# Kinetic energy of a pendulum in motion

1. Dec 31, 2013

### CAF123

1. The problem statement, all variables and given/known data
A trolley consists of an enclosed carriage of mass M supported by 4 wheels each of mass m/4. A simple pendulum consists of a bob of mass μ on a light rod of length l supported at a pivot attached to the roof; its motion is confined to a plane. a) Write down expressions for the kinetic and potential energies in terms of an angle θ to the vertical and its time derivative.
b)Suppose now that the carriage is in motion on a horizontal surface. The plane of swing of the pendulum includes the direction of motion of the carriage). Let y(t) be the displacement of the carriage. Show that the total kinetic energy is $$T = \frac{1}{2}\left[\left(M+\frac{3}{2}m\right)\dot{y}^2 + \mu(\dot{y}^2 + \ell^2 \dot{\theta}^2) \pm 2\mu \ell \cos \theta \dot{y}\dot{\theta}\right],$$ where the sign of the last term is dependant on whether $\theta$ is measured in the same or opposite sense to y.

2. Relevant equations
Results all derived from usual definitions of kinetic and potential energy.

3. The attempt at a solution
In another problem with the surface at an angle $\phi$ (unrelated as far as the pendulum is concerned), the acceleration of the carriage (without the pendulum) was found to be $(M + \frac{3}{2}m)\ddot{x} = (M+m)g\sin\phi$. So the kinetic energy of the carriage alone is $T_c = \frac{1}{2} (M+\frac{3}{2} m) \dot{y}^2$. In part a), the kinetic energy of the pendulum is $T_p = \frac{1}{2} \mu \ell^2 \dot{\theta}^2$. So this accounts for two of the terms. I cannot get the last term and my reasoning for the second term is that the bob's added kinetic energy in the x direction is simply $1/2 \mu \dot{y}^2$ since the pendulum is attached to the carriage which has a constant horizontal velocity of $\dot{y}$. Is that reasonable? The vertical direction is $\ell \cos \theta$, but I am not sure how to get the coupled term $\dot{\theta}\dot{y}$.
Many thanks.

2. Dec 31, 2013

### voko

Let the velocity of the pendulum's attachment be $\vec v$. Let the velocity of the pendulum's bob be $\vec u$ with regard to the attachment. What is the total velocity of the bob?

3. Dec 31, 2013

### CAF123

Hi voko, happy new year.
The velocity of the bob wrt the attachment is $\vec{u}$. The velocity of the attachment wrt an inertial frame is $\vec{v}$. Hence $\vec{u} = \vec{V} - \vec{v}$ where $\vec{V}$ is the velocity of bob wrt inertial frame : $\vec{u} + \vec{v}$

4. Dec 31, 2013

### voko

Happy new year to you, too.

If you knew the magnitudes of $\vec v$ and $\vec u$ and the angle between them, what would the magnitude of their sum be?

5. Dec 31, 2013

### CAF123

$\vec{v}$ is equivalent to $\vec{\dot{y}}$ and so is horizontal. The velocity of the bob wrt the attachment is tangential to a segment of a circle with magnitude $\ell \dot{\theta}$. I see also that the angle between them is $\theta$.
Rewrite $\vec{u} = u\cos\theta \hat{x} + u\sin\theta \hat{y}$ and $\vec{v} = \dot{y}\hat{x}$ and so $\vec{u} + \vec{v} = (\ell \dot{\theta} \cos \theta + \dot{y} ) \hat{x} + \ell \dot{\theta} \sin \theta \hat{y}$ which means the magnitude of the total velocity is $\sqrt{\ell^2 \dot{\theta}^2 + \dot{y}^2}$
I can then obtain the correct answer, thanks!

I have a question about the acceleration of the carriage I posted in the OP. This was derived by considering energy conservation to derive an expression for $\dot{x}$ before using the chain rule to determine $a(x)$. The result was $(M+\frac{3}{2}m)\ddot{x} = (M+m)g\sin\phi$. Usually by considering point masses on a frictionless slope, the result is $m\ddot{x} = mg\sin \phi \Rightarrow \ddot{x} = g \sin \phi$ but in this case, the masses on each side of the equation are different. What is the physical reasoning for this? Given that the masses of the 4 wheels are m/4, the total mass is intuitively M + 4(m/4) = M + m, but instead there is this factor of 3/2.

Last edited: Dec 31, 2013
6. Dec 31, 2013

### voko

I do not see how the component-wise representation results in your final expression. Surely the square of the x-component is a bit more complex than $l^2 \dot \theta^2 \cos ^2 \theta + \dot y^2$.

7. Dec 31, 2013

### voko

Ah, so you figured that out. Good.

