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Consider a mass $m$ connected to an oscillating pivot by a massless rod of fixed length $l$. The pivot is located at $(x_0(t), 0)$, where $x_0(t)=A_0\cos(\omega_0t)$. The $y$-axis is positive downwards. The angle between $y$ and the rod is $\theta(t)$. $\theta$ is the only generalized coordinate.

I have a hard time deriving the kinetic energy, which should be $T=\frac{1}{2}m\left(l^2\dot{\theta}^2+2l\dot{\theta}\dot{x_0}\cos\theta+\dot{x}_0^2\right)$.

Here's my thinking:

$$v_{rod}=l\dot{\theta}$$ $$v_{pivot}=\dot{r}_{pivot}$$

$r_{pivot}=x_0\implies\dot{r}_{pivot}=\dot{x}_0$ So $$v_{total}^2=\left(l\dot{\theta}+\dot{x}_0\right)^2$$

But something doesn't add up. I think that something is wrong with the velocity equation for the pivot, because if $\vec{r}_{pivot}=r\hat{r}$ then it somehow depends on $\theta$, but the angle of the moving pivot with the $y$-axis is always fixed. I'm quite confused. Would appreciate some help.

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  • $\begingroup$ "if $\vec{r}_{pivot}=r\hat{r}$" ... What are $r$ and $\hat r$ (this is the first mention of either of those symbols), and why should $\vec{r}_{pivot}$ equal their product? $\endgroup$ – David K Nov 13 '16 at 13:59
  • $\begingroup$ This is just a standard convention for polar coordinates $(r,\theta)$. $r$ is $\lVert \vec{r} \rVert$, $\hat{r}$ is the unit base vector, hence $\vec{r}=r\hat{r}$. $\endgroup$ – sequence Nov 13 '16 at 16:48
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For me, the first step is to make a good sketch, then write the position vector and check it against the sketch. Carefully checking it for various angles is so important. When $\vec{r}(t)$ is correct, then take the derivative.
$$\vec{r} = \vec{r}_{pivot} + \vec{r}_{rod} \\ \vec{r}(t) = x_0(t) \hat{i} + (L\sin\theta(t) \hat{i} + L\cos\theta(t) \hat{j}) \\ \vec{v}(t) = [\dot{x}_0(t) + L\dot{\theta}\cos\theta(t)] \hat{i} - L\dot{\theta}\sin\theta(t) \hat{j} \\ \vec{v}\cdot\vec{v} = (\dot{x}_0 + L\dot{\theta}\cos\theta)^2 + L^2\dot{\theta}^2\sin^2\theta \\ v^2 = \dot{x}_0^2 + 2L\dot{x}_0\dot{\theta}\cos\theta + L^2\dot{\theta}^2$$

Having a good expression for the position vector also comes in handy for writing the potential energy function, $PE=-\vec{F}_{gravity}\cdot\vec{r}$.

Clarifications:

The position of the mass $m$ is given by $\vec{r}$, which always points from the origin to the mass at the end of the pendulum rod. There is also a vector $\vec{r}_{pivot}$ that always points from the origin to the pivot point of the pendulum. The position of the pivot is given as always on the $x$-axis at a distance $x_0(t)$ from the origin, so $\vec{r}_{pivot}=x_0(t)\hat{i}$

The vector $\vec{r}_{rod}$ always points along the pendulum rod from the pivot point to the mass no matter what angle, $\theta$, the rod makes with the vertical. The components of $\vec{r}_{rod}$ are found by projecting $\vec{r}_{rod}$ onto the horizontal and vertical axes. The length of $\vec{r}_{rod}$ is always $L$. By our choice of $\theta$, the vertical projection is $L\cos\theta$ and the horizontal projection is $L\sin\theta$.

$\theta$ is the angle that the rod makes with the vertical line through the pivot point. This is an important point for two reasons. First, $\theta$ is often used as the polar coordinate about the origin, but here it is a generalized coordinate that indicates the rotation of the pendulum rod about its pivot point. Second, we have chosen $\theta$ so that it is zero when the pendulum is in equilibrium. We could have used a different angle, such as the angle $\phi$ between the rod and horizontal, say, but then we would have a different expression for $v^2$.

