Pendulum with changing length

Apr 2008
7
4
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.
 

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topsquark

Forum Staff
Apr 2008
3,106
659
On the dance floor, baby!
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
 
Apr 2008
7
4
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
Apr 2008
3,106
659
On the dance floor, baby!
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
\(\displaystyle cos(\phi _0) = \frac{(l + b) - h)}{l + b}\)
leading to
\(\displaystyle h = (l + b)(1 - cos(\phi _0))\)

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
 

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