Beads on Semi-Circular wire

ehild

It is ingenious, Infinitum!

My method is much more complicated.
The speed of the bead on the vertical line is dx/dt=u.
The speed of the bead on the circle is v=Rdβ/dt=Rω

From the cosine law in the tringle AOB,

[tex]L^2=R^2+x^2+2Rxcos(\beta)[/tex]

With implicit differentiation,

[tex]0=x u+R u cos(\beta)-Rx\sin(\beta) ω[/tex]

isolating v=Rω

[tex]v=u\frac{(1+\frac{R}{x}cos(\beta))}{\sin(\beta)}[/tex]

Form the sine law, [tex]\frac{R}{x}=\frac{\sin(\alpha)}{\sin(\beta-\alpha)}[/tex]

Plugging in and simplifying, we get:

[tex]v=u\frac{cos(\alpha)}{\sin(\beta-\alpha)}[/tex]

ehild

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Pranav-Arora

It's really complicated.
Thanks for an alternative solution.