My question is motivated from a question from another user. You can see the configuration of the rotating system here: https://physics.stackexchange.com/q/143377/. enter image description here

I am not interested in all the complicated arguments of his question, but only of the expression for the total kinetic energy. My answer was that the rotational KE can be expressed as the addition of the KE of the center of mass plus the KE relative to the center of mass, which results in this expression:


(Notice that this result is independent of the sign of $\omega_2$).

But the original OP claims that the right expression is $$E_k=\frac{1}{2}md^2\omega_1^2+\frac{1}{2}mr^2(\omega_1-\omega_2)^2,$$ based on answers obtained on other forums (which I checked) and even the moderators in those forums seem to agree with it. The OP itself does not know enough physics to come up with its own answer, but still does not believe mine for the reasons given above.

So, my question is: I am missing something pretty obvious here? which of the expressions is the correct one (if any?) Thanks!


The answer depends on what the symbols mean. The question does not make it clear how the symbols are defined. The most confusing quantity is $\omega_2$. How is this defined? Is it the angular velocity of the disc relative to the fixed lab axes or relative to the axle about which it is rotating (where this axle itself will be rotating at $\omega_1$)? Also what is the sign convention for $\omega_1$? The problem states that $\omega_1$ is rotating clockwise, so is a positive $\omega_1$ supposed to mean clockwise rotation, or counterclockwise rotation? We will see that different answers to these questions give different expressions for the kinetic energy---one gives your answer and the other gives his answer. Thus I think the ultimate reason for disagreement is confusion about what the symbols mean.

Let us first solve the problem using one choice of meaning for the symbols, and obtain the expression for the kinetic energy, then we will see how the expression changes when we use different meanings for the symbols. I will use "your" definition, where positive $\omega_1$ and $\omega_2$ both indicate rotations in a counterclockwise direction, and the meaning of $\omega_2$ is the angular velocity of the disc about its center of mass relative to the fixed lab frame.

Then we know the center of mass of the disc has a speed $\omega_1 d$, and the disc rotates at an angular velocity $\omega_2$ with respect to the lab axes. Then we can obtain the kinetic energy using the following result: the kinetic energy of a rigid object is the sum of a translational piece $T_{\textrm{trans}}$ given by $T_{\textrm{trans}}=\dfrac{1}{2} m v_{cm}^2$ and a rotation piece $T_{\textrm{rot}}$ given by $T_{\textrm{rot}}=\dfrac{1}{2}I \omega^2$ where $m$ is the mass of the object, $v_{cm}$ is its center of mass speed, $\omega$ is its angular velocity about its center of mass, and $I$ is its moment of inertia about its center of mass (this is shown in the appendix). Using $v_{cm} = \omega_1 d$, $\omega = \omega_2$, and $I=\frac{1}{2}mr^2$, we find the total kinetic energy $T$ is given by $T=\dfrac{1}{2}m \omega_1^2 d^2 + \dfrac{1}{4} m \omega_2^2 r^2$. This is the expression you have.

Now let's see what happens if we instead express the kinetic energy not in terms of the absolute disc angular velocity $\omega_2$, but the angular velocity $\omega'_2$ of the disc relative to the axle. Then the angular velocity of the disc relative to the lab frame is the angular velocity of the disc relative to the axle plus the angular velocity of the axle relative to the disc. That is, $\omega_2 = \omega'_2 + \omega_1$. Plugging this in to our equation for kinetic energy, we get $T=\dfrac{1}{2}m \omega_1^2 d^2 + \dfrac{1}{4} m (\omega'_2 + \omega_1)^2 r^2$. Now suppose we want to express this kinetic energy in terms of $\omega'_1$, where $\omega'_1$ is positive for clockwise rotations and negative for counterclockwise rotations; that is, $\omega'_1= -\omega_1$. Then the expression for the kinetic energy becomes $T=\dfrac{1}{2}m \omega_1^2 d^2 + \dfrac{1}{4} m (\omega'_2-\omega'_1)^2 r^2$. This is now his expression. So it seems the only difference was what you meant by your variables


$\newcommand{\r}{\mathbf{r}}$ $\newcommand{\v}{\mathbf{v}}$ $\newcommand{\w}{\boldsymbol\omega}$ Here I will explain the only physics in the problem, which is that the kinetic energy of a rigid body can be decomposed into a part that gives the translation energy of the center of mass and a part for rotation about the center of mass. At some point in time this object is described by a spatial density profile $\rho(\r)$, and a spatial velocity profile $\mathbf{v}(\r)$. Since the body is rigid, the velocity profile $\mathbf{v}(\r)$ must have the form $\mathbf{v}(\r) = \mathbf{v}_0 + \w \times \r$, where $\w$ is the angular velocity. The kinetic energy is given by $T=\dfrac{1}{2}\int \rho(\r) v^2(\r) d\r$.