Thus, the vector from the origin to the mass is the sum of the vector from the origin to the pivot and the vector from the pivot to the mass.

As for the formula for the potential energy, let $\vec{F}$ be the gravitational force on mass $m$ at position $\vec{r}$ in a uniform gravitational field. We can calculate the force two ways. One is simply $\vec{F}=mg\hat{j}$, with the unit vector $\hat{j}$ pointing down along the positive $y$-axis. The other is $\vec{F}=-\vec{\nabla}{V(\vec{r})}$, where $V(\vec{r})$ is the potential energy function. Since the gravitational field is uniform, $\vec{F}$ has no $\vec{r}$ dependence. It is not difficult to let $V(\vec{r})=-\vec{F}\cdot\vec{r}$ and verify that $$-\vec{\nabla}{V(\vec{r})} = -\vec{\nabla}{(-\vec{F} \cdot \vec{r})}=(\vec{F} \cdot \vec{\nabla})\vec{r} = \vec{F} $$ Therefore $V(\vec{r}) = -\vec{F} \cdot \vec{r}$ satisfies our one requirement for it to be the potential energy function. When we have a complicated expression for $\vec{r}$ and we need the potential energy, this is a very handy formula. Two warnings, however: First, our force $\vec{F}$ must be the same everywhere, no $\vec{r}$ dependence, which is often the case, for this formula to be valid. Second, the potential energy is not a unique quantity, so we might not get the exact same function as someone else, the difference usually being an additive constant. We are free to add or subtract any constant we want to our potential energy.

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  • $\begingroup$ Can you please clarify why $\vec{r}=\vec{r}_{pivot}+\vec{r}_{rod}$? What does this $\vec{r}$ mean and how is it justified? Also, why is $PE=-\vec{F}_{gravity}\cdot\vec{r}$? $\endgroup$ – sequence Nov 14 '16 at 6:18
  • $\begingroup$ Thank you for your detailed reply. Why is $(\vec{F}\cdot\nabla)\vec{r}=\vec{F}$? $\endgroup$ – sequence Nov 15 '16 at 19:06
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    $\begingroup$ The simplest way to see it is to use the usual (x,y,z) coordinates: $(\vec{F}\cdot\nabla)\vec{r}=(F_x\frac{\partial}{\partial x} + F_y\frac{\partial}{\partial y} + F_z\frac{\partial}{\partial z})(x\hat{i} + y\hat{j} + z\hat{k})=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}=\vec{F}$ $\endgroup$ – LouisB Nov 15 '16 at 20:05
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The x coordinate of the mass: $$x(t)=x_0(t)+l\sin\theta$$ The y coordinate $$y(t)=l\cos \theta$$

So $$\dot x(t)=\dot x_0(t)+l\dot \theta\cos\theta $$ $$\dot y(t)=-l\dot \theta\sin \theta$$

$$T=\frac{1}{2}m(\dot x^2+\dot y^2)$$

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The definition of velocity has to allow an object (in this case the pivot) to move only along a straight line through the origin. In fact you have the correct magnitude for the velocity of the pivot; it is $\dot x_0$. Both the position and the velocity of the pivot do indeed have directions as well as magnitudes, but the directions are always horizontal.

To properly understand the other angles and directions in this problem, however, it pays to have a picture. The figure below is consistent with the way this problem apparently was intended to be solved.

pendulum at angle theta

In this figure the $x$ axis is directed to the right and the angle between the $y$ axis and the pendulum is measured counterclockwise from the $y$ axis. The pendulum is shown as a solid red line segment, with a large black dot at the end representing the mass.

