Re: How do I find the potential for this unusual pendulum?

[Dustinsfl]

An unusual pendulum is made by fixing a string to a horizontal cylinder of radius $R$, wrapping the string several times around the cylinder, and then tying a mass $m$ to the loose end.
In equilibrium the mass hangs a distance $l_0$ vertically below the edge of the cylinder.
Find the potential energy if the pendulum has swung to an angle $\phi$ from the vertical.

The definition for potential energy is
$$
U(\mathbf{r}) = -W(\mathbf{r}_0\to\mathbf{r}) = -\int_{\mathbf{r}_0}^{\mathbf{r}}\mathbf{F}(\mathbf{r}')\cdot d\mathbf{r}'
$$
How do I find the potential for this unusual pendulum?

 

[Ackbach]

Re: potential energy

Actually, $W=-\Delta U$. Absolute potential energy has no physical meaning, only a change in potential energy. The only forces on the mass are gravity and the tension in the string. The trick with this problem is that when the mass swings, the cylinder is either taking up more string, or releasing string. So the length of the pendulum is changing.

Let us define $y$ positive up, $x$ positive to the right, and $\phi$ positive in the counter-clockwise direction from the $+x$ axis, as usual. Let us assume that the cylinder's center is at the origin, and that the string is hanging in equilibrium from a point of tangency that is on the positive $x$ axis.

Several key concepts here: the potential energy due to the gravitational force is given by $mgh$, where $h$ is the height above some zero point for the energy. When the mass swings, the string is always tangent to the cylinder, and hence the angle $\phi$ that the pendulum has swung is also the angle $\phi$ which the string has either wrapped more around the cylinder or less. Therefore, the arc length of additional string taken up is given by $s=\phi R$. So the length of the pendulum is always $l_{0}+s$. If we can find the point of tangency as a function of $\phi$, and then go $l_{0}+s$ along the string, we'd arrive at the mass.

Then the vector to the point of tangency $P_{t}$ we can write as
$$\mathbf{P}_{t}=R \, \langle \cos( \phi), \sin( \phi) \rangle.$$
From the point of tangency, the mass is then located a distance $l_{0}+ \phi R$ away, at an angle of $\phi$ from the vertical. Hence, the vector $\mathbf{r}$ from the point of tangency to the mass is given by
$$\mathbf{r}=(l_{0}+ \phi R) \, \langle \sin( \phi), -\cos( \phi)\rangle.$$
The minus sign on the $y$ component reflects the fact that we've defined $y$ positive down. Hence, for any angle $\phi$, the $y$-coordinate of the mass is given by the sum of the $y$ components of these two vectors, or
$$y=R \sin(\phi)-(l_{0}+ \phi R) \cos( \phi).$$

We can just define our zero point to be when $\phi=0$, in which case $y=-l_{0}$. Therefore, the height $h$ in the formula $U=mgh$ is given by $U=mg(l_{0}+y)$, since
$l_{0}>0$ and $y<0$. Therefore, the potential energy is
$$U=mg\left[l_{0}+R \sin( \phi)-(l_{0}+ \phi R) \cos( \phi)\right].$$