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A smooth wire in the shape of the helix $\vec{r} = a (\cos θ)\vec{i}+ a (\sin θ)\vec{j}+ cθ\vec{k}$, where $a$ and $c$ are positive constants is fixed with the $z$-axis pointing vertically downwards. A bead of mass $m$, free to slide along the wire, is released from rest at the point A where $θ = 0$. Show that the vertical component of the reaction is constant.

I applied $\vec{f}=m\vec{a}$ to the particle vertically so I got

$$f=ma=R_{k}-mg=mc\ddot{\theta}$$ My textbooks solution is $R_{k}=mc\ddot{\theta}$

Is there anything wrong what I did?

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The bead slides down a slope with a horizontal distance of 2πa and a vertical distance of 2πc. The cosine from the vertical is c/a. The acceleration down the slope is aα and the vertical component is cα. (α being the angular acceleration.) Then taking + down, mg – R = mcα. Where R (the vertical component of the reaction force) is a constant (But it is not equal to mcα).

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  • $\begingroup$ Thank you I mistook something $\endgroup$ – D ake Jan 5 at 20:09

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