# Does rotational kinetic energy affect the height of a ball thrown into the air?

• Nick89
In summary: So in summary, a rotating ball has a rotational kinetic energy of K_{rot} = 1/2 I \omega^2, right? Ignoring air resistance, a ball will reach the same height regardless of whether it also spins.
Nick89
Hi,

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 $K_{rot} = 1/2 I \omega^2$, 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:
$$\frac{1}{2}mv^2 = mgh$$ for a non-rotating ball (no friction) thrown into the air with speed v.

$$\frac{1}{2}mv^2 = mgh + \frac{1}{2} I \omega^2$$ 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 ^^

Well I may be missing something here, but for Newtonian mechanics translational and rotational motion can be thought of as mathematically separate when describing an objects motion. Therefor unless the rotational motion of the object affects directly the ability to throw an object (disregarding friction) there should be no reason for it the affect the translational parabolic motion of the object.
I think this would especially hold true for an object with regular geometry such as a sphere.

Someone with more expertise should be along shortly to clear this up for sure.

You're probably right, but how can this be understood with the energy balances I used as an example?

I think it might also have something to do with the cause of the rotation. How and why would a ball start to rotate when you throw it in the air? Without friction it certainly won't do that...
It will only rotate if you give it an initial rotation together with it's initial velocity, and then the two rotational kinetic energies cancel out maybe?

A ball whose center of mass is moving at some velocity will rise to the same height regardless of whether it also spins. (Ignore air resistance for simplicity.) Since gravity does not exert a torque about the center of mass, it cannot affect the rotational motion.

Energy is conserved, like always. It's just that the rotational KE doesn't change.

So the answer is that, to get the ball to start spinning you need to add the same amount of energy that the ball loses while spinning, because nothing can change the spinning motion of the ball once it is in flight?

To get the ball spinning you must exert a torque on it and give it rotational kinetic energy. Once the ball is in flight the only force acting on it is gravity--we ignore air resistance--then its rotational motion about its center of mass cannot change since gravity exerts no torque on the ball.

The force of Gravity acts only on a point mass at an objects geometric center of mass. Think about how two massive stars would orbit each other...

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