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 Quote by Infinitum An easier approach is equating the velocities of A and B along the rod. Let the velocity of B be v, and then try to solve the problem.
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,

$$L^2=R^2+x^2+2Rxcos(\beta)$$

With implicit differentiation,

$$0=x u+R u cos(\beta)-Rx\sin(\beta) ω$$

isolating v=Rω

$$v=u\frac{(1+\frac{R}{x}cos(\beta))}{\sin(\beta)}$$

Form the sine law, $$\frac{R}{x}=\frac{\sin(\alpha)}{\sin(\beta-\alpha)}$$

Plugging in and simplifying, we get:

$$v=u\frac{cos(\alpha)}{\sin(\beta-\alpha)}$$

ehild
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 Quote by ehild 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, $$L^2=R^2+x^2+2Rxcos(\beta)$$ With implicit differentiation, $$0=x u+R u cos(\beta)-Rx\sin(\beta) ω$$ isolating v=Rω $$v=u\frac{(1+\frac{R}{x}cos(\beta))}{\sin(\beta)}$$ Form the sine law, $$\frac{R}{x}=\frac{\sin(\alpha)}{\sin(\beta-\alpha)}$$ Plugging in and simplifying, we get: $$v=u\frac{cos(\alpha)}{\sin(\beta-\alpha)}$$ ehild
It's really complicated.
Thanks for an alternative solution.