Conceptual question about torque translation

[Ertosthnes]

Homework Statement

I was just thinking about the fact that torque = rxF = I*alpha when there is no translation. Suppose that I had a lever (length 2r) with a pivot in the middle, and I attached a weight of mass m on the end. The torque would be rmg and the angular acceleration would be rmg/I, right? But now suppose that I dropped the weight from some height. One would expect that the angular acceleration of the lever would be greater for the weight that was dropped from the height. Yet, as far as I can tell, the force on the weight is still mg! Does this mean that angular acceleration does not depend on the momentum of the weight?

 

[Dr.D]

Looks like you are presuming that your (frictionless) bar is initially level, although I don't see that in your problem statement.

In the first case, the torque would be r*mg = I*alpha1

In the second case, there are two phases to be considered. During the first phase, the mass falls freely through a distance h about the bar, while the bar simply sits still. During this phase, energy is conserved, so that
mg*h = (1/2)*m*v^2
v = sqrt(2*g*h)
for the velocity of the mass just prior to impact with the bar.

If we assume a perfectly plastic impact, so that the mass sticks to the bar on impact, then we can say that, by conservation of system angular momentum about the pivot for the bar,
r*m*v = (I+m*r^2)*w
where w is the angular velocity just after impact. Note that the bar has an instantaneous (discontinuous) change in angular velocity.

From that point on, we are back to the original problem, with
r*mg = I*alpha2
until such time passes that the geometry changes and that has to be taken into account with the cross product. But the thing that is different is the initial condition on the movement of the bar; in the second case it starts with a jump, where as in the first case it starts from zero.

 

[Ertosthnes]

Ah, yes I was assuming that the (frictionless) bar was level.

I'm pretty sure I follow your reasoning in the second paragraph: You have L = rmv = Iw +mwr^2. But v = rw, so then you have rmv = Iw +rmv, which doesn't make sense unless the velocity of the mass changes after it hits (and sticks) to the bar, in which case you have rmv = Iw + rmv', which is fine.

In your third paragraph, however, I don't understand how you got the new equation for torque. Did you differentiate? Also, you have rmg = I*alpha2, but earlier you wrote rmg = I*alpha1, so does that mean alpha1 = alpha2?

 

[Dr.D]

Well, if the mass hits and STICKS to the bar, then the mass and the bar have to travel together which means that they have the same velocity. The velocity of the bar at the tip, the place where the mass sticks is r*w, so that has to be the velocity of the mass when it sticks at that point also.

You asked how I got the final equation of motion and asked if I differentiated. No, I simply did a sum of torques again. The sum of torques is, again, m*g*r, so the left side is mg*r. On the right side, I wrote I*alpha2 simply to denote that this was the angular acceleration for the second case that was being analyzed. As you noted, it is exactly the same as in the first case.

The key point, however, was the final sentence: But the thing that is different is the initial condition on the movement of the bar; in the second case it starts with a jump, where as in the first case it starts from zero. That is where the effect of the falling mass is taken into account in this problem.

 

[Ertosthnes]

Okay; so the torques WILL be the same in both cases? The only difference between the two cases is that in the second case, the bar will start with a larger angular velocity, though the angular acceleration is the same. Is that correct?

Also, is there any way to quantify how much larger the angular velocity will be in the second case, say, through a given rotation or period of time?

 

[Ertosthnes]

Regarding my previous post, I'm thinking that the answer to my last question is that w2 = w1 + (r*mv)/I, where v = sqrt(2gh). That would mean that w(f) = (w1 + r*mv/I) + alpha*t, which would answer the question for a given period of time.

The given rotation part seems messier. Basically, r*theta = t(v + r*w(f))/2, which would then be solved for t and plugged into w(f) = (w1 + r*mv/I) + alpha*t, whereupon the quadratic equation would be solved for w(f). Can anyone confirm this?

 

[Dr.D]

You did not bother to define w1 or w2, so I have no way to know what you mean here (mind reading is not something I do; gave it up a long time ago).

The notation w(f) is pretty fuzzy; what is the f supposed to mean?

You are forgetting that we have not accounted for the cross product which will but a sine of the angle into the equation. Then the equation becomes nonlinear and is somewhat more difficult to integrate. It can still be done, but it is a harder problem.

