# Kinetic energy, angular momentum of a rotating body

We have a disk rotating about an arbitrary axis, and we can supposedly quantify the kinetic energy of such a disk by $$K = \frac 12Iw^2$$.

Now, it is also true that, as the disk is rotating, each elemental mass, $$dm_j$$, of the body possesses an instantaneous tangential/linear velocity, $$v_j$$. Does this not mean that the kinetic energy of the disk is also characterized by some form of $$K = \frac 12mv^2$$?

I am confused as to several things. Firstly, how do we know that we are not double counting the kinetic energy of the body, by supposing it is given by $$K = \frac 12Iw^2$$ and $$K = \frac 12mv^2$$? Second, if this were true, how can we find such a velocity, $$v$$, given that it varies throughout the body?

I also wish to understand how this could bode well with the angular momentum of the disk. I am certain, in this case, that the body would well need have some linear momentum, given that $$L$$ depends upon the existence of $$p$$, by $$L = r × p$$. So, I suppose that each elemental mass has a certain $$dp$$, that then "sums up" to the body's overall $$p$$... Yet somehow, this conclusion seems funny as well...

• I believe rotational kinetic energy should be $\frac{Iw^2}{2}$ – Bob D Jan 30 at 23:01
• @BobD: correct. – Gert Jan 30 at 23:04

The equation $$1/2I\omega^2$$ IS the result of adding all the mass elements with their respective $$v$$'s. There is no single $$v$$, but there is a single $$\omega$$. Same with the angular momentum, every $$dm$$ has a different $$p$$, but they end up cancelling each other so that the total momentum $$p=0$$, unless there is a translation of the center of mass.

• Thank you for your response. Two follow-up questions: (1) Could you explain as to how the momenta's cancel? (2) If we have a point mass rotating about an axis, then will it have a non-zero $p$? So, then, the "cancelling" effect is just a property of extended objects, correct? – Julia Kim Jan 30 at 23:31
• it is easier to see for a rotating disk, for each dm moving in one direction you will have a dm moving in the opposite direction that cancels it. For a general result, see en.wikipedia.org/wiki/… the answer to the second question is yes. Notice that in this case the CM of the point mass is moving – Wolphram jonny Jan 30 at 23:41
• As a last confusion to clear up (I promise)... my professor asked the following: if we have a rod, attached to a pivot at one end, and allowed to fall, how does $K_{rot} = Iw^2/2$ compare with $K_{cm} = Mv_{cm}^2/2$? (Ans provided: $K_{rot}$ > $K_{cm}$). I was wondering how, in this case, we recognized that the rod had two separate kinetic energies? By your answer, shouldn't $K_{rot}$ calculation have already accounted for $K_{cm}$? And, how do the momenta, $dm$ cancel here? – Julia Kim Jan 30 at 23:50
• it depends what axis of rotation you are considering. It would be $1/2I\omega^2$ if you consider an axis on the pivot, but you can also decompose that and have $1/2I_{CM}\omega^2$ around the center of mass plus the kinetic energy of the center of mass, which is what your professor wants\ – Wolphram jonny Jan 30 at 23:58
• in this case the momenta do not cancel because the CM is moving – Wolphram jonny Jan 31 at 0:02

If the disc is rotating about axis $$O$$ with angular speed $$\omega$$ then a small element of the disc of mass $$m_{\rm i}$$ travelling at a speed $$v_{\rm i}$$ at distance $$r_{\rm i}$$ from the axis of rotation has a kinetic energy $$\frac12 m_{\rm i} v^2_{\rm i}$$.

Now $$v_{\rm i}= r_{\rm i}\omega$$ so the kinetic energy of the element is $$\frac12 m_{\rm i} r^2_{\rm i}\omega^2$$

Summing the elements for the whole disc $$\sum\limits_{\rm i} \frac12 m_{\rm i} r^2_{\rm i}\omega^2= \frac 12 I_{\rm o} \omega^2$$ where $$I_{\rm o} =\sum\limits_{\rm i} m_{\rm i} r^2_{\rm i}$$

You can do a similar analysis for the angular momentum by summing the linear momentum $$m_{\rm i} v_{\rm i}$$ times the radius $$r_{\rm i}$$ over the whole disc.

Does this not mean that the kinetic energy of the disk is also characterized by some form of $$\frac{mv^2}{2}$$?

Yes

The relationship between angular velocity and tangential velocity is

$$ω=\frac{v}{r}$$

The moment of inertia of a disc of uniform thickness is

$$I=\frac{mr^2}{2}$$

The kinetic energy of the disc is

$$KE=\frac{Iω^2}{2}$$

Putting them together

$$KE=\frac{Iω^2}{2}=\frac {mr^2}{2}\frac{v^2}{r^2}=\frac{mv^2}{2}$$

Farcher’s answer, which I agree with, shows how summing up the kinetic energy contributions of the individual elements of mass add up to the total kinetic energy.

Hope this and Farcher's answer together helps

• What would $v$ be? – Wolphram jonny Jan 31 at 0:01
• v is the tangential velocity at the distance r. – Bob D Jan 31 at 0:31
• that is why your last equation does not make sense, the disk has many $r$'s, and so many $v$'s – Wolphram jonny Jan 31 at 0:34
• @Wolphramjonny Are you taking issue with my last equation? – Bob D Jan 31 at 0:36
• yes, I cannot make sense of it – Wolphram jonny Jan 31 at 0:40

Let me explain by a general case. Consider a rigid body with so many particles in it. Suppose all these particles move with certain velocities. When you sum up all the kinetic energy of all the particles,

       Total kinetic energy = (1/2)M(V^2) + (1/2)I(W^2)


where M is the total mass,V is the velocity of centre of mass,I is the momenta of inertia about an axis passing through centre of mass and W is the angular speed.

The combined effect of kinetic energy of all the particles is equivalent to a body both translating and rotating.