How can I find the height for a looping pendulum? [closed]

Question

I have a ball attached to a string, hanging on a rod placed $L$ meters above the floor. I release the ball and string at a 90 degree angle and let it loop around another horizontal bar after it reaches the bottom. How can I determine the height of the middle rod (distance between top rod and middle rod) where the ball will loop around the rod with a tension of 0? I know it comes out to $h=(3/5)L$ but I'm not really sure where that comes from.

I tried using the work-kinetic energy theorem, but I may be interpreting the energies wrong.

Trip down: $$ \begin{aligned} K_i + U_{gi} &= K_f + U_{gf}\\ 0 + U_{gi} &= K_f + 0\\ U_{gi} &= K_f\\ mgL &= \frac{1}{2}mv_f^2\\ L &= \frac{v_f^2}{2g} \end{aligned} $$

Trip up: $$ \begin{aligned} K_i + U_{gi} &= K_f + U_{gf}\\ K_i + 0 &= K_f + U_{gf}\\ \frac{1}{2}mv_i^2 &= \frac{1}{2}mv_f^2 + mgh\\ h &= (v_i^2-v_f^2)/2g \end{aligned} $$

I'm not really sure how to get $L$ and $h$ in the same equation.

Answer 1

You're on the right track here. I'm going to use a bit of different notation, because "initial" and "final" aren't unique if you're breaking the path up into two segments like you are. So let's define the following:

• Point A: release point for the pendulum, with the string horizontal
• Point B: bottom of the swing
• Point C: top of the swing, after the string has wrapped around the peg

Your first step was to calculate the velocity at the bottom of the swing, i.e., point B. Your derivation for this was completely correct; I've just copied it over below with the new notation, using "A" and "B" instead of initial and final: \begin{aligned} K_A + U_{gA} &= K_B + U_{gB}\\ 0 + U_{gA} &= K_B + 0\\ U_{gA} &= K_B\\ mgL &= \frac{1}{2}mv_B^2\\ L &= \frac{v_B^2}{2g} \end{aligned} For the subsequent trip up (from point "B" to point "C"), you have to be a little more careful. The final height of the bob will not be $h$, but will instead be $2(L - h)$; this is because the bob is swinging in a circle of radius $(L- h)$ now. So you have \begin{aligned} K_B + U_{gB} &= K_C + U_{gC}\\ K_B + 0 &= K_C + U_{gC}\\ \frac{1}{2}mv_B^2 &= \frac{1}{2}mv_C^2 + mg(2(l-h))\\ 2(L - h) &= (v_B^2-v_C^2)/2g \end{aligned} You already found the relationship between $v_B^2$ and $L$ in the previous set of equations, though; so you can just plug that in to get $$ 2L - 2 h = L - \frac{v_C^2}{2g}. $$

The last piece you'll need is to figure out what $v_C$ needs to be in order for the tension to go to zero at the top of the loop. I would be very surprised if you hadn't done this question already, either in your textbook or on a homework assignment, so you might want to go back and take a look at that. But if not, here's how to get started on it: The bob is moving in a circle, so the net force on it must be acting inwards and have a particular magnitude. Now, what forces are be acting on the bob at this point in time? Which directions do they have?

As an aside: we didn't really have to break up the pendulum's trip into two segments this way; I did it because you seemed to be thinking about it that way. But it would be equally valid to treat Point A as the "initial" position, Point C as the "final" position, and ignore Point B altogether.

Answer 2

Initially, just before release, the system has potential energy $U$:

$$T=U=mgL$$

As the string hits the horizontal bar, it enters into a different 'orbit', this one with radius $L-H$. At the highest point of this loop, the mass is at height $2(L-H)$. Assuming full conservation of energy, then the new total energy is:

$$T=2mg(L-H)+K$$

Where $K$ is the kinetic energy at the top of the loop. So:

$$mgL=2mg(L-H)+K$$ So that:

$$K=mg(2H-L)$$

With $v$ the tangential velocity at the top of the loop:

$$K=\frac12 mv^2=mg(2H-L)$$

So:

$$v^2=2g(2H-L)$$

The centripetal force $F_c$ is:

$$F_c=\frac{mv^2}{R}=\frac{mv^2}{L-H}$$

For there not to tension in the string at the top of the loop:

$$mg=\frac{mv^2}{L-H}$$

$$g=\frac{2g(2H-L)}{L-H}$$

So I get:

$$L-H=2(2H-L)$$

$$H=\frac35 L$$