# Help with a dynamics homework problem

1. Nov 27, 2007

### jdonn987

i was wondering if anyone could help me with this problem i got for homework. i'm not completely sure where to start with this, so any advice is appreciated. the closest problem i've done with this is just finding the angular velocities and accelerations of the other rods, and i don't know if its the weights thrown in there that is confusing me, but again, any help is appreciated.

The pins in the system shown are currently located at the following coordinates (in meters): A: (0,0); B: (0,1) ; C: (1,0); D: (0.5,0). Rod AB always rotates at 0.5rad/s CCW and accelerates at 0.1rad/s2 CW. Rod CD rotates at a constant 2 rad/s CCW. The collar has a mass of 1200grams. BC is one rigid body with mass of 2kg and radius of gyration of 0.3m.

a. Through what angle (radians) will AB have rotated after 42 seconds?
b. What are the accelerations and net forces on the collar at this instant? Show all steps.
c. What is the net moment applied to BC about its center of gravity?

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2. Nov 28, 2007

### kataya

a) You don't need the masses for this. The expression:

$$\theta - \theta_{o}= \frac{1}{2}\alpha^{2}t + \omega_{o}t$$

will give you the angle you need.

b) You can find the acceleration of the collar through rigid body dynamics:

$$a_{A} = a_{B} + \alpha\times r_{B/A} - \omega^{2}r_{B/A}$$

Where $$r_{B/A}$$ is the vector from point b to point a, and alpha and omega are the angular acceleration and angular velocity, respectively.

Then use the mass of the body in question to solve for the acceleration using newton's laws. Keep in mind that the angular accelerations are about the body's center of mass, not necessarily the point at which they are rotating.

c) Since you know the angular acceleration of the body about its center of mass, you can use its moment of inertia to solve for the moment.

Last edited: Nov 28, 2007
3. Nov 28, 2007

### jdonn987

so for part a do i disregard the forces that BC and CD put on AB that would counter its own motion? thanks for the reply...i've been working on this problem for a few hours now and i'm still not really getting anywhere

4. Nov 28, 2007

### kataya

One problem you might be having is that in your problem description you wrote:

Rod AB always rotates at 0.5rad/s CCW and accelerates at 0.1rad/s2 CW

This does not make any sense the way it is written, because the rod cannot "always" rotate one direction at a given speed while it is accelerating in another direction. Which of these are instantaneous? That makes a big difference, because if the rod will always accelerate at a given rate you can predict its angle without any of the force analysis.

5. Nov 28, 2007

### jdonn987

that was a typo in the original problem; the initial velocity is instantaneous, while the acceleration is constant, so at that instant, it is rotating at .5 rad/s, but slowing down because of the negative acceleration

6. Nov 28, 2007

### kataya

Okay then for a constant acceleration you can find the angle it rotates after a given time without having to take the forces into account.

7. Nov 28, 2007

### paul34

I have a question furthering this: for part B, which omega is the appropriate one for the equation? Is it the omega of member CD or member BC?

Can we assume that for *this instant*, member BC is only in translation (hasn't started rotating yet since its still perpendicular to member AB), and thus, only member CD really has an omega?

But then if that assumption is made, then where does part C come in - since there shouldn't be a moment if its just translating.

Also, for part C, how does one go about finding the center of an awkward body like that?

8. Nov 28, 2007

### jdonn987

from what understand about this problem so far, i'm trying to use that equation for rigid body motion stating with the given $$\alpha_{AB}$$ . as for the second part, i'm not sure if BC is rotating or translating, but i'm leaning more towards translating since its perpendicular to AB at this instant. and for the last part, i'm assuming that its center of gravity has something to do with its radius of gyration, where i was looking at it as $$M=I \times \alpha_{BC}$$

9. Nov 28, 2007

### jdonn987

i think i need more help with the second part of this question, because i've tried solving it a few different ways, and the answers that i'm getting aren't working out. i've tried saying that AB is a vector, CD is a vector, and that piece BC is essentially $$r_{C/B}$$, and trying to solve for $$\omega_{BC}$$ , but nothing is cooperating with the math i'm doing. any other suggestions that can put me on the right path?

10. Nov 28, 2007

### paul34

here's what I'm thinking (feedback please on my thinking).

Part A is self-explanatory and can be found using that simple equation in post #2 by kataya.

Then, the part C can be found by finding the acceleration at point B (using given info, should be doable by alpha*r on member AB), then it's just a matter of finding the radius from point B to the center of grav. of member BC, then you can find the alpha of that member.

The problem is, what *is* the radius to the center of gravy on member BC? I assume its somewhere not actually on the member itself. Do you just draw a line between B and C and just choose the center point? that can't be right.

11. Nov 28, 2007

### kataya

I don't know if that would work. How could you find the center of mass if you don't know the dimensions of body BC? They only give you the location of its endpoints.

The way that I had in mind was to find the acceleration of points B and C and then solve for the angular acceleration of body BC based on the formula given. I don't think you can assume that body BC is in translation. You will have to solve for the angular acceleration of BC by finding the velocity at points B and C (from the rotation of the two rods). The only problem with this is that you would have to leave the moment of inertia of the body as an unknown. Thus your solution to part C would only be partially complete.

12. Nov 28, 2007

### jdonn987

from asking my teacher about this problem, he said that the the shape of BC doesn't matter in the whole scheme of the problem. you can assume that BC is a slender rod, and use the mass moment of inertia for it (1/12)mL^2, and use the angular acceleration for BC that is found from part B, with L being the radius of gyration

13. Nov 28, 2007

### kataya

That seems strange to me, but maybe he is right. If the two pieces of the rod are of equal length, then it would have the same moment of inertia, but if one is longer than the other (like a distorted V shape) then I don't think that it would be the same.

The moment of inertia is a measure of how hard it is to rotate a solid about a given axis, and if you change where the mass is located (as long as it is not an equivalent configuration) then the moment of inertia will change. Anyways, for your purposes, this problem is solved.

14. Nov 28, 2007

### jdonn987

thanks so much for all your help - you've been a lifesaver

15. Nov 28, 2007

### paul34

Well, I think you should be able to figure it out. If I did the math correctly, the first, straight section is 0.5 m (given), and by way of the pyth. thm., the second section should be ~1.12 m

16. Nov 28, 2007

### fredboy

I am stumped on part B. Could you please further explain how to apply the equation presented in post #2? I'm not sure whether to relate the equation to acceleration of CD or AB. Also, how does the angular acceleration of BC get determined? Thanks for your help.

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