- #1

Nick89

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This is probably quite a silly question, someone asked it to me but I have to admit I didn't know what to answer lol...

The question is:

A rotating ball has a rotational kinetic energy of [itex]K_{rot} = 1/2 I \omega^2[/itex], right?

If you throw this ball (while rotating) up into the air, will it reach a smaller height than a ball that is not spinning?

In other words, should you account for this rotational energy if you set up the energy balance? For example:

[tex]\frac{1}{2}mv^2 = mgh[/tex] for a non-rotating ball (no friction) thrown into the air with speed v.

[tex]\frac{1}{2}mv^2 = mgh + \frac{1}{2} I \omega^2[/tex] for a rotating ball?

If we disregard friction with the air I think this does not make sense at all intuitively... Why should a rotating ball reach a smaller height?

Is it because the rotational kinetic energy is just another form of normal kinetic energy and is therefore somehow hidden inside the original equation, maybe?

I'm clueless on this lol... But it doesn't seem like a very hard question at all... I guess I just never studied this properly ^^