Sliding Down a Helix

Classical Mechanics Level 4

A bead slides from rest down a wire that's bent into a helix, which can be parametrized in the following way: { x = cos ⁡ ( θ ) y = sin ⁡ ( θ ) z = θ . \begin{cases} x = \cos(\theta) \\ y = \sin(\theta) \\ z = \theta. \end{cases} Find the magnitude of the bead's vertical acceleration.


Assumptions: The bead slides without friction and there is a uniform, gravitational field − g z ^ . -g\,\hat{\mathbf{z}}.

g 4 \frac{g}{4} g 2 \frac{g}{2} g 3 \frac{g}{\sqrt{3}} g 2 \frac{g}{\sqrt{2}}

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Steven Chase by Steven Chase , 37, USA

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3 solutions

Steven Chase
Aug 5, 2017

Write the ( x , y , z ) (x,y,z) velocities:

x ˙ = − sin ⁡ ( θ ) θ ˙ y ˙ = cos ⁡ ( θ ) θ ˙ z ˙ = θ ˙ \dot{x} = -\sin(\theta) \, \dot{\theta} \\ \dot{y} = \cos(\theta) \, \dot{\theta} \\ \dot{z} = \dot{\theta}

Square of the velocity magnitude:

v 2 = x ˙ 2 + y ˙ 2 + z ˙ 2 = 2 θ ˙ 2 v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2 = 2 \dot{\theta}^2

Set the kinetic energy equal to the change in potential energy (assuming the bead starts at a particular height / angle):

m g ( θ 0 − θ ) = 1 2 m 2 θ ˙ 2 g ( θ 0 − θ ) = θ ˙ 2 mg (\theta_0 - \theta) =\frac{1}{2} m 2 \dot{\theta}^2 \\ g (\theta_0 - \theta) = \dot{\theta}^2

Time-differentiate both sides:

− g θ ˙ = 2 θ ˙ θ ¨ − g = 2 θ ¨ θ ¨ = − g 2 -g \dot{\theta} = 2 \dot{\theta} \ddot{\theta} \\ -g = 2 \ddot{\theta} \\ \boxed{\ddot{\theta} = -\frac{g}{2}}

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The problem becomes a lot easier when you straighten out the wire: the bead simply slides down a 4 5 ∘ 45^\circ gradient. The centripetal force component (in the ( x , y ) (x,y) plane) does not affect the acceleration along the path at all.

Arjen Vreugdenhil - 3 years, 9 months ago

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Yes, indeed. It's basically a ramp problem in disguise. Although I'm starting to develop a taste for scalar-based physics (Lagrange / energy) as opposed to vector-based physics (Newton / force). With the former, the behavior simply comes out of the math, without requiring a bunch of reasoning about how things "should" behave.

Steven Chase - 3 years, 9 months ago

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The energy approach (or Lagrange or Hamilton) is very nice if the system is conservative-- but that does not always work!

Arjen Vreugdenhil - 3 years, 9 months ago

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@Arjen Vreugdenhil – That's right. That's the Achilles heel.

Steven Chase - 3 years, 9 months ago

Analysis with forces: the net force is a combination of the centripetal force F c p F_{cp} and a force along the wire F p a r F_{par} . It is the result of a normal force F n F_n perpendicular to the wire and the gravitational force F g F_g . We have

F c p = F n , i n ; F p a r = F g sin ⁡ 4 5 ∘ ; 0 = F n , u p − F g cos ⁡ 4 5 ∘ . F_{cp} = F_{n,in};\ \ \ F_{par} = F_g\sin 45^\circ;\ \ \ 0 = F_{n,up} - F_g\cos 45^\circ.

The acceleration along the wire (i.e. rate of increase of speed) is a p a r = F g sin ⁡ 4 5 ∘ m = g 2 . a_{par} = \frac{F_g\sin 45^\circ}m = \frac g{\sqrt 2}. The vertical component of this acceleration is a z = a p a r sin ⁡ 4 5 ∘ = g 2 ⋅ 1 2 = g 2 . a_z = a_{par}\sin 45^\circ = \frac{g}{\sqrt 2}\cdot \frac 1{\sqrt 2} = \frac g 2. The centripetal acceleration points toward the axis of the spiral and does not contribute to the vertical component.

