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This might be a stupid question or has already been asked before, so please refer me to the answer if it has.

Let's say we have a rolling ball, which is purely rolling, and there is no friction. (Is that even possible?) What I'm having trouble with is that the total kinetic energy of the ball is the translational kinetic energy and the rotational kinetic energy. The question is why? Doesn't the rotational kinetic energy cause the translational movement? So aren't we adding the "same" energy twice? The cause for translational movement is from the rolling.

I will give another example: let;s say we have this object which we can make rotate really fast (above ground) and so it has a lot of rotational kinetic energy but no translational kinetic energy. We then put it on the ground, and it starts moving forward, it can't suddenly have more energy just because it is moving forward! So why is it that now when it is purely rolling we account for both?

Maybe the question my teacher gave is just not realistic? Maybe at every moment rotational energy will be lost to translational kinetic energy (and heat)? And that's how it works?

Anyways as you can see I am confused...

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  • $\begingroup$ basics.altervista.org/test/Physics/Me/… Here the derivation of the expression of the kinetic energy for discrete rigid bodies, to avoid the integrals that occur in continuous body (but almost nothing changes): you can write the kinetic energy taking every point as a "reference point", so that it makes no sense to me to define translational and rotational contribution of kinetic enery, unless you use the center of mass G as the reference point (and you get the sum of two distinct contributions in correspondence one with tranls and one with rot.) $\endgroup$
    – basics
    Feb 20 at 8:00
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    $\begingroup$ If there is no friction energy is not converted between rotation and translation $\endgroup$
    – my2cts
    Feb 20 at 12:26
  • $\begingroup$ Note that if there no friction, the first rolling ball might rotate, e.g. East to West and move North. This may provide intuition as to that the kinetic energies are additive and not doubly counting. $\endgroup$
    – TAR86
    Feb 20 at 17:50

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lets say we have this object which we can make roll really fast (above ground) and so it has a lot of rotational kinetic energy but no translational kinetic energy.

Great example.

We then put it on the ground

Let's first imagine that we put it on a super-slippery ice rink. The object continues to spin, but without friction, it can't turn that rotation into motion. It just sits there spinning. So the energy hasn't changed when it reached the ground.

But now let's imagine a normal ground with friction. The friction between the two creates a force on the object. This force has two effects:

  • it accelerates the object (linearly) in one direction
  • it creates a torque on the rotation of the object which slows down the spin

These two effects oppose each other. The translational energy goes up as the rotational energy goes down. (In addition, some energy will also be lost as the object skids on the ground).

maybe at every moment rotational energy will be lost to translational kinetic energy (and heat)? and that's how it works?

Correct.

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These are not silly questions and have puzzled many students in the past and probably will so in the future.
The idea of splitting the kinetic energy into two parts is to help with the analysis of rotating systems.

When a body is rolling (without slipping) every particle which make up the object have a velocity as shown by the blue arrows in the right-hand diagram, thus, in principle, one could evaluate the kinetic energy of each of the particles and sum them to find the total kinetic energy.

enter image description here

However this complex system is much easier to deal with by combining the contributions from rotation about the centre of mass and the translation of the centre of mass.
A proof of the total kinetic energy about any point being the sum of the rotational kinetic energy and the translational kinetic energy of the centre of mass is shown in many texts, for example, 20.4 Kinetic Energy of a System of Particles.

As to your second query, $\dots$ it can't suddenly have more energy just because it is moving forward!
Indeed it cannot.

When the object is dropped on the ground the speed of its centre of mass, $v$, is less than the speed of the object, $r\omega$, at the point of contact.

enter image description here

Because there is relative motion between the object and the ground a frictional force $F$ is exerted on the object.
The friction force accelerated the centre of mass to the right (increasing $v$) and provides a torque about the centre of mass which provides a counterclockwise angular acceleration (decreasing $\omega$) until such a time as the no slipping condition $v=r\omega$ is reached and then there is no frictional force.
During that time the total kinetic energy of the object decreases as some of its mechanical energy is dissipated as heat due to the frictional force acting on it.

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  • $\begingroup$ wait shouldn't friction be against motion? how is it that the friction is in the direction of motion. $\endgroup$
    – Beast
    Feb 20 at 12:02
  • $\begingroup$ Friction is reducing the relative motion between the object and the ground by making the object spin slower and the centre of mass of the object move faster. Rotational dynamics can often be counter-intuitive! $\endgroup$
    – Farcher
    Feb 20 at 13:01
  • $\begingroup$ @Beast Technically, F goes in BOTH directions - often ignored in examples like this, but the Earth (or whatever the ground plane is in the example) gets accelerated equally in the opposite direction - that is the basic principle behind ball-throwing machines etc. $\endgroup$
    – MikeB
    Feb 21 at 10:49
  • $\begingroup$ If the object was not spinning initially and was given a forward velocity the frictional force would be in the direction opposite to the forward velocity to decrease the forward velocity of the centre of mass and increase the angular velocity of the object. $\endgroup$
    – Farcher
    Feb 21 at 10:56
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Imagine a wheel with radius R that is rolling along in the vacuum of free space and advancing in a straight line by $2 \pi R$ for every complete revolution, just like a wheel rolling along a road, but there is no road. The wheel will continue to roll and translate at the same rate forever. The rolling motion is not causing the linear velocity and linear velocity is not causing the rotation. The two are independent and the imaginary road it is rolling along in space, is the perfect frictionless road. This should convince you friction is not required for constant rolling motion. Our wheel in space could be rotating backward or not rotating at all and this would have no influence on the linear motion of the wheel.

