball is rolling so i know:
\[v=R\omega\]=>
\[v ^{2}=(R\omega) ^{2}\]

I could then insert this into
\[K _{rot}=\dfrac{R^{2}\omega ^{2}}{6}\]
So i'd have:
\[K _{rot}=\dfrac{v^{2}}{6}\]
And i know \[v=1,2m/s\]
So i'd get:
\[K _{rot}=\dfrac{(1,2m/s )^{2}}{6}=0,24J\]
Would that make sense ?

Answer 3

It appears to make sense. =)

Answer 4

If you calculate the translational kinetic energy you'd get \[K _{trnas}=\dfrac{1}{2}\cdot\dfrac{1}{2}kg\cdot (1,2m/s)^{2}=0,3J\]
And since the ball is rolling \[v=R\omega\] is true.
Which means that the rotational energy should more or less equal to the the translational energy.

Answer 5

Should be more or less equal to the translational energy*