To see how to break this up into center of mass quantities, we will need to define the center of mass quantites. Let us define the total mass $m$ of the object to be $m=\int \rho(\r)d\r$, and let us define the center of mass position $\r_{cm}$ by $\r_{cm} = \dfrac{1}{m} \int \r \rho(\r) d \r$. Let us define the center of mass velocity $\v_{cm}$ to be the derivate of the center of mass position, which is given by the expression $\v_{cm} = \dfrac{1}{m} \int \v(\r) \rho(\r) d \r$.

First we will prove an intuitive result, which is that $\v(\r_{cm}) = \v_{cm}$. This is intuitive because the body is rigid so the center of mass has to move with the object. To prove this just see that $\begin{equation} \begin{aligned} \v_{cm} &= \dfrac{1}{m} \int \v(\r) \rho(\r) d \r \\ &= \dfrac{1}{m} \int (\mathbf{v}_0 + \w \times \r) \rho(\r) d \r \\ &= \dfrac{1}{m} \int \mathbf{v}_0 \rho(\r) d \r + \dfrac{1}{m} \int \w \times \r\rho(\r) d \r \\ &= \mathbf{v}_0 \dfrac{1}{m} \int \rho(\r) d \r + \w \times \dfrac{1}{m} \int \r\rho(\r) d \r \\ &= \mathbf{v}_0 + \w \times \r_{cm} \\ &= \v(\r_{cm}) \end{aligned} \end{equation}$

Now the first we want to do is re-express $\v$ in terms of $\r_{cm}$ and $\v_{cm}$. We have that $\begin{equation} \begin{aligned} \mathbf{v}(\r) &= \mathbf{v}_0 + \w \times \r \\ &= \mathbf{v}_0 + \w \times \r - \v(\r_{cm}) + \v(\r_{cm}) \\ &= \mathbf{v}_0 + \w \times \r - (\mathbf{v}_0 + \w \times \r_{cm}) + \v(\r_{cm}) \\ &= \w \times (\r - \r_{cm}) + \v(\r_{cm}) \\ &= \w \times (\r - \r_{cm}) + \v_{cm} \end{aligned} \end{equation}$

Now let us plug this form of $\v(\r)$ into our formula for the kinetic energy $T$: $\begin{equation} \begin{aligned} 2T&= \int \rho(\r) v^2(\r) d\r \\ &= \int \rho(\r) (\w \times (\r - \r_{cm}) + \v_{cm})^2 d\r \\ &= \int \rho(\r) ((\w \times (\r - \r_{cm}))^2 + 2 (\w \times (\r - \r_{cm}))\cdot\v_{cm} + v^2_{cm}) d\r \\ &= \int \rho(\r) (\w \times (\r - \r_{cm}))^2 d\r + \int \rho(\r) 2 (\w \times (\r - \r_{cm}))\cdot\v_{cm} d\r + \int \rho(\r) v^2_{cm} d\r \end{aligned} \end{equation}$

Let's look at this last line term by term. Let's start with the first term. $\begin{equation} \begin{aligned} \int \rho(\r) (\w \times (\r - \r_{cm}))^2 d\r &= \omega^2 \int \rho(\r) (\hat{\w} \times (\r - \r_{cm}))^2 d\r \end{aligned} \end{equation}$ Now the integral on the right had side is the weighted sum of the square of the component perpendicular to $\w$ of displacement from the center of mass. That is, it is the scalar moment of inertia for rotations about the axis parallel to $\w$ passing through the center of mass. Denoting this moment of inertia $I$, we find the first term is $I \omega^2$.

Moving on to the second term for the kinetic energy, we see $\begin{equation} \begin{aligned} \int \rho(\r) 2 (\w \times (\r - \r_{cm}))\cdot\v_{cm} d\r &= 2 \v_{cm} \cdot \left( \w \times \int \rho(\r) (\r - \r_{cm}) d\r \right) \\ &= 2 \v_{cm} \cdot \left(\w \times \left( \int \rho(\r) \r d\r - \int \rho(\r)\r_{cm} d\r \right)\right) \\ &=2 \v_{cm} \cdot \left(\w \times (m \r_{cm} - m \r_{cm}) \right) \\ &=0 \end{aligned} \end{equation}$

Finally let's look at the third term. The third term is

$\begin{equation} \begin{aligned} \int \rho(\r) v^2_{cm} d\r &= v^2_{cm} \int \rho(\r) d\r \\ &= m v^2_{cm} \end{aligned} \end{equation}$

Putting the three terms back together, we find that $2T = I \omega^2 + m v_{cm}^2$, or $T=\dfrac{1}{2}I \omega^2 +\dfrac{1}{2} m v_{cm}^2$ where $I$ is the moment of inertia about the center of mass. This is what we needed to show.

