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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?

ALMEra
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1 Answers1

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There is a laborious way to show that. The movement equations can be obtained from energy conservation considerations.

Calling $p = (x(t),y(t),z(t))$ the movement lagrangian is

$$ L = \frac 12 m (\dot p\cdot \dot p) - m g z(t) + \lambda(x^2(t)+y^2(t)-a^2)+\mu\left(z(t)-c \arctan\left(\frac{y(t)}{x(t)}\right)\right) $$

so the movement equations are given by

$$ \left\{ \begin{array}{rcl} 2 \lambda x(t)-m x''(t)+c \frac{\mu y(t)}{x(t)^2+y(t)^2} &=&0\\ 2 \lambda y(t)-m y''(t)-c \frac{\mu x(t)}{x(t)^2+y(t)^2} &=&0\\ -g m-m z''(t)+\mu &=&0\\ z(t)-\arctan\left(\frac{y(t)}{x(t)}\right)&=&0\\ x^2(t)+y^2(t)-a^2&=&0\\ \end{array} \right. $$

Deriving twice regarding $t$ the last two equations and solving for $z''(t)$ over the full set of equations, we obtain

$$ z''(t) = -\frac{c \left(x(t)^2 \left(c g+2 x'(t) y'(t)\right)+y(t)^2 \left(c g-2 x'(t) y'(t)\right)+2 x(t) y(t) \left(y'(t)^2-x'(t)^2\right)\right)}{\left(x(t)^2+y(t)^2\right) \left(c^2+x(t)^2+y(t)^2\right)} $$

now substituting

$$ \cases{ x(t) = a\cos(\theta)\\ y(t) = a\sin(\theta) } $$

we obtain

$$ z''(t) = -\frac{c^2g}{a^2+c^2} $$

Cesareo
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