8. Dec 31, 2013

### voko

$M + m$ does appear in the RHS, where it is responsible for the force of gravity.

The LHS is trickier. That probably takes into account the moments of inertia of the wheels. I have not tried myself to obtain kinetic energy for the trolley, I just took your earlier result for granted :)

9. Dec 31, 2013

### CAF123

Indeed I did take into account the moment of inertia of the wheels, but I do not understand why this changes the mass on the LHS. Qualitatively, the moment of inertia of a body about an axis is a measure of how difficult it is to rotate a body about that axis. I guess I don't see why this would 'add' mass.

10. Dec 31, 2013

### voko

You could take a simpler example of a solid disk rolling horizontally without slipping. You should see how the "effective" mass in the KE expression is greater than the "true" mass.

11. Jan 1, 2014

### CAF123

Take the rotational axis to be through the centre of the disk perpendicular to its motion horizontally. Then the kinetic energy is T = (3/4)Mv2, if M is the 'true' mass of the disk.

So the kinetic energy of a rigid body may be found by summing the kinetic energy of the C.O.M (which in this case is intersected by the rotational axis) and the kinetic energy due to the rotational motion relative to the C.O.M. (Morin, P313)

If the mass distribution of a planar object is somewhat complicated, then the C.O.M may not be at all obvious. To find the total kinetic energy of the object, (suppose it is rotating and translating), I should first find the location of the C.O.M and then find the moment of inertia of the object about the 'correct' axis (the one with which the object is rotating about) intersecting the C.O.M?

12. Jan 1, 2014

### voko

Indeed, the full kinetic energy of a system is the kinetic energy of its COM plus the kinetic energy of the motion in the COM frame. More generally, you could use the following relationship: $T = T' + \boldsymbol V \cdot \boldsymbol P' + M \dfrac {V^2} {2}$, where $T$ is the kinetic energy in the laboratory frame, $\boldsymbol V$ is the velocity of some other frame w.r.t. the lab frame, $T'$ and $\boldsymbol P'$ are the kinetic energy and the total momentum, respectively, in the "other frame", and $M$ is the total mass of the system. In the COM, $\boldsymbol P' = 0$, which gives the simple rule mentioned above.

Note that the derivation of the kinetic energy of the pendulum on the trolley could have used that equation, where the dot-product term would be responsible for the cosine term.

13. Jan 1, 2014

### CAF123

Many thanks, I see this derivation in my notes and book too. Could you check this analysis? Applying this equation here, $$\frac{1}{2}Mv^2 = \frac{1}{2}M\dot{y}^2 + 4\left(\frac{1}{2}\left(\frac{m}{4}\right)\dot{y}^2\right) + \frac{1}{2}\mu \dot{y}^2$$ which is the translational kinetic energy in frame S.

Frame S' is a frame co-moving with the carriage with a constant velocity $\dot{y}$ so the only contribution from T' comes from the rotational motion of the bob and wheels. To describe the rotational motion I believe we choose the attachment as the origin of this frame for mathematical convenience, is that right? So, $$T' = 4 \left( \frac{1}{2} \left( \frac{ma^2}{8}\right) \left( \frac{\dot{y}}{a}\right)^2\right) + \frac{1}{2}\mu \ell^2 \dot{\theta}^2$$ where ma2/8 is the MoI of each wheel.

Finally, $\mathbf{V} = \dot{y}\hat{x}$ and $\mathbf{P'}$ is the momentum of the system in S'. The total momentum in the x direction is zero relative to S'because the frame co-moves with the carriage. $\mathbf{P'} = \mu \mathbf{V'} = \mu \ell \dot{\theta}$. Hence $\mathbf{V} \cdot \mathbf{P'} = \dot{y}\mu \ell \dot{\theta} \cos \theta$

I believe I have accounted for all the terms and I was wondering if I have the correct terms in each part of the expression you posted in #12.

Is it also correct to say that if the carriage was accelerating then I would not be transforming between inertial frames and so the method here and before would be invalid?

Many thanks.

14. Jan 1, 2014

### voko

This is somewhat cryptic. You have three major subsystems in the co-moving frame. One is the trolley itself. It is obviously at rest w.r.t. itself. Then four wheels: even though they are in motion, for every point on a wheel there is another one on it with the exactly opposite velocity, so their momentum in the carriage frame is zero. The only remaining part in the carriage frame is the pendulum, so you only have its momentum to account for. Which you did.

The derivation of the energy transformation equation does not assume inertial frames (check it out, it is very simple). So even if the trolley were accelerating, your result would still be correct.

15. Jan 2, 2014

### CAF123

In S, the wheels would have a non zero linear momentum $(=\frac{m}{4} \dot{y})$ per wheel but wouldn't the argument about opposite velocity vectors be valid in this frame too?

