Rolling Ball III B

Classical Mechanics Level 3

A solid ball with a moment of inertia of $I=\frac{2}{5}mr^2$ is initially moving with a velocity of $v_i$ and it is sliding with no rotational motion at $t=0$ on flat ground (which you can assume to be extended to infinity).

What is the ratio of the final total KE to the initial total KE?

If the ratio is $\frac{a}{b}$ where $a$ and $b$ are coprime integers,

find $a+b$ .

note: (assume the ball to be rigid and the ground too, so you don't have to consider rolling friction. KE is Kinetic Energy.)

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The answer is 12.

1 solution

Pierre Carrette
Aug 16, 2019

At the end, the ball will roll without sliding. This means that it would have transfer a part of its KE into rotational energy. How much? As much as needed for satisfying a no sliding condition, i.e. the contact point as no velocity. The no velocity contact condition is $v=r\omega$ , as if the ball was unrolling itself at the bottom. Rotational energy is $I\omega^2/2$ .

So, Final Total Energy (TEfinal) $=\frac{mv^2}{2}+\frac{I\omega^2}{2} = \frac{mr^2\omega^2}{2}(1+\frac{2}{5})$ with the first term being the final KEfinal. From total energy conservation, TEfinal=KEinitial. So, the ratio KEfinal/KEinital=KEfinal/TEfinal=1/(1+2/5)=5/7.

So a=5 and b=7. Then a+b=12.

I am a bit puzzled by this solution . . . Since the ball was sliding for some period of time, during which energy is lost due to kinetic friction. How was that included in the argument above? In the final part of your solution did you take into account that the velocity has changed?

Max Yuen - 1 year, 2 months ago

Rolling Ball III B

The answer is 12.

1 solution

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