Your last observation appears to contradict the notion that angular momentum is conserved independently of linear momentum. When we drop the ball on the surface of the Earth, the ball imparts angular momentum to the Earth and angular momentum is conserved, but the increase in the angular velocity of the Earth is too small for us to measure. Its the same for the linear momentum. The initial linear momentum is zero and after the ball accelerates its forward momentum is matched by the rearward momentum imparted to the Earth, but again the change in linear momentum of the Earth is too small to actually measure. If we dropped the ball onto for example an efficient skateboard, we would see the rearward linear momentum imparted to the skateboard.

Now back to kinetic energy, which was what the focus of your question is about. Unlike momentum, angular kinetic energy and linear kinetic energy are not separately conserved. It is the total energy of a closed system that is conserved. In your example, angular kinetic energy is indeed converted to linear kinetic energy. The total energy is conserved, but due to the friction required to accelerate the ball linearly, some of the energy will have also been converted to heat but the total energy (angular + linear + heat + sound) is conserved. In the case of the rolling wheel in space, the linear kinetic energy is not just the angular kinetic energy looked at in a different way. The wheel will travel linearly through space just fine without rotation. We could have two wheels with the same linear velocity moving through space, one rotating and the other not. The rotating wheel will have a greater total energy. We can also have a wheel that is rotating in space, that is not moving linearly and it will have its own rotational energy independent of the fact it has no linear kinetic energy. The two forms of energy do not require each other to exist, but they can be converted one to the other.

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Lets say we have a rolling ball, which is purely rolling, and there is no friction (is that even possible lol)

Actually, this is not possible, no :) Rolling requires static friction with the surface. Otherwise your ball would simply slide. But anyways, that's not relevant for your question.

doesn't the rotational kinetic energy cause the translational movement? so aren't we adding the "same" energy twice? the cause for translational movement is from the rolling.

Those two types of energy are fully separable. Imagine a bowling ball sliding down an icy slope. The slope is so smooth that it doesn't "grap" the bowling ball at all, and thus it does not cause it to start spinning. The bowling ball will simply slide - so translate in a linear motion - without spinning in this ideal scenario. Clearly, it does have kinetic energy because it is moving with some speed $v$:

$$K=\frac12 mv^2.$$

Now imagine a flat, horizontal icy lake. You spin your bowling ball and place it on the smooth surface, just like you would place a yo-yo for instance. Because of the smoothness, the surface again does not "grap" the bowling ball, so it simply spins in its place - it rotates without translation. Clearly, it does have kinetic energy because of the spinning with some angular speed $\omega$:

$$K=\frac 12 I\omega^2.$$

My point is that these two types of kinetic energy are not just different versions of each other. They can be present independently of one another. In your scenario they just so happen to be present simultaneously. But still, they are not the same thing; they are still the kinetic energies associated with two distinct, independent types of motion. So, there is no problem in adding them together:

$$K=K_{translation}+K_{rotation}=\frac12m v^2+\frac12I\omega^2.$$

I will give another example: lets say we have this object which we can make roll really fast (above ground) and so it has a lot of rotational kinetic energy but no translational kinetic energy. We then put it on the ground and it starts moving forward, it cant suddenly have more energy just because it is moving forward! so why is it that now when is is purely rolling we account for both?

In the case of rolling, then the translational speed must perfectly match the rotation speed, so you might say that there is some link between the two types of motion. But you can think of that as just a coincidence. There is no link in general, and these two types of energy can be treated separately.

In the example that you describe here where you spin the ball and then place it on the ground, and you then see that the ball starts rolling on the ground, there, in fact, some of the rotational energy is being converted into translational energy. Meaning, the ball will in fact slow down its spinning speed in order for it to start moving. But still, the total kinetic energy is just the sum of both types at any moment.

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Aren't we adding the same energy twice ?

No, we are not - rotational kinetic energy is intentionally defined so that this does not happen.

Let's consider a small part $P$ of the ball. $P$ has a velocity $\vec v_P$ which we can divide into two parts - one part is $P$'s velocity relative to the centre of the ball, and the other is the velocity of the centre of the ball itself, which is the translational velocity of the ball as a whole. Similarly we can divide $P$'s kinetic energy into two parts - one part due to the translational velocity of the ball as a whole, which we will call $KE_1$, and the other part is whatever is left over due to $P$'s motion relative to the centre of the ball, which we will call $KE_2$. If $P$ has mass $m_P$ and the speed of the centre of the ball is $v$ then $KE_1 = \frac 1 2 m_Pv^2$

When we add up the kinetic energy of all of the parts of the ball, we get one term that is the sum of all the $KE_1$ parts, which add up to $\frac 1 2 mv^2$ where $m$ (which is the sum of all the $m_P$s) is the mass of the ball as a whole, and another term which is the sum of all the $KE_2$ parts. So

$\displaystyle KE_{ball} = \Sigma_P (KE_1) + \Sigma_P (KE_2) = \frac 1 2 mv^2 + \Sigma_P (KE_2)$

The first term, $\frac 1 2 mv^2$, is the translational kinetic energy of the ball. And the rotational kinetic energy of the ball is defined to be equal to the second term $\Sigma_P (KE_2)$, just so that we do not count the translational kinetic energy of the ball twice.

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