The Physics of Rotation: Why Does It Happen?

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Discussion Overview

The discussion revolves around the physics of rotation, specifically exploring why objects rotate when forces are applied, the relationship between translational and rotational motion, and the work involved in these processes. Participants examine theoretical concepts and seek to understand the physical mechanisms behind rotation, including the role of torque and energy transfer.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question why rotation occurs when a force is applied off-center, noting that it causes both translation and rotation.
  • Others argue that applying a force at the center of mass results in maximum linear velocity, while off-center application leads to less linear velocity and additional rotational motion.
  • There are claims that rotational motion does require work, as energy is transformed into angular momentum when a force is applied off-center.
  • Some participants propose that the work done is related to the distance through which the force acts, suggesting that off-center pushes require more work due to greater movement at the point of contact.
  • There is a discussion about the necessity of knowing the mass distribution of an object to calculate its resulting motion after a force is applied.
  • Participants express confusion about the relationship between applied force, linear acceleration, and the effects of applying force at different points on an object.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relationship between applied force and the resulting linear and rotational motion. There are competing views regarding the energy required for rotation and the effects of force application on linear acceleration.

Contextual Notes

Some limitations in the discussion include assumptions about energy transfer and the need for clarity on the definitions of terms like torque and work. The discussion also highlights the complexity of calculating motion in systems with both translational and rotational dynamics.

Who May Find This Useful

This discussion may be of interest to individuals studying physics, particularly those focused on mechanics, dynamics, and the principles of motion in rigid bodies.

Red_CCF
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Sorry if this sounds like a dumb question, but I can't understand why something would rotate. Ex. with a two particles connected by a rod, if a force is applied anywhere other than its center of mass, the same force would cause both translation and rotation.

Also, how come if the same force is acted onto the center of mass of the object described above, it would move at the same rate linearly as if the force was applied anywhere else, so it seems the rotational motion doesn't require any work, what's the explanation for this?

Thanks.
 
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bp_psy said:

I know about torque and moment but that's not what I'm getting at. I'm trying to understand the physics behind the concept of rotation, like what's actually going on physically not a concept that scientists invented to describe it. For example, what kind of internal interaction are going on to make something rotate?
 
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Think of the rod as a bunch of individual masses connected by springs. Your initial force pushes on one of the masses. You can see that it wants to drag its neighbors, pulling in a direction associated with that spring's connection. If you model that, you'll see that rotation and overall movement both emerge from the initial push.
 
Red_CCF said:
Also, how come if the same force is acted onto the center of mass of the object described above, it would move at the same rate linearly as if the force was applied anywhere else, so it seems the rotational motion doesn't require any work, what's the explanation for this?
The center of mass may move the same, but the point of contact will move more if the force is off-center. It's the same force acting through a greater distance that provides the extra work for the rotation.
 
Your assumption is the original post is incorrect...as Doc Al explains...why would you assume rotation requires no energy but translation motion does??

Ever try to crank an engine by hand? or, equivalently, that's why you have a battery (electrical energy) to start your car...work is required to rotate the engine...and work is required in all rotations. Or think about windmill power generation...those huge blades all over the place don't turn by themselves...the wind does work to cause them to rotate and produce electrical power...no wind and they sit motionless, right??

In a system with the combined translational and rotational movement, there is KE in both the both movements...the rotational velocity is v = wr and so if you can visualize velocity v requiring energy it's clear the w also does.
 
Red_CCF said:
Sorry if this sounds like a dumb question, but I can't understand why something would rotate. Ex. with a two particles connected by a rod, if a force is applied anywhere other than its center of mass, the same force would cause both translation and rotation.

Also, how come if the same force is acted onto the center of mass of the object described above, it would move at the same rate linearly as if the force was applied anywhere else, so it seems the rotational motion doesn't require any work, what's the explanation for this?

Thanks.

Actually, it would not move at the same rate linearly as if the force were applied off-center. You're putting energy into the two-particle system. If you hit it center of mass, it'll move with the greatest linear velocity possible, but if you hit it off-center, it'll move with less velocity. The remaining energy is transformed into angular momentum (spin).
 
Thanks for the responses.
Doc Al said:
The center of mass may move the same, but the point of contact will move more if the force is off-center. It's the same force acting through a greater distance that provides the extra work for the rotation.

Sorry I didn't really quite get what you mean, can you expand on that?

mugaliens said:
Actually, it would not move at the same rate linearly as if the force were applied off-center. You're putting energy into the two-particle system. If you hit it center of mass, it'll move with the greatest linear velocity possible, but if you hit it off-center, it'll move with less velocity. The remaining energy is transformed into angular momentum (spin).

