1
$\begingroup$

This problem is from Hounser & Hudson, Applied Mechanics: Dynamics (1959):

Weight lifted from cable wound on a cone

The answer given by the book is: $$a = \frac{Dd \omega^2}{4 \pi h}$$

I was trying to solve it using polar coordinates (solution (2) ahead). First I'll present a typical solution, as it has been given to me by colleagues and then ask what's wrong with solution (2):

Solution 1: no cylindrical polar coordinates

An infinitesimal displacement of cable is $$ds = r d \phi$$ Differentiating twice, which is the acceleration of the cable, one gets: $$\ddot{s} = \dot{r} \omega \tag{1}$$ The relationship between horizontal displacement $z$ and $r$ is (triangle geometry): $$r = \frac{Dz}{2h}$$

relationship between r and z

and the relationship between $z$ and $\phi$ is given by pitch: $$\Delta z = \frac{\phi d}{2 \pi}$$ Now we have: $$\Delta r = \frac{dD \phi}{4 \pi h} \Rightarrow \dot{r} = \frac{dD \omega}{4 \pi h}$$ Replace it in $(1)$ and there it is.

Solution 2: cylindrical polar coordinates

We assume an inertial coordinate system lying on the surface of the cone in the position where the cable starts bending from the linear lifting. Using polar coordinates: $$\mathbf{r} = r \hat{\mathbf{e_r}} + z \hat{\mathbf{e_k}} \Rightarrow \mathbf{\dot{r}} = \dot{r} \hat{\mathbf{e_r}} + r \omega \hat{\mathbf{e_\phi}} + \dot{z} \hat{\mathbf{e_k}} \Rightarrow \mathbf{\ddot{r}} = -r \omega ^2 \hat{\mathbf{e_r}} + 2 \dot{r} \omega \hat{\mathbf{e_\phi}} + \ddot{z} \hat{\mathbf{e_k}}$$

Where it has been assumed $\omega$ (as given by the exercise) and $\dot{r}$ (from the expression given in Sol. 1) are constant.

Now the point is: could I assume the tangential acceleration ($\hat{\mathbf{e_\phi}}$) is the same as the one lifting the cable linearly ? There's no curvature immediately before contact, therefore no centripetal acceleration. Also, the horizontal acceleration should be neglected, as instructed. That is:

$$\ddot{s} = 2 \dot{r} \omega$$

If that is so, this result is exactly double that given as the answer (see $(1)$).

How can I reconcile these results?

I've noticed that the second "parcel" of this value is given by the differentiation of the coordinate vector ($\hat{\mathbf{e_r}}$) on an inertial reference frame. I'm unable to decide why it should be discarded, if the textbook answer is correct.

$\endgroup$
4
  • $\begingroup$ "The relationship between horizontal displacement 𝑧 and 𝑟 is (triangle geometry): 𝑟=𝐷𝑧2ℎ" Why is horizontal displacement relevant? Surely the turns of the cable are close-packed along the surface of the cone, which is not horizontal. $\endgroup$ Dec 19, 2020 at 19:49
  • $\begingroup$ By horizontal displacement I meant the distance from the cone's tip along the axis. The radius of the cone is proportional to the distance from the tip. The turns are close-packed, but at every turn there is an increase of $~d$ in the value of $z$ $\endgroup$
    – Petrini
    Dec 19, 2020 at 19:54
  • $\begingroup$ "but at every turn there is an increase of 𝑑 in the value of 𝑧" I think not. There is an increase of d in the 'slope distance' occupied by the turns of cable along the cone's surface. This corresponds to a smaller increase in z. $\endgroup$ Dec 19, 2020 at 19:56
  • $\begingroup$ You're correct, assuming the cable section as perfectly round, it should be an increase of about $d \cos (\theta)$ where $\theta = \arctan \left( \frac{D/2}{h} \right)$. But let's assume then $\theta$ is "small". I believe even if this angle wasn't small, it would be simply then a matter of a constant correction correction factor in these answers $\endgroup$
    – Petrini
    Dec 19, 2020 at 20:08

1 Answer 1

1
$\begingroup$

The second method is incorrect. If you calculate how fast the constantly changing contact point is moving, it is only moving horizontally, and the horizontal acceleration can be ignored. If you follow the instantaneous motion of the current contact point, you should get only the centripetal acceleration (it is rotating so the acceleration is orthogonal to the tangential direction), which shows that you got one sign mistake and got $2\dot{r}\omega$ instead. The acceleration of the weight really comes from the increased speed of the shortening of the cable.

$\endgroup$
4
  • $\begingroup$ I did not understand how a single sign mistake could produce the Coriolis term, since it's on a different vector of the basis and was produced by vector differentiation. There's also the notion that even if we disregard the horizontal displacement, the radius of the cone at the contact point is increasing: that's also part of the forementioned term. $\endgroup$
    – Petrini
    Dec 19, 2020 at 22:59
  • 1
    $\begingroup$ The Coriolis term would come only if you are moving at a nonzero speed in the rotating coordinate system. $\endgroup$
    – C Tong
    Dec 19, 2020 at 23:12
  • 1
    $\begingroup$ Think of the acceleration of the weight as v(c(t+dt))-v(c(t)), where c(t) is the current contact point, not the material reference frame of the contact point. $\endgroup$
    – C Tong
    Dec 19, 2020 at 23:13
  • 1
    $\begingroup$ Oh I see what you meant, you are saying the horizontal $\dot{r}$ produces a Coriolis term, but that's the direction of angular momentum. You can only get a term in $\phi$ direction if you have a radial motion. $\endgroup$
    – C Tong
    Dec 19, 2020 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.