Arjen Vreugdenhil - 3 years, 9 months ago
Ujjwal Rane
Aug 25, 2017

A helix is nothing but an inclined plane that is wound around a vertical axis! So Let's unwide a turn or a complete rotation 2 π 2\pi of this helix.

Since the radius is 1 unit, the horizontal distance covered = 2 π = 2\pi this will result in vertical rise of 2 π 2\pi . Hence the angle of this inclined plane must be 45°

Since the acceleration of a body sliding down an inclined plane in absence of friction is g sin ⁡ θ g \sin \theta its vertical component will be g sin ⁡ θ × cos ⁡ θ g \sin \theta \times \cos \theta

Substituting θ = 45 ° \theta = 45° we get vertical acceleration = g sin ⁡ 45 ° × cos ⁡ 45 ° = g 2 = g \sin 45° \times \cos 45° = \frac{g}{2}

26 Helpful 0 Interesting 0 Brilliant 0 Confused

Exceptional! Lovely way of looking at the problem. Kudos!

Aniruddh Patil - 3 years, 9 months ago

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Thank you! :-)

Ujjwal Rane - 3 years, 9 months ago

um, the vertical component would be g sin ⁡ θ × sin ⁡ θ g \sin \theta\times \sin \theta according to where θ \theta is on the picture, it's just that in this case sin = cos.

Duke Garland - 3 years, 9 months ago
Oliver Piattella
Aug 25, 2017

I will use Newton's second law projected along the constraint (the wire). The position of the particle is:

r = ( cos ⁡ θ , sin ⁡ θ , θ ) \textbf{r} = (\cos\theta,\sin\theta,\theta)

its velocity is

r ˙ = ( − sin ⁡ θ , cos ⁡ θ , 1 ) θ ˙ \dot{\textbf{r}} = (-\sin\theta,\cos\theta,1)\dot{\theta}

where the dot denotes derivation with respect to the time, and its acceleration is

r ¨ = ( − cos ⁡ θ , − sin ⁡ θ , 0 ) θ ˙ 2 + ( − sin ⁡ θ , cos ⁡ θ , 1 ) θ ¨ \ddot{\textbf{r}} = (-\cos\theta, -\sin\theta, 0)\dot{\theta}^2 + (-\sin\theta, \cos\theta, 1)\ddot{\theta}

The vertical acceleration is θ ¨ \ddot{\theta} and the force field is F = − m g z ^ \textbf{F} = -mg\hat{\textbf{z}} . The direction of the constraint (the wire) is given by the versor

r ˙ ^ = 1 2 ( − sin ⁡ θ , cos ⁡ θ , 1 ) \hat{\dot{\textbf{r}}} = \frac{1}{\sqrt{2}}(-\sin\theta,\cos\theta,1)

and the projected Newton's second law reads:

F ⋅ r ˙ ^ = m r ¨ ⋅ r ˙ ^ \textbf{F}\cdot \hat{\dot{\textbf{r}}} = m\ddot{\textbf{r}}\cdot\hat{\dot{\textbf{r}}}

Substituting the components determined earlier, we get:

− m g 2 = m 2 [ − sin ⁡ θ ( − cos ⁡ θ θ ˙ 2 − sin ⁡ θ θ ¨ ) + cos ⁡ θ ( − sin ⁡ θ θ ˙ 2 + cos ⁡ θ θ ¨ ) + θ ¨ ] -\frac{mg}{\sqrt{2}} = \frac{m}{\sqrt{2}}\left[-\sin\theta(-\cos\theta\dot{\theta}^2 - \sin\theta\ddot{\theta}) + \cos\theta(-\sin\theta\dot{\theta}^2 + \cos\theta\ddot{\theta}) + \ddot{\theta}\right]

and finally

θ ¨ = − g 2 \boxed{\ddot{\theta} = -\frac{g}{2}}

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