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The following is a question from a past exam paper that I'm working on, as I have an exam coming soon. I would appreciate any help.


A fairground ride takes the form of a hollow, cylinder of radius $R$ rotating about its axis. People stand against the cylinder wall when it is stationary and its axis is vertical. The rotation is then started, and once the cylinder has reached its operating angular velocity $\omega$, its axis, and the people, are gradually rotated to lie horizontally,as shown in the figure below.

Figure

A person called Alice who has mass $m$, is indicated by $A$ in the figure at the moment when she subtends angle $\theta$ relative to the downwards vertical direction, as seen from the cylinder’s axis. Assuming that Alice does not slide relative to the cylinder wall, and neglecting her size compared to the radius $R$.

  1. Show that the component of force upon Alice from the cylinder wall normal to its surface is given by $N = mg\cos\theta +m\omega^{2}R$.

    • The acceleration is towards the center of the cylinder, hence $$\sum{F}=ma \Rightarrow N - mg\cos\theta = ma \Rightarrow N = mg\cos\theta + m\omega^{2}R.$$
  2. Show that the component of force upon Alice from the cylinder wall parallel to its surface is given by $F = mg\sin\theta$.

    • This follows from the diagram.
  3. Derive an expression for the minimum $\omega$ required to ensure that Alice remains in contact with the cylinder wall at all times.

  4. Show that in order for Alice to avoid sliding along the cylinder wall, $\omega^{2} \geq \frac{g\sin\theta}{\mu_{S}R} - \frac{g\cos\theta}{R}$, where $\mu_{S}$ is the coefficient of static friction between her and the wall.

    • $f_{S} \leq\mu_{S}N \Rightarrow mg\sin\theta \leq \mu_{S}(mg\cos\theta +m\omega^{2}R).$ Rearranging, we get $$\omega^{2} \geq \frac{g\sin\theta}{\mu_{S}R} - \frac{g\cos\theta}{R}.$$
  5. Use the result of question 4 to show that Alice will not slip for any $\theta$ if $\omega^{2} \geq \frac{g}{R}\sqrt{1 + \frac{1}{\mu_{S}^{2}}}$.


Question

I would like to know if the working for the questions above that I have answered are correct. I'm not sure what concept underlies question 3. For question 5, I put $\theta = \pi$, the angle that subtends when she is at the top but that doesn't work.

I'm looking for hints to be able to answer questions 3 and 5. Thank you for your time.

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I am a bit confused with the orientation of the ride. Do I understand correctly that the axis of the cylinder (and thus the rotation axis), is initially aligned with the direction of gravity? And then after some time, this axis is rotated over 90 degrees, to go from vertical to horizontal wrt ground level? –  Bernhard Apr 21 '13 at 13:36
    
For 3. What is $N$ in this situation? –  Bernhard Apr 21 '13 at 13:41
    
@Bernhard The axis of the cylinder is the dot in the middle of the picture (in the direction out of the page). The cylinder rotates on that axis. No translations, just rotational motion, something like a hamster on a wheel, only this time the hamster is stationary on a certain point in the inside of the wheel, and the wheel rotates with a certain speed. For question 3, I would say $N = mg\cos\theta + m\omega^{2}R$, as it is asked in question 1 beforehand. –  user4167 Apr 21 '13 at 13:57
    
ok, so the story is there to confuse people. :) –  Bernhard Apr 21 '13 at 13:58
    
It would be good if OP (or somebody else?) could endow the question (v4) with a more informative title than the current Mechanics question. –  Qmechanic May 10 '13 at 11:16
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2 Answers 2

I can offer a partial answer. Sorry if it's too late.

For question 3, the minimum $\omega$ is when the normal force $N$ is $0$. If $N$ is non-zero toward the center, then the wall is keeping her in. If $N$ is non-zero away from the center, our conditions are broken and little Alice is falling, because no normal force will pull her towards the wall. So set $N=0$ and $\theta=\pi$ and you should find that $\omega_{min}=\sqrt\frac{g}{R}$.

For question 5, I've got nothing. You have to show by some trickery that $$\frac{\sin\theta}{\mu_S}-\cos\theta\le\sqrt{1+\frac{1}{\mu_S^2}}.$$ I tried and I don't know really where to go on that. In any event, your exam is probably past (passed?). If you figured 5 out, please share.

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I understand it's a bit late, but it might still help somebody else in the future.

For question 5:

Solve $\frac{d\omega}{d\theta}=0$ to find the minimum value of $\omega$

$$\frac{d\omega}{d\theta}=\frac{\cos \theta}{\mu_s}+\sin\theta=0$$ gives $\tan\theta= \frac{-1}{\mu_s}$ or $\theta=\arctan\left(\frac{-1}{\mu_s}\right)$

Substitute this in the left part of the last equation given by krs013 gives:

$$\frac{\sin\left(\arctan\left(\frac{-1}{\mu_s}\right)\right)}{\mu_S}-\cos\left(\arctan\left(\frac{-1}{\mu_s}\right)\right)$$

To handle double trigonometric functions, I usually find it useful to draw the actual triangle (note that the -1 side of the triangle should actually be down, but we might as well have defined downwards to positive): Triangle of $\arctan\left(\frac{-1}{\mu_s}\right)$

Now you can see that the $\sin\left(\arctan\left(\frac{-1}{\mu_s}\right)\right)$ is actually equal to $\frac{-1}{\sqrt{\mu_s^2+1}}$ whereas $\cos\left(\arctan\left(\frac{-1}{\mu_s}\right)\right)$ is equal to $\frac{\mu_s}{\sqrt{\mu_s^2+1}}$

Substituting these simplfications in the equation leads to:

$$\frac{-1}{\mu_s\sqrt{\mu_s^2+1}}-\frac{\mu_s}{\sqrt{\mu_s^2+1}}$$

or

$$\frac{-\left(\mu_s^2+1\right)}{\mu_s\sqrt{\mu_s^2+1}}$$

Which simplifies to

$$-\frac{\sqrt{\mu_s^2+1}}{\mu_s}$$ And writing everything in square roots gives

$$-\frac{\sqrt{\mu_s^2+1}}{\sqrt{ \mu_s^2}}=-\frac{\sqrt{\mu_s^2+1}}{\sqrt{ \mu_s^2}}=-\sqrt{1+\frac{1}{\mu_s^2}}$$

For some reason the minus remained, which seems unlogical, as negative frequencies do not make sense. However, I did not manage to find the error, so if somebody sees my mistake, please help.

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