Help With AP Physics

[Luke Skywalker]

I'm self studying the AP Physics C: Mechanics Exam at my high school, and I have difficulty understanding the difference between translational and angular motion.

So from what I understand, an object has translational motion if its axis is moving. So, if you held a rod pinned at one end horizontally and let it drop, all of the rod's potential energy would be converted into rotational kinetic energy. But that brings up a few questions (in my mind):

1. Are rotational and translational motion instrinsically seperate, or can you convert them at will?
2. Is there a net force on the rod at any point in time? If not, is this because the rotational axis isn't moving?
3. Is rotational kinetic energy the same as translational kinetic energy, just expressed differently mathematically?
4. If there is zero net force on the rod and zero translational kinetic energy, would it not have zero linear momentum? But if there was a ball in the rod's way, that ball would gain linear momentum...how does this work?
5. If you can convert rotational and translational KE, then how can there be kinetic energy without any force having been applied to the rod...?

Any help would be appreciated.

[Hawkwings]

Translation is sliding without rotating. Rotation is obviously rotating. Traditionally both translation and rotation are measured in reference to the center of mass. So in your example, there is both translation and rotation going on. The rod is rotating 90 or more degrees, and the center of mass of the rod is translating downwards in an arc.

Another way to think about it is this. Any object has six degrees axes of movement. They are: slide in x, slide in y, slide in z, pitch, roll, and yaw. Any motion or an object can be broken up into these six components of motion (some of which may be zero). Translation is the sliding 3, and rotation is the pitch/roll/yaw. So yes, they are intrinsically different and you cannot mathematically convert translation into rotation. Or x-axis translation to y-axis translation.

For #2, draw a free body diagram and look at it.
#3: kinetic energy is simply energy of motion. It can be organized into categories of motion, but it is all motion.

[Terralthra]

#4 The rotational motion of the rod is moving through the ball's center of mass.

[Luke Skywalker]

Thanks for your help.

Traditionally both translation and rotation are measured in reference to the center of mass.

But, you can't define the reference point as any arbitrary axis? Would that mean that the translation of the rod would depend on your reference point, since if you make the reference axis the rotational axis, or in this case where the pin is (this is what my prep book does), the translational motion is zero, since the axis here isn't moving?

But, to my knowledge, net force is simply the resultant of the sum of the x and y (and z) components of the forces acting on the object, with no regard to distance from the pivot. How is this compatible with the idea that the rod has translational motion in reference to one but not in others...?

So in your example, there is both translation and rotation going on. The rod is rotating 90 or more degrees, and the center of mass of the rod is translating downwards in an arc.

I learned that all of the rod's potential energy is converted into rotational kinetic energy. But if the two types of KE are not interchangeable, I don't see how the rod can have translational acceleration.

Another way to think about it is this. Any object has six degrees axes of movement. They are: slide in x, slide in y, slide in z, pitch, roll, and yaw. Any motion or an object can be broken up into these six components of motion (some of which may be zero). Translation is the sliding 3, and rotation is the pitch/roll/yaw. So yes, they are intrinsically different and you cannot mathematically convert translation into rotation. Or x-axis translation to y-axis translation.

OK then, so what does the formula v=wr mean conceptually, if the two types of motion cannot be converted?

#3: kinetic energy is simply energy of motion. It can be organized into categories of motion, but it is all motion.

Right, but if linear and translational acceleration are intrinsically different, would the two KE's not be conceptually separate?

[Simon_Jester]

Hi. I don't teach AP Physics, but I could and I kind of wish they'd let me. I do teach AP Calculus.

Thanks for your help.

Traditionally both translation and rotation are measured in reference to the center of mass.

But, you can't define the reference point as any arbitrary axis?

You can. This does not affect whether the rod is translating. Translational motion, "sliding," is motion that has the same magnitude and direction for all parts of the object. It also does not affect whether the rod is rotating.

Rotational motion means that different parts of the object are moving in different directions and/or different speeds, so that the object is in motion relative to its own center of mass. If the object is moving relative to its center of mass, it is also moving relative to any other arbitrary point on the object that we nail our frame of reference to.

The catch, really, is that we mostly care about rotational motion in the context of angular momentum, moment of inertia, and torque. All of which can become a bitch to calculate for an object if we don't put the origin of our coordinate frame at the center of mass. For example, it is easy to do a three-dimensional integral to get the moment of inertia of a cylinder about its long axis. It is quite a lot less easy to do the same integral about a line parallel to the long axis, but offset from the centerline of the cylinder.

So for practical purposes we measure these 'rotational' properties (moment of inertia, torque, angular momentum, etc.) with respect to the center of mass. It's easier, and it's faster, and that WILL matter on an AP exam because you NEVER want to be stuck working on the same integral or modeling problem for thirty minutes on an exam that only gives you a few hours in the first place.

