# Mechanics - particle in a half cylinder

1. Dec 6, 2009

### Pagan Harpoon

1. The problem statement, all variables and given/known data

A point particle of mass m is released at point A on the rim of a half cylinder of radius R, it is hit sharply such that it acquires initial velocity v0 directly downwards. There is nonzero friction between the particle and the cylinder, Mu. The particle's motion is always in a plane perpendicular to the central axis of the cylinder. Calculate, in terms of the given unknowns, the minimum value of v0 that allows the particle to reach the point B on the other side of the cylinder. I have made a diagram,

2. Relevant equations

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3. The attempt at a solution

Looking at the forces acting on the particle, it's pretty clear that, at any particular point, the component of weight that acts along the tangent line to the surface of the half cylinder is $$mgsin\theta$$. Also, the normal force is $$mgcos\theta+mR\dot{\theta}^2$$, so the total tangential force is $$mgsin\theta-\mu(mgcos\theta+mR\dot{\theta}^2)$$. Now, my (apparently bad) idea was to just integrate that across $$\theta$$ from $$\theta=\pi/2$$ to $$\theta=-\pi/2$$. I hoped that that would provide an expression for the work done on the particle (the energy that leaks out of the system) in terms of the various unknowns, including v0 and I would tune v0 such that the initial kinetic energy is equal to the energy that it loses to friction. However, integrating that involves integrating a $$\dot{\theta}$$ term, it would seem to me that in order to do that, I would need to have $$\dot{\theta}$$ (and then easily v) as a function of $$\theta$$, which would enable me to just set it equal to 0 at $$\theta=-\pi/2$$ and the problem would be solved.

Any help is greatly appreciated.

There is also the option of setting $$mgsin\theta-\mu(mgcos\theta+mR\dot{\theta}^2)=mR\ddot{\theta}$$ and trying to solve that differential equation, but I'm pretty sure that that is a bit beyond my current ability, and presenting nasty differential equations is not the usual style of my mechanics homework, so I suspect that there is something better to do.

Last edited: Dec 6, 2009
2. Dec 6, 2009

### kuruman

Do it by using the work energy theorem. Note that the work done by gravity from A to B is zero, the change in kinetic energy is -(1/2)mv02 and the net work is the "work done by friction" which you have to find by doing the appropriate integral.

3. Dec 6, 2009

### Pagan Harpoon

Alright, so I can throw away the gravity part of the integral, but that still leaves me with

$$\int^{-\pi/2}_{\pi/2} fR d\theta = m\mu\int^{-\pi/2}_{\pi/2}(gcos\theta+R\dot{\theta}^2)Rd\theta = \frac{1}{2}mv_0^2$$

Is this the appropriate integral to which you refer? The problem with $$\dot{\theta}^2$$ seems to persist, is there some way to rewrite this to make the integration possible without explicitly having $$\dot{\theta}^2$$ as a function of $$\theta$$? If so, I don't see it. I get the feeling that there is a whole side to this that I am missing, as $$\dot{\theta}$$ will obviously depend on v0 somehow, even aside from th relationship stated above (that is just the special case where the particle just about makes it to B).

Last edited: Dec 6, 2009
4. Dec 6, 2009

### Zorba

It think you're missing an extra $$R$$ there since the direction vector for the change is $$d\textbf{\vec{r}}=0 \hat{r} + Rd \theta \hat{ \theta}$$ ? Otherwise I can't contribute much, the integral has stumped me also, I assume it just involves some clever substitution for $$d\theta$$ though I can't figure it out.

Last edited: Dec 6, 2009
5. Dec 6, 2009

### Pagan Harpoon

Ah, yes, thank you. I had started working in linear distance, to avoid any funny business with angles, hence the vestigial l on the diagram. Realising that was unnecessary, I switched back too hastily.

6. Dec 6, 2009

### kuruman

What expression are you using for the force of kinetic friction? It should be

f = μkN, where N = normal force.

7. Dec 7, 2009

### Pagan Harpoon

Yes, and the normal force is $$mgcos\theta+mR\dot{\theta}^2$$ because it first cancels out the perpendicular component of weight and then provides the $$mR\dot{\theta}^2$$ to keep the particle in circular motion, so friction is $$\mu$$ times that, in accordance with the integral I wrote.