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  • $\begingroup$ I have a doubt in the fourth paragraph of your answer: How do you even define the angular velocity $\omega'_1$ of the disk about the axle since it's apparent that every point on the disk has a different $\frac{v\bot}{r}$ ratio with respect to the axle compared to a rigid body which has the same ratio throughout its body ? However, 'angular velocity' is defined only for 'rigid' bodies about an axle and in this case, the rotating disk does not satisfy that condition . $\endgroup$ – Gaurav Nov 3 '14 at 12:38
  • $\begingroup$ @Simha I am not sure I understand your question. Is your quesiton "How can a disk rotating about a moving axle have a well difined angular velocity?" $\endgroup$ – Brian Moths Nov 3 '14 at 17:54
  • $\begingroup$ Yes, that's the question. $\endgroup$ – Gaurav Nov 3 '14 at 17:56
  • $\begingroup$ Just view the motion in the frame whose origin is moving with the axle. Then there will be a well-defined $v_1/r$. $\endgroup$ – Brian Moths Nov 3 '14 at 17:59
  • $\begingroup$ Now there is an additional quesiton of should the frame of the axle also rotate with the axle or should the coordinate axes be always aligned with the lab frame. The difference between these two choices is what makes up my answer to the OP's question. $\endgroup$ – Brian Moths Nov 3 '14 at 18:00

Take the reference frame as centered in the fixed axis. The $R$ that connects the origin to the centre of the spinning disk forms an angle $\phi$ with the horizontal. Now, inside the disk of radius $r$, the angle of a certain point mass is given by the angle it forms inside the spinning circle, which we'll call $\theta$.

Now take as generalised coordinates those angles and write the radius vector as a decomposition in $\mathcal{X}$ and $\mathcal{Y}$:

$$\mathcal{X}: r_x = R \cos \phi + r\cos \theta \\ \mathcal{Y}: r_y = R \sin \phi - r \sin \theta ,$$

which means that

$$\dot{r}_x = -R \dot{\phi} \sin \phi - r \dot{\theta} \sin \theta \\ \dot{r}_y = R \dot{\phi} \cos \phi - r \dot{\theta} \cos \theta .$$

Summing up the squares,

$$ \dot{r}^2_y + \dot{r}^2_x = R^2 \dot{\phi}^2 \cos^2 \phi + r^2 \dot{\theta}^2 \cos^2 \theta - 2 rR\dot{\theta} \dot{\phi} \cos \theta \cos \phi \\ + R^2 \dot{\phi}^2 \sin^2 \phi + r^2\dot{\theta}^2 \sin^2 \theta + 2 rR\dot{\theta} \dot{\phi} \sin \theta \sin \phi \\ = R^2 \dot{\phi}^2 + r^2\dot{\theta}^2 - 2Rr\dot{\phi} \dot{\theta} \cos(\phi + \theta) . $$

The kinetic energy will then be

$$T = \frac{m}{2}[ R^2 \dot{\phi}^2 + r^2\dot{\theta}^2 - 2Rr\dot{\phi} \dot{\theta} \cos(\phi + \theta)] .$$

My answer is coming out pretty different from both the ones you're aiming for, and I'm also using generalized coordinates and Lagrangian mechanics instead of the Newtonian one (so I'm not integrating anything, really). I think this extra term that appeared term does have a meaning, because if you consider the disk spinning in one direction and being rotated in another, then the kinetic energy should decrease in some specific configuration. If you only keep the squared angular velocity terms, then this is impossible.

P.S.: I can surely be doing something VERY wrong here.

