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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 [tex]\phi_0[/tex]. While he is moving to the right when he pass the vertical point 1, the length changes to lb. Calculate the maximum angle [tex]\phi_f[/tex] where the child is on the highest point of the orbit.
It is given that for small angles [tex] (1cos(\phi_0)=\frac{{\phi_0}^2}{2} [/tex] .Also [tex]\frac{b}{l}<<1[/tex] and
[tex] (1+\frac{b}{l})^3 \approx (1+3\frac{b}{l}) [/tex]
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
[tex] mgy=\frac{1}{2}mu^2[/tex]
From point 1 to 3 we have
[tex] mgy'=\frac{1}{2}mu^2[/tex]
And because the second parts are equal so are the first parts.
[tex] gy=gy' \Rightarrow gxtan\phi_0 = gx' tan \phi_f [/tex]
Because the angles are small [tex]tan\phi \approx sin \phi[/tex]
Also [tex] x=(l+b)sin\phi x'=(lb)sin\phi [/tex]
Therefore
[tex] gxtan\phi_0 = gx' tan \phi_f \Rightarrow (l+b) sin^2 \phi_0=(lb) sin^2 \phi_f \Rightarrow [/tex]
[tex]\frac{1cos2\phi_0}{2} (l+b) = \frac{1cos2\phi_f}{2} (lb) \Rightarrow [/tex]
[tex] \cdots \Rightarrow \phi_f=\sqrt{\frac{\phi_0^2(l+b)}{lb}}[/tex]
Is everything ok?
I also have to point out that he cannot use Lagrange dynamics.
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 [tex]\phi_0[/tex]. While he is moving to the right when he pass the vertical point 1, the length changes to lb. Calculate the maximum angle [tex]\phi_f[/tex] where the child is on the highest point of the orbit.
Homework Equations
It is given that for small angles [tex] (1cos(\phi_0)=\frac{{\phi_0}^2}{2} [/tex] .Also [tex]\frac{b}{l}<<1[/tex] and
[tex] (1+\frac{b}{l})^3 \approx (1+3\frac{b}{l}) [/tex]
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
[tex] mgy=\frac{1}{2}mu^2[/tex]
From point 1 to 3 we have
[tex] mgy'=\frac{1}{2}mu^2[/tex]
And because the second parts are equal so are the first parts.
[tex] gy=gy' \Rightarrow gxtan\phi_0 = gx' tan \phi_f [/tex]
Because the angles are small [tex]tan\phi \approx sin \phi[/tex]
Also [tex] x=(l+b)sin\phi x'=(lb)sin\phi [/tex]
Therefore
[tex] gxtan\phi_0 = gx' tan \phi_f \Rightarrow (l+b) sin^2 \phi_0=(lb) sin^2 \phi_f \Rightarrow [/tex]
[tex]\frac{1cos2\phi_0}{2} (l+b) = \frac{1cos2\phi_f}{2} (lb) \Rightarrow [/tex]
[tex] \cdots \Rightarrow \phi_f=\sqrt{\frac{\phi_0^2(l+b)}{lb}}[/tex]
Is everything ok?
I also have to point out that he cannot use Lagrange dynamics.
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