Question

A slender bar rests on the surface of a fixed half-cylinder of radius r, as shown bellow. If the lower end of the bar slides to the right with speed v, determine the angular velocity θ˙ of the bar in terms of x?

http://imgur.com/TD7rLaM

Any ideas, im struggling?

Answer 1

I came up with \[\tan(\theta) = {r \sin(90 - \theta) \over x}\] for the kinematic relationship. The derivative of this is ugly. http://www.wolframalpha.com/input/?i=d%2Fdt+%28r+sin%2890+-+x%28t%29%29%2F%28y%28t%29%29

Answer 2

Thanks eashmore. Can I ask how you got to this solution?

Answer 3

I don't agree with eashmore's answer.

Try drawing a triangle with vertexes at the center of the semi-circle, the point where the bar touches the semi-circle, and the point where the bar touches the ground.

Answer 4

(Hint: you should get a right triangle)

Answer 5

Ok, still rather lost as to how to proceed?

Answer 6

I misread the question. Looking at it again.

Answer 7

I think you're right about the right triangle. The whole triangle isn't fixed, but the smallest side of it is.

Answer 8

The pole will always be normal to the half cylinder.

Answer 9

Yes, and so you can find a function of theta in terms of x:\[\theta = \arcsin(\frac{r}{x(t)})\]But trying to differentiate to find the angular velocity leaves some problems, unless you don't worry about finding a function of velocity in terms of x (which may be the case, since there isn't enough information to do that).

Answer 10

Well, v is constant, so:\[v(x)=v\]I guess it's about that simple. Nevermind.

Answer 11

Velocity is constant unless otherwise specified, I thought, so the angular velocity is all that would be changing

Answer 12

If you wanted to fuss about with time you can change your time differential to position

Answer 13

For a given velocity, each set position of x would have a set angular velocity of thetadot associated with it. But my differentials keep getting wonky

Answer 14

:/

Answer 15

So yeah, simple taking the derivative of theta with respect to t (keeping in mind that x is an implicit function of t) should give you an answer.

I get:

\[\theta = \arcsin(\frac{r}{x(t)})\]\[\frac{d \theta}{dt}=\frac{1}{\sqrt{1-\frac{r^2}{x^2(t)}}}\cdot \frac{d}{dt}(\frac{r}{x(t)})\]Which is where we run into issues, since we don't know x's explicit function in terms of t.

What about:\[x=vt+v_0\]Assume v_0 is 0, and so:\[x=vt\]\[\frac{d \theta}{dt}=\frac{1}{\sqrt{1-\frac{r^2}{x^2(t)}}}\cdot \frac{d}{dt}(\frac{r}{vt})=\frac{1}{\sqrt{1-\frac{r^2}{x^2(t)}}}\cdot \frac{r}{v}\cdot \frac{d}{dt}(t^{-1})\]and so on.

I don't have a lot of faith in this method, so take it at face value.

Answer 16

Hmm, I would not have got to this point. Thanks both :)

Answer 17

Ended up with (and it's totally wrong, so ignore this)

sin(theta)=r/x t = (x-r)/v

d/dt [ sin(theta)=r/v ] dt = (dx +x - r)/v

(d(theta)\dt) cos (theta) = -(r/x^2)dx/dt

thetadot cos (theta) = -v (r/x^2)

and it's messy and kinda makes logical sense, because then thetadot (d (theta)/dt) is getting more negative faster when v increases. And the units work. But it looks like vomit and I dunno if it simplifies...

More face value from this direction :)

Answer 18

My solution (attached) matches what you'd expect the model, and so does yours (but they are different)

The problem with mine is that it's not explicitly in x, like yours. I remember doing problems like this in calc 1, but haven't touched related rates problems in a good while.

Answer 19

And actually, on closer inspection, the graphs look very similar except mine is angular speed vs. time and yours is angular position vs. position.

Answer 20

It just feels like anything time related should cancel out before it gets too complicated, since everything is really linear, and within extremely narrow angle margins. I dunno. More ponderin' is needed!

Answer 21

Mine's actually angular speed verse position. I didn't do the whole substitution because I got lazy. You can make the d theta / dt in mine completely just d theta/dx if you pop in that dt transformation, I just didn't think that I did it right. I gotta run thought. I'll check back later. :)

Answer 22

@z61850 Sorry if any of this confused you d;. Your best bet is to wait for someone who knows what he/she is doing to get here (or eashmore to come back and explain his answer). Cheers (:.

Answer 23

I thought about my solution some more. It is flawed. I was relating different triangles.

Let me revisit it.



The quantity \(h\) will increase with increasing \(x\) and decreasing \(\theta\).

\(h\) can be expressed as\[\sin(90 - \theta) = {h \over r}\]Note that \(r\) is fixed.

We can relate \(x\) to \(\theta\) as\[\tan(\theta) = {h \over x - r \cos(90 - \theta)}\]

where \(r \cos(90 - \theta)\) is as shown

Substituting for h yields, \[\tan(\theta) = {r \sin(90 - \theta) \over x - r \cos(90 - \theta)}\]