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  • $\begingroup$ I could not follow your derivation. First, I cannot imagine what is $\theta$ that appears subtracted in the definitions of $r_x$ and $r_y$. Second, the derivatives are wrong, you forgot to derive the sines and cosines, but the most indicative issue about something being wrong is that you start your derivation for a single point on the disk, and get that the result is independent of the angles, which do not make sense at all (each point should have different kinetic energy). Then, you integrated in your last step assuming is a ring, because you multiplied by mass assuming r as constant? $\endgroup$ – Wolphram jonny Oct 29 '14 at 22:42
  • $\begingroup$ Last, if you do the derivatives right, the last two terms are not equal and do not cancel, so you get a dependence on position (angle) $\endgroup$ – Wolphram jonny Oct 29 '14 at 22:58
  • $\begingroup$ You're quite right. I'll correct my erros now. $\endgroup$ – QuantumBrick Oct 30 '14 at 2:21
  • $\begingroup$ @julianfernandez I'm not really integrating anything, I'm just using Lagrangian mechanics. The angles are taken as generalized coordinates, and I'm only using a single point to write them. $\endgroup$ – QuantumBrick Oct 30 '14 at 2:48
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    $\begingroup$ That's quite offensive. I'll step off of answering and flag your post. Besides, the Lagrangian always needs to be a function of the generalized coordinates, even if I'm only writing the kinetic part. I don't believe I was rude to you at any time. $\endgroup$ – QuantumBrick Oct 30 '14 at 21:15

This is simple, take the linear and angular velocity of the center of mass (point B) and combined them with the inertial properties

  1. Linear Velocity of B : $\vec{v}_B = (0,d \omega_1,0)$
  2. Angular Velocity of B : $\omega_B = (0,0,\omega_2-\omega_1)$
  3. Mass of disk $m$
  4. Mass Moment of Inertia of Disk $I_{zz} = \frac{m}{2} r^2$, where $r$ is the radius of the disk.
  5. Kinetic energy is $T = \frac{1}{2} m (\vec{v}_B \cdot \vec{v}_B) + \frac{1}{2} I_{zz} ( \vec{\omega}_C \cdot \vec{\omega}_C) = \frac{m}{2} d^2 \omega_1^2 + \frac{I_{zz}}{2} (\omega_2-\omega_1)^2 $

$$ \boxed{ T = \frac{m}{2} \left( d^2 \omega_1^2 + \frac{r^2}{2} (\omega_2-\omega_1)^2 \right) } $$


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The key here is the phrase in the original question:

$\omega_2$ is relative to the black arm

This means that in the lab frame of reference, the disk is spinning with angular velocity $\omega_2-\omega_1$. And with that piece of knowledge, the answer follows what you already know: the total energy is the energy of rotation of the center of mass plus the energy of rotation of the disk ($\frac12 I \omega^2$, where $I=\frac12 m r^2$ for a disk), that is

$$\begin{align}E_{com} &= \frac12 m (\omega_1 d)^2\\ E_{disk} &= \frac14 m (\omega_2 - \omega_1)^2 r^2\\ E_{total}&=\frac12 m (\omega_1 d)^2 + \frac14 m (\omega_2 - \omega_1)^2 r^2 \end{align}$$

The only "thing you were missing" was the definition of $\omega_2$. Your expression would have been correct if the velocity was defined in the lab frame. Note that if the velocities had not been defined (again, by the drawing in the original question) as being in opposite directions, then you would have used $\omega_1 + \omega_2$ in the second term. Note that since you square the quantity, it doesn't matter what the order of the subtraction is.

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  • $\begingroup$ @NowIGetToLearnWhatAHeadIs - you are right, I was sloppy. I have fixed / clarified, I think. Thanks. $\endgroup$ – Floris Oct 31 '14 at 20:10
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    $\begingroup$ you are right! (as always) $\endgroup$ – Wolphram jonny Nov 1 '14 at 1:21

enter image description here

For a ring:

$Ke = \frac{1}{2}m ( \int_0^{2\pi} (Rw_1-rw_1cos(\theta)+rw_2cos(\theta))² +(rw_2abs(sin(\theta))² d\theta) $

$Ke = \frac{1}{2}m \int_0^{2\pi} R²w_1²+r²w_1²cos²(\theta)+r²w_2²cos²(\theta)-2Rw_1²rcos(\theta)+2Rw_1rw_2cos(\theta)) -2rw1cos(\theta)rw_2cos(\theta)+r²w_2²sin²(\theta) d\theta$

$Ke = \frac{1}{2}m (R²w_1²+\frac{1}{2}r²w_2²+\frac{1}{2}r²w_2²-r²w_1w_2) $

$Ke = \frac{1}{2}mR²w_1² + \frac{1}{2}mr²(w1-w2)² $


$abs(W_1) > abs(W_2)$

$w_2<0$ and $w_1>0$

The result take in account the sign of $w_2$, kinetics energy increase if $abs(w_2)$ decrease.

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  • 1
    $\begingroup$ Could you avoid making minor edits to your answer please? $\endgroup$ – David Z Nov 6 '14 at 5:36
  • $\begingroup$ Ok, it's because I forgot to say it's only for a ring not a disk. It's a problem to edit for minor edits ? $\endgroup$ – Sx7 Nov 6 '14 at 6:37

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