Last edited: Dec 7, 2009
8. Dec 8, 2009

### Pagan Harpoon

Fresh ideas, anyone?

9. Dec 8, 2009

### ehild

I have got a crazy idea: It is possible to find a differential equation for the normal force N as function of the angle theta. (I feel it easier to use the angle with respect to the horizontal: $\Phi$ .

$$N-mgsin\Phi=mv^2/R$$

for the normal component of the acceleration and

$$mgcos\Phi-\mu N=m\dot{v}$$

for the tangential component, and taking the time derivative of the first equation:

$$\frac{dN}{dt}=(\frac{dN}{d\Phi}-mgcos\Phy)\frac{v}{R} =2m\frac{v}{R}\dot{v}$$

we can eliminate $m\dot{v}$, obtaining a first order linear differential equation for $N(\Phi)$

$$\frac{dN}{d\Phi}+2\mu N=3mgcos(\Phi)$$

This equation can be solved.

ehild

10. Dec 8, 2009

### kuruman

Not so crazy an idea. I wish I had thought of it.

However, I think there is still a a problem. I solved the differential equation to get N(Φ) in terms of the given quantities. The problem is that one can find the starting value of the normal force in terms of the given quantities by setting N(0) = mv02/R in the starting equation, and this determines v0 in terms of the given quantities. However, v0 is an independently defined constant. I checked ehild's derivation and could find no error, so ...

Good problem though. Made me look at the normal force in novel ways.

11. Dec 8, 2009

### Pagan Harpoon

I don't think there's any problem with it at all. Setting $$N(0)=mv_0^2/R$$ gives an expression for v0 in the given variables and an integration constant. The value for v0 can be changed by adjusting the constant, then by setting $$N(\pi)=0$$ the correct value for the constant that just about brings it to the rim can be found.

I think that is correct. Thanks a lot, chaps.

12. Dec 9, 2009

### kuruman

If you get a chance, can you show me a few details? I must have a blind spot in my reasoning. Thanks.

13. Dec 9, 2009

### ehild

I am not sure if my solution is correct, but it will do for illustration.

$$N(\Phi)=3mg\frac{sin\Phi+2\mu cos\Phi}{1-4\mu^2}+Cexp(-2\mu \Phi)$$
Substituting N back into the first equation and taking $\Phi=0$,

$$C=mv_0^2/R-3mg\frac{2 \mu}{1-4 \mu ^2}$$,

Substituting this expression for C into N:

$$N(\Phi)=3mg\frac{sin\Phi+2\mu cos\Phi)}{1-4\mu^2}+(mv_0^2/R-3mg\frac{2 \mu}{1-4 \mu ^2})exp(-2\mu \Phi )$$

At Phi=pi, v=0, and N should be zero, too, according to the first equation.

$$N(pi)=3mg\frac{-2\mu }{1-4\mu^2}+(mv_0^2/R-3mg\frac{2 \mu}{1-4 \mu ^2})exp(-2\mu \pi )=0 \rightarrow$$

$$v_0^2=\frac{6Rg \mu}{1-4 \mu ^2}(1+exp(-2\mu \pi))$$

I can not promise that all this is correct.

ehild

14. Dec 9, 2009

### kuruman

Thanks. It turns out I did have a blind spot handling the homogeneous solution. The correct form of the general solution has 1 + 4μ2 in the denominator.

15. Dec 9, 2009

### Pagan Harpoon

The differential equation turns into

$$\frac{d}{d\phi}e^{2\mu\phi}N=e^{2\mu\phi}3mgcos\phi$$

Integrate that for

$$N(\phi)=\frac{3mg}{4\mu^2+1}(2\mu cos\phi+sin\phi)+\frac{c}{e^{2\mu\phi}}$$

Then setting $$N(0)=mv_0^2/R$$ gives an expression for $$v_0$$ in terms of the given variables and some constant, c. It is not defined by the other variables, because the varying of c can make it any value that we want it to be. Having done this, setting $$N(\pi)=0$$ gives the correct value for c that satisfies the desired condition - it reaches point B with 0 velocity.

Edit - Yes... same as above.

Another edit - My solution for v0 has $$e^{2\mu\pi}$$ rather than $$e^{-2\mu\pi}$$, which I think is correct.

Last edited: Dec 9, 2009
16. Dec 9, 2009

### ehild

Yes... There were three "-" instead of two....

ehild