 

[Ertosthnes]

Sorry about the notation. Okay, for the first case with a weight hanging from the bar:

[tex]\Delta[/tex]L = Rx[tex]\Delta[/tex]P
Iw1 = rmvsin[tex]\theta[/tex]
=>w1 = rmvsin[tex]\theta[/tex]/I

For the second case with the weight that fell from a height h:

[tex]\Delta[/tex]L = Rx[tex]\Delta[/tex]P
[tex]\Delta[/tex]L = rm(v+[tex]\sqrt{2gh}[/tex]-0)sin[tex]\theta[/tex]
Iw2 = rm(v+[tex]\sqrt{2gh}[/tex]-0)sin[tex]\theta[/tex]
=>w2 = {rm(v+[tex]\sqrt{2gh}[/tex])sin[tex]\theta[/tex]}/I

So w2 - w1 = {rm[tex]\sqrt{2gh}[/tex]sin[tex]\theta[/tex]}/I, which tells us the difference in speed between the two cases when the bar moves through a rotation theta. Is that correct?

 

[Ertosthnes]

If that's all right, then to determine the difference in speeds after a given time, I think that using the equation:

[tex]\theta = \omega t + (1/2) \alpha t^2 [/tex]

would do the trick, after substitution into [tex]w2 - w1 = {rm \sqrt{2gh}sin \theta}/I [/tex].

Since the initial w is zero, and [tex]\alpha = rmg/I[/tex], [tex]\theta = (1/2I)rmgt^2[/tex]

So the difference in speed after a time t would be:
[tex]w2 - w1 = {rm \sqrt{2gh} sin((1/2I)rmgt^2)}/I[/tex]

 

[Dr.D]

Do you recall how you very carefully wrote that the governing equation was rxF = I alpha in your original post? Now you are neglecting that cross product.

The equation of motion, in either case, with the cross product taken into account, is:

I*dd(theta) = m*g*r*cos(theta)

where theta(t) is measured from the horizontal, and dd(theta) is alpha. Notice that this is not a constant acceleration problem, so your simple constant acceleration formula does not apply. This is why I said that it is more difficult to integrate, meaning that you have to solve the differential equation.

 

[Ertosthnes]

Oh, I see. I guess the acceleration is non-constant because the force of the weight does not remain perpendicular to the bar? I don't think I know how to solve differential equations like dd(theta)-(rmg/I)cos(theta) = 0. I entered it into mathematica and got this:

[tex]\left\{\left\{\theta[t]\to \frac{1}{2} \left(\pi -4 \text{JacobiAmplitude}\left[\frac{1}{2} \sqrt{(-2 i \text{rmg}+C[1]) (t+C[2])^2},\frac{4 \text{rmg}}{2 \text{rmg}+i C[1]}\right]\right)\right\}[/tex]

[tex]\left\{\theta[t]\to \frac{1}{2} \left(\pi +4 \text{JacobiAmplitude}\left[\frac{1}{2} \sqrt{(-2 i \text{rmg}+C[1]) (t+C[2])^2},\frac{4 \text{rmg}}{2 \text{rmg}+i C[1]}\right]\right)\right\}\right\}[/tex]

Then differentiating,

[tex]\theta'[t] = \left((-2 i \text{rmg}+C[1]) (t+C[2]) \text{JacobiDN}\left[\frac{1}{2} \sqrt{(-2 i \text{rmg}+C[1]) (t+C[2])^2},\frac{4 \text{rmg}}{2 \text{rmg}+i C[1]}\right]\right)/\left(\sqrt{(-2 i \text{rmg}+C[1]) (t+C[2])^2}\right)[/tex]

Seems like an incredibly complicated answer for such a simple question.

 

[Dr.D]

Well, that depends. It gave you the answer in terms of Jacobian elliptic functions which is not surprising at all. You thought it was a simple question, but then you did not know how to set it up, so how are you able to judge that it was a simple question? This answer has been known for several hundred years, so in that respect it is not a "hard problem" but the answer is not elementary either.

Mathematica in your hands is most likely a dangerous weapon. It will solve all manner of problems, but do you know what you have? Considering your inability to set up the problem, I would suggest that you lay Mathematica aside for several years until you have learned to have a proper use for it. Right now, it is simply dangerous for you.

 

[Ertosthnes]

Well in this case, I don't know how to interpret the answer. I only used Mathematica in this case because I had a suspicion that I didn't know how to solve that type of differential equation, but I was curious as to what the answer might look like. If the problem had involved an equation that I knew how to solve, then I would not have used it.

By the way, does my answer mean that my work in post #8 was also wrong?

 

[Dr.D]

Rather than me try to figure out what you meant there, I'm going to let you try to figure out what you meant there. It is correct if it leads in a rational fashion to the differential equation that I gave you; otherwise, it is incorrect.