I have checked it out, but the very first line is 'Suppose frame S' moves at a constant velocity wrt frame S'. Also in the first method where I found the total velocity vector for the bob, to obtain $\vec{u} + \vec{v}$ I was doing a Galilean transformation between one frame and another attached to the pivot of the pendulum. By definition a Galilean transformation takes place between inertial frames, so that is why I wondered about the added constraint of a constant velocity S' relative to S.

16. Jan 2, 2014

### voko

No, it would not. Consider, for example, the wheel's centers.

But is that really used in the derivation? Assume $V$ is not constant. You still have $$T = \sum {m_i v_i^2 \over 2} = \sum {m_i (\boldsymbol V + \boldsymbol v_i') \cdot (\boldsymbol V + \boldsymbol v_i') \over 2} \\ = \sum {m_i v_i'^2 \over 2} + \sum m_i \boldsymbol V \cdot \boldsymbol v_i' + \sum {m_i V^2 \over 2} = T' + \boldsymbol V \cdot \boldsymbol P' + {M V^2 \over 2}$$

As above, this restriction is unnecessary here. If you had been considering forces in an accelerating frame, that would have been different.

17. Jan 2, 2014

### CAF123

Sorry, I don't think I phrased my question correctly. Each point on the circumference of a wheel has a corresponding point diametrically opposite it with an opposite velocity vector. So all velocity vectors tangentially to the wheel cancel in frame S. The $\frac{m}{4}\dot{y}$ is the linear momentum of the wheel center and is the only contribution of the momentum for each of the wheels in S. Do I have that correct?

Isn't the term $(\mathbf{V} + \mathbf{v_i'})$ (1) only valid for a relative constant velocity between the frames? Let the position vector in S be $\mathbf{r}$. The position vector in S' is given by $\mathbf{r_i} = \mathbf{r_i'} + \mathbf{V}t + \mathbf{q}$, where $\mathbf{q}$ is an offset. Hence $\mathbf{v_i} = \mathbf{v_i'} + \mathbf{V}$, which is (1). However, this assumes the displacement of S' from S is $\mathbf{V}t$ and if the frame S' was accelerating horizontally, then it's displacement would be $\mathbf{q} + \mathbf{V}t + \frac{1}{2}at^2$.

Did I make an error?

18. Jan 2, 2014

### voko

No, this not correct. It is obvious the the central point violates this. But a single central point cannot have the entire momentum of the wheel. Other points violate that, too. The lowest point of the wheel, under the condition of pure rolling, is stationary in S. The highest point is obviously not stationary, so their velocities do not cancel out.

I honestly do not see why $\boldsymbol V$ has to be constant. $\boldsymbol r = \boldsymbol R + \boldsymbol r'$, so $\boldsymbol v = \boldsymbol V + \boldsymbol v'$, and this is true without any further assumptions on $\boldsymbol V$.

19. Jan 2, 2014

### CAF123

Another two points are to the left and right of the wheel centre. To the right, the velocity vector points at an angle -45 degrees to the horizontal and to the left +45 degrees to the horizontal. (Assuming rotation clockwise). These do not cancel out. In fact, the velocity vectors nearer the bottom of the wheel will be smaller than those nearer the top so is there any points on the wheels for which the vectors cancel? The MoI of the wheels given was ma2/8 and on the assumption that this corresponds to a uniform mass density, then the C.O.M of each wheel lies in the centre of the wheel with momentum $\frac{m}{4}\dot{y}$.

This makes sense - In $\boldsymbol r = \boldsymbol R + \boldsymbol r'$, if $\boldsymbol R$ represents the relative position vector of S' and S, then the time derivative is $\boldsymbol V$, the relative velocity of the two frames. This has the form $\boldsymbol V = \boldsymbol u$ in the question and $\boldsymbol V = \boldsymbol u + \boldsymbol{a}t$ for a horizontally accelerating frame, which is ultimately the same result in my post.

I still do not see why an accelerating system would have the same K.E as the constant velocity one. $\boldsymbol V$ is different in each case. Edit: I misread one of your comments previously - the form of the expression for $T$ is the same, but the actual value after taken into consideration the form of $\boldsymbol V$ is different. This part is clear now.
Many thanks.

Last edited: Jan 2, 2014
20. Jan 2, 2014

### voko

We know that in the co-moving frame, velocities do cancel out: $\sum v_i = 0$. In the stationary frame, their velocities are $V + v_i$, so $\sum m_i (V + v_i) = \sum m_i V = MV$. So the total momentum of a wheel is just its translational velocity times its mass, as expected.