So if I hit the rod off center, then I can't simply get its linear acceleration based on the applied force divided by mass of rod? How would I determine the linear acceleration of this system then knowing only the duration and magnitude of the applied force and the mass of the rod?
 
Naty1 said:
Your assumption is the original post is incorrect...as Doc Al explains...why would you assume rotation requires no energy but translation motion does??

Ever try to crank an engine by hand? or, equivalently, that's why you have a battery (electrical energy) to start your car...work is required to rotate the engine...and work is required in all rotations. Or think about windmill power generation...those huge blades all over the place don't turn by themselves...the wind does work to cause them to rotate and produce electrical power...no wind and they sit motionless, right??

In a system with the combined translational and rotational movement, there is KE in both the both movements...the rotational velocity is v = wr and so if you can visualize velocity v requiring energy it's clear the w also does.

Yeah I realize that now, thanks. I initially assumed that the applied force would cause the same linear acceleration no matter where it is applied (at the CoM or off-center), which led to my confusion.
 
  • #10
Red_CCF said:
So if I hit the rod off center, then I can't simply get its linear acceleration based on the applied force divided by mass of rod? How would I determine the linear acceleration of this system then knowing only the duration and magnitude of the applied force and the mass of the rod?

You must also know something about the distribution of mass throughout the rod in order to use the equations of rotational http://en.wikipedia.org/wiki/Dynamics_(mechanics)" required to calculate the resulting motion (both linear and rotational) of the rod after you push on it.

http://en.wikipedia.org/wiki/Rigid_body_dynamics" with the equations and information you will require to comput both rigid body linear momentum as well as rigid body angular momentum.
 
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  • #11
mugaliens said:
Actually, it would not move at the same rate linearly as if the force were applied off-center. You're putting energy into the two-particle system. If you hit it center of mass, it'll move with the greatest linear velocity possible, but if you hit it off-center, it'll move with less velocity. The remaining energy is transformed into angular momentum (spin).
If you apply the same force to a body, its center of mass will have the same linear acceleration regardless of where that force is applied, off-center or not. That's just Newton's 2nd law.
 
  • #12
Red_CCF said:
Sorry I didn't really quite get what you mean, can you expand on that?
What determines the work done is not simply the applied force, but the movement of that applied force through some distance. If you push something off-center, the motion of the point of contact will be greater than if you pushed it on-center. That means more work is required to sustain that motion; that extra work goes into rotational energy.

So if I hit the rod off center, then I can't simply get its linear acceleration based on the applied force divided by mass of rod?
Sure you can.

Red_CCF said:
Yeah I realize that now, thanks. I initially assumed that the applied force would cause the same linear acceleration no matter where it is applied (at the CoM or off-center), which led to my confusion.
No, that's not the source of your confusion. You were correct about that!
 
  • #13
Doc Al said:
What determines the work done is not simply the applied force, but the movement of that applied force through some distance. If you push something off-center, the motion of the point of contact will be greater than if you pushed it on-center. That means more work is required to sustain that motion; that extra work goes into rotational energy.

Ah okay, thanks for clearing that up; I'm always confused when energy is involved.

So the same force acting at a different position would add more energy to the system (however the work needed for its linear motion is the same) and I can calculate the rotational work by using Ek= 1/2Iw^2?

Also, the point of contact may move forward more, but other part of the rod also moves backwards due to the same force, how come the motion of those parts don't need to be considered?

Lastly back to my original question, how does the particles that make up the rigid body interact when the force is applied off center to make it rotate?
 
  • #15
JDługosz said:
Think of the rod as a bunch of individual masses connected by springs. Your initial force pushes on one of the masses. You can see that it wants to drag its neighbors, pulling in a direction associated with that spring's connection. If you model that, you'll see that rotation and overall movement both emerge from the initial push.

I think I'm getting the general idea, but how come they follow a circular motion instead of just moving in the direction of the applied force?
 
  • #16
Bob S said:
Read about center of percussion at

http://en.wikipedia.org/wiki/Center_of_percussion

This describes how an off-center impulse causes both linear acceleration and rotation.

Bob S

I actually have an additional question regarding the article, on the diagram, the article mentions that the beam hanging off the wire would rotate about the center of gravity? Why would it not pivot around the connection point between the wire and the beam? Is there a definition that would help us identify the pivot in a rotating object?
 