Would that mean that the translation of the rod would depend on your reference point, since if you make the reference axis the rotational axis, or in this case where the pin is (this is what my prep book does), the translational motion is zero, since the axis here isn't moving?

Nope. Translational force is, pretty much by definition, force that causes all parts of the object to accelerate at the same rate, in the same direction.

Motion of a rod pivoting from "sideways" to "down" is the superposition of two separate motions.

ONE is a translation of the center of mass, from being L/2 meters to the (left/right) of the pivot point, to being L/2 meters below it, where L is the length of the rod.

The other is a rotation of 90 degrees about the center of mass, in the plane of the page you're drawing the diagram in.

If you superimpose the consequences of those two motions ("shift whole thing left and down" plus "pivot whole thing 90 degrees") you will reproduce the location of every element on the rod- and conveniently separate the rotational and translational components of the motion.

I learned that all of the rod's potential energy is converted into rotational kinetic energy. But if the two types of KE are not interchangeable, I don't see how the rod can have translational acceleration.

The two types of kinetic energy can and do interchange- physically, it's all atoms moving in straight lines, possibly with normal forces deflecting them. Think about objects that move powered by flywheels, or for that matter cars- the pistons in the car engine are pure translational motion, which is then used to drive a crankshaft in pure rotational motion, which drive wheels in pure rotational motion... which then make the car translate forward fast.

It is not unreasonable to think of the rod as rotating about one end- fortunately, "rod rotating about its own endpoint" is one of the few rotational problems that IS easy to solve even when you aren't using the center of mass as the origin of the coordinate system, since it boils down to

[integral of](m(r))*(r^2)dr,

with r being the distance you travel down the rod away from its endpoint.

In that context you CAN argue that it's all rotational kinetic energy. Here's the difference:


From the pivot's point of view, the rod has large moment of inertia. Calculated about the pivot point at one end, a rod of mass M and length L will have moment of inertia... [rederives in head] (M*L^2)/3.

Getting this to rotate with a fixed angular speed takes a large amount of energy, so ALL the energy released by dropping the rod goes into that.


From the rod's center of mass point of view, the rod's moment of inertia is lower- because the center of mass is closer to the 'far' end of the rod. Remember that moment of inertia is greatly increased when you have a lot of mass located far from the axis of rotation. In the center-of-mass frame, the rod has moment of inertia (M*L^2)/12.

But it's still rotating at the same angular speed, with a quarter the moment of inertia. So it only needs 1/4 as much rotational kinetic energy as measured in the center-of-mass frame.


Since both frames agree that the original potential energy release from dropping the rod is the same, the extra energy as measured in center-mass frame has to go somewhere. Where? Answer: into a translational motion of the rod in the center-of-mass frame.

OK then, so what does the formula v=wr mean conceptually, if the two types of motion cannot be converted?

The physical velocity of a hunk of stuff at THIS distance r from the center of mass equals the distance, times the angular velocity at which the hunk of stuff is spinning about the axis.

Right, but if linear and translational acceleration are intrinsically different, would the two KE's not be conceptually separate?

They're not- they're tracked differently for our convenience.

It takes energy to get a big heavy flywheel spinning. It also takes energy to take that flywheel and accelerate it to some translational speed so you can carry it across the room.

If you track only the rotational energy, you ignore the fact that the rod is moving- which can be important. Consider the tiny rotational energy of a spin-stabilized rifle bullet, versus the massive deadly translational kinetic energy!

If you track only the translational energy, you ignore other important facts and can get systematically wrong results- this is a common problem when calculating things like "ball rolling down ramp."

So you have to track both. But if you use identical common notation for both... well you CAN but it turns all motion problems involving rotation into messy things with a lot of integral calculus in. It's much neater and cleaner to track them separately, and doing so can give you very useful insights into things like orbital mechanics.

[will keep explaining if things are still messy]

[Hawkwings]

v=wr (which is actually lowercase omega times r) means that the instantaneous tangential velocity of a certain point is the rotational velocity (omega) multiplied by the distance the point is from the center of rotation (r). Typically the center of rotation is some axis through the center of mass, so things work out neatly.

You seem to be referring to the translational axis being the same as the rotation axis a lot, and I just want to make sure I understand what you're saying. What you're saying is that if you have a rod, and you spin it so that the rotational axis is through the long dimension of the rod (like spinning a pencil on its end, or spinning a rifle bullet as it flies through the air), if you use that axis as the translational axis as well, then the translation is the rifle bullet moving forward through the air.

Are there example problems online that you are looking at? Perhaps having an example to work through together would help you more than these conceptual things.

[Luke Skywalker]

Holy shit, the exam was brutal.

But thanks for the help.

[MrDakka]

Hope you get a 5. BTW do they still charge an arm and a leg for taking the test?

[Simon_Jester]

~80 dollars I think, state-defrayed in some places (for better or for worse), and when you think just how much cheaper that is than the equivalent college tuition it's a steal.