Rotational dynamics

[Kerbox]

Imagine a jojo. You hold the end firm, and let it go. It falls down, gaining linear speed and angular speed. Forces acting on it is G due to gravity, and F due to the string tension. What I am wondering is this:

How do you prove that the F in the equation "F*r = I*\omega" is of the same magnitude as the F in "G-F = m*a"? Since F act normal to the radius, and not through the center of gravity, I dont find it that obvious. Should it be obvious?

 

Thanks

[[Tycho?]]

Its called a "yoyo"

 

Other than that I can't help you, I havn't learned this yet.

[mezarashi]

Hmmm, for this question, I don't think there is any logical reason to actually assume that. It would be safer just to analyze it using the tools you've been taught, a la free body diagrams.

 

If you enclose the falling yoyo in a box and look at the FBD, then there are basically two forces, the string Fstring, and the force of gravity, Fgrav. If you look at it from reference to the yoyo itself, then you have Fgrav acting at its center of mass, and a force Fstring, acting tangentially to the body surface. Though obviously, the yoyo will not fall as fast as it would in complete freefall, because some of the potential energy is not only being converted into translational kinetic energy, but also rotational inertia through Fstring which is a tangential force, thus as we know, causes rotation then using the I*omega formula.

 

I'm not really sure what you're confused or up to, if you could give me more specific calculations or answers you are looking for maybe I could help you better.

[Kerbox]

When setting up the equations:

G-F_1 = m*a

F_2*r = I*alpha

we are assuming that F_1=F_2. The basis of my problems with this is that we say that G acts on the yoyo as if it were acting on a pointmass m, at the center of mass. Doesnt the first equation suggest that F does the same thing? But since it acts tangential to the yoyo, something doesnt quite seem right to me. Is it the fact that the linear acceleration HAS to equal the angular acceleration multiplied by the radius that makes the equality F_1=F_2 hold up?

 

EDIT: Yea, I ment alpha, not omega Sorry.

[mezarashi]

Okay, I see where you are getting confused. Dynamics, I remember this was the horry of my 1st year college course. Anyway, to clarify, this doesn't assume anything. The answer simply lies with proper understanding of how to use FreeBody Diagrams. When you analyze the free body diagram of the entire yoyo, you draw a box around the yoyo. The point being is that you do not care where the forces act on the yoyo. You are dealing with external forces. Draw a box around it, then add the force vectors of any that extrude or intrude the box, in this case you have Fgrav and Fstring. The resultant force, Fnet will act on *the center of mass* of the yoyo. Yes.

 

The rotational analysis is done using the yoyo itself as the reference frame. For obvious reasons, when we do the rotational analysis, we see again that the tangential force is indeed that save force Fstring. There's nothing arcane about that

[Kerbox]

Arent you assuming that a force acting tangential to a body has the same effect as the same force acting with a line of action going through the center of mass, as far as linear motion is concerned? Is this a general property when we deal with rigid bodies?

[mezarashi]

The law of Newtonian mechanics holds for any reference frame you see. *IF* your reference is the entire yoyo. That is you enclose the yoyo in a box, then the net force will act on the center of mass of the box/yoyo. Regardless of where the force acts on the yoyo, as it is an internal system. This is a relatively fundamental understanding you need in not just dynamics, but statics. For example, imagine firing a canon ball into the air at an angle, the canon ball flies up, and say it explodes in mid air. The center of mass of the exploded particles will continue to travel in a clean particle trajectory because there was no external force. That's just to illustrate the concept of internal/external forces.

 

Back to your yoyo problem. Now imagine that the yoyo is on the table, and you wind two strings around it in opposite directions and pull with the same strength. It will start spinning right? Now take the freebody diagram of the BOXED yoyo. Two forces in opposite directions, the net force is zero, the center of mass of the yoyo won't move, and that is exactly the case. The yoyo will start rotating, but the center of mass will remain the same. The yoyo will remain undisplaced. So reiterating my first statement. It is important to consider your FRAME OF REFERENCE. Physical laws apply equally to all inertial frames, and indeed that is fundamental, even in relativistic mechanics.

[Kerbox]

For example' date=' imagine firing a canon ball into the air at an angle, the canon ball flies up, and say it explodes in mid air. The center of mass of the exploded particles will continue to travel in a clean particle trajectory because there was no external force. That's just to illustrate the concept of internal/external forces.

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Isnt this so because gravity acts equally on every particle in the system, and if integrated over all particles, the gravity would be AS IF it acted on the center of mass? But say you have a wind column acting on only a part of the debree that is moving through the trajectory. The whole system will not accelerate upward as if it acted through the center of mass. But then again, its not a rigid body once exploded, so I think the example might be a bad one.

 

But lets imagine the yoyo floating in space, without any gravity. The only force is a constant force F on the string. Will the linear acceleration then be a = F/m, according to your argument?

[mezarashi]

Hi again. Yes, that example is not directly relevant to your question. I was merely illustrating the concept of internal/external forces and how they don't affect each other in predicting the motion of objects. If the concept of external forces on a rigid body still isn't clear to you from the yoyo example. I have yet one more. Imagine a rigid ruler. If you push it at its center of mass with force F1, it will accelerate. If you push it with double the force, 2*F1, it will accelerate twice as fast. Now consider if you pushed it with F1 each at both its ends. Would it accelerate at the same rate? The answer would be yes. It doesn't matter where you apply a force to a rigid body as long as what concerns you is the motion of its center of mass. That's atleast one thing easy about dynamics

[Kerbox]

But still, in my case, the force would be at just one end. There would be no F2.

[mezarashi]

Hmmm, I guess you're not understanding what I'm saying. Maybe it's better off we discuss this somewhere like on the IRC channel. There is no F2 because you are looking at it from the reference of the entire yoyo. In that frame, there is on ly Fgrav and Fstring. If you look at it using yoyo's body as the reference, then there is an F2, which is coincidentally also the Fstring. Sorry, but I don't think I can explain it better than this

[Kerbox]

Maybe Im missing some fundemental understanding, but just answer me this:

What would the linear and angular acceleration be if the yoyo was in space with only the string tention acting on it, with a constant force F?

[mezarashi]

Ok, so if the yoyo was in space, then we again apply basic Freebody analysis. There is no external force acting on the yoyo other than the Fstring, so the acceleration of its center of pass is indeed a = F/m. It will also gain rotational momentum, with an accelerated rate of alpha = Torque/I. It would be as simple as that. It's also important to note that this may appear to be in violation of the laws of conservation of energy as in, how can you get acceleration and rotation. But consider the yoyo, as you pull the string, the string unwinds. In effect, to pull its center of mass a distance x, you need to pull the string a bit longer than distance x due to the unwinding string. I hope you can visualize all this.

[Kerbox]

I think Im beginning to get it now.

Thanks for your help, I appreciate it

[mezarashi]

Haha, alright. Glad you're making progress