# Pendulum with changing length

#### zenctheo

A friend of mine asked me to help him with this exercise. He has to deliver it tomorrow:
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 $$\displaystyle \phi_0$$. While he is moving to the right when he pass the vertical point 1, the length changes to l-b. Calculate the maximum angle $$\displaystyle \phi_f$$ where the child is on the highest point of the orbit.

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

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
$$\displaystyle mgy=\frac{1}{2}mu^2$$
From point 1 to 3 we have
$$\displaystyle mgy'=\frac{1}{2}mu^2$$
And because the second parts are equal so are the first parts.

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

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

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

Therefore

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

Is everything ok?

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

#### topsquark

Forum Staff
A friend of mine asked me to help him with this exercise. He has to deliver it tomorrow:
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 $$\displaystyle \phi_0$$. While he is moving to the right when he pass the vertical point 1, the length changes to l-b. Calculate the maximum angle $$\displaystyle \phi_f$$ where the child is on the highest point of the orbit.

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

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
$$\displaystyle mgy=\frac{1}{2}mu^2$$
From point 1 to 3 we have
$$\displaystyle mgy'=\frac{1}{2}mu^2$$
And because the second parts are equal so are the first parts.

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

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

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

Therefore

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

Is everything ok?

I also have to point out that he cannot use Lagrange dynamics.
Why are you taking an approximation? We can get an exact answer:
$$\displaystyle \phi = cos^{-1} \left ( 1 - \frac{l + b}{l - b}~(1 - cos(\phi _0) ) \right )$$
is what I get.

-Dan

#### zenctheo

Why are you taking an approximation? We can get an exact answer:
$$\displaystyle \phi = cos^{-1} \left ( 1 - \frac{l + b}{l - b}~(1 - cos(\phi _0) ) \right )$$
is what I get.

-Dan

Can you please show me how? you you replace the $$\displaystyle tan=\frac{sin}{cos}$$?

Is everything else ok?

#### topsquark

Forum Staff
Can you please show me how? you you replace the $$\displaystyle tan=\frac{sin}{cos}$$?

Is everything else ok?
I never used the tangent function. As far as I can tell you did everything correctly.

Here's my derivation.

From 0 to 1 I am setting the GPE 0 point at the bottom of the swing.

So
$$\displaystyle mgh = \frac{1}{2}mv^2$$

h is the change in the vertical position from 0 to 1, so
$$\displaystyle cos(\phi _0) = \frac{(l + b) - h)}{l + b}$$
$$\displaystyle h = (l + b)(1 - cos(\phi _0))$$

So
$$\displaystyle mg(l + b)(1 - cos(\phi _0)) = \frac{1}{2}mv^2$$

$$\displaystyle v = \sqrt{2g(l + b)(1 - cos(\phi _0 ))}$$

Assuming this kinetic energy is transferred to the l - b part of the motion without loss we can do the same problem backward to obtain
$$\displaystyle \frac{1}{2}m \cdot 2g(l + b)(1 - cos(\phi _0 )) = mg(l - b)(1 - cos(\phi))$$

My result follows.

-Dan