At the instant of time represented in this figure, $t$, the pivot is at position $(x_0(t))\hat x$ relative to the origin. Now consider what happens during a very short period of time $\Delta t$ after time $t$. In that time the pendulum rotates through an angle $\Delta\theta$; let $\Delta t$ be small enough so that $\frac{\Delta\theta}{\Delta t} \approx \dot\theta$, and so that $\Delta\theta$ also is small enough to say that this rotation displaces the mass by a distance of approximately $\ell \Delta\theta$, approximately perpendicular to the pendulum rod. That displacement vector is represented by the black arrow whose tail is at the mass.

In addition to the displacement of the mass corresponding to the rotation of the pendulum, however, the pivot also moves a distance $\Delta x_0$ to the right, where $\frac{\Delta x_0}{\Delta t} \approx \dot x_0$. Since we have already supposed that the pendulum will have rotated to angle $\theta + \Delta\theta$ at time $t + \Delta t$, the motion of the pivot means the entire pendulum (including the mass at the end) moves a distance $\Delta x_0$ to the right. This movement is represented by the two horizontal black arrows, one along the $x$ axis and the other whose tail is at the head of the arrow we obtained from the pendulum's rotation.

The movement of the mass from time $t$ to time $t + \Delta t$ is the composition of these two motions, that is, the vector sum of the motion due to the rotation (a vector of length approximately $\ell \dot\theta \Delta t$ at an angle approximately $\theta$ counterclockwise from the $x$ axis) and the motion due to the moving pivot (a vector of length $\Delta x_0$ parallel to the $x$ axis). The vector sum is shown as a longer purple vector with its tail at the position of the mass.

You could also get the same vector sum by moving the entire pendulum $\Delta x_0$ to the right and then rotating it $\Delta \theta$.

As we let $\Delta t$ shrink toward zero, the approximations in the figure become better. In the limit, the average velocity of the mass during this timestep approaches the instantaneous velocity of the mass, which is the vector sum of two vectors, one of magnitude $\ell\dot\theta$ at angle $\theta$ with the $x$ axis, and one of magnitude $\dot x_0$ parallel to the $x$ axis.

When actually working a problem like this, I would usually skip all the "small timestep" and "approaching zero" business and simply write the velocity of the mass as the vector sum described at the end of the previous paragraph. All the $\Delta$s and dashed red lines are merely there to try to make the leap to that vector sum a little less mysterious.

The magnitude of the mass's velocity according to your computation, $v_\text{total} = \ell\dot\theta + \dot x_0$, is actually the correct value when $\theta = 0$, because in that case the two vectors in the vector sum are parallel. In every other case, however, the magnitude of the velocity will be less than $\ell\dot\theta + \dot x_0$. You can find the correct magnitude by applying the Law of Cosines to the triangle representing the summation of the motion vectors, keeping in mind that $\theta$ is the exterior angle of the sides of length $\ell\dot\theta$ and $\dot x_0$.

Alternatively, you can work out the $x$ and $y$ components of the rotational part of the velocity, $\vec v_\text{rod} = \ell\dot\theta(\cos\theta)\hat x - \ell\dot\theta(\sin\theta)\hat y$, add this componentwise to the velocity of the pivot, $\dot x_0 \hat x$, and then find the magnitude of the resulting vector from its $x$ and $y$ components.


There is actually a bit of ambiguity in the problem statement. We are not told which way the $x$ axis is directed (left or right), nor the direction in which the angle $\theta$ is measured. In the figure, I chose to orient $\hat x$ and $\theta$ in such a way that when $x_0$ and $\theta$ simultaneously increase, the mass moves farther than it would by making the same increase in either coordinate without changing the other. If I had reversed the direction of the $x$ axis or the direction in which $\theta$ is measured (but not both), an increase in $x_0$ and an increase in $\theta$ would have partially canceled each other; the kinetic energy then would have been $\frac12 m(\ell^2\dot\theta^2 - 2\ell\dot\theta\dot x_0\cos\theta + \dot x_0^2)$ instead of $\frac12 m(\ell^2\dot\theta^2 + 2\ell\dot\theta\dot x_0\cos\theta + \dot x_0^2)$. The difference is that one of the factors $\dot\theta$ or $\dot x_0$ in the middle term would have changed sign.

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