Mechanics - particle in a half cylinder

[Pagan Harpoon]

Homework Statement

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,

Homework Equations

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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.

 

[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)mv_0² and the net work is the "work done by friction" which you have to find by doing the appropriate integral.

 

[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).

 

[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.

 

[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.

 

[kuruman]

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

f = μ_kN, where N = normal force.

 

[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.

 

[Pagan Harpoon]

Fresh ideas, anyone?

 

[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 .
Start with the equations

<br /> N-mgsin\Phi=mv^2/R<br />

for the normal component of the acceleration and

<br /> mgcos\Phi-\mu N=m\dot{v}<br /> <br />

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

<br /> \frac{dN}{dt}=(\frac{dN}{d\Phi}-mgcos\Phy)\frac{v}{R} =2m\frac{v}{R}\dot{v}<br />

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

 

[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.

 

[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.

 

[kuruman]

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.

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

 

[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

 

[kuruman]

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

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.

 

[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.

 

[ehild]

The correct form of the general solution has 1 + 4μ² in the denominator.

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

ehild