# Pendulum with changing length

A friend of mine asked me to help him with this exercise. He has to deliver it tomorrow:

## Homework Statement

A small child with mass equal to m is on a swing with length l and has the ability to change the length of the swing l. In the beginning he starts from the point 0 (see in the figure) where the length of the swing is l+b and the angle is $$\phi_0$$. While he is moving to the right when he pass the vertical point 1, the length changes to l-b. Calculate the maximum angle $$\phi_f$$ where the child is on the highest point of the orbit.

## Homework Equations

It is given that for small angles $$(1-cos(\phi_0)=\frac{{\phi_0}^2}{2}$$ .Also $$\frac{b}{l}<<1$$ and
$$(1+\frac{b}{l})^3 \approx (1+3\frac{b}{l})$$

## The Attempt at a Solution

This is my solution:

Since there is no information about the mass of the swing I assume its weight is negligible. Therefore the only returning force is the component of the weight.
The point 1 is where the potential energy is zero.
From the 0 point to 1 we have
$$mgy=\frac{1}{2}mu^2$$
From point 1 to 3 we have
$$mgy'=\frac{1}{2}mu^2$$
And because the second parts are equal so are the first parts.

$$gy=gy' \Rightarrow gxtan\phi_0 = gx' tan \phi_f$$

Because the angles are small $$tan\phi \approx sin \phi$$

Also $$x=(l+b)sin\phi x'=(l-b)sin\phi$$

Therefore

$$gxtan\phi_0 = gx' tan \phi_f \Rightarrow (l+b) sin^2 \phi_0=(l-b) sin^2 \phi_f \Rightarrow$$
$$\frac{1-cos2\phi_0}{2} (l+b) = \frac{1-cos2\phi_f}{2} (l-b) \Rightarrow$$
$$\cdots \Rightarrow \phi_f=\sqrt{\frac{\phi_0^2(l+b)}{l-b}}$$

Is everything ok?

I also have to point out that he cannot use Lagrange dynamics.

#### Attachments

• 4.4 KB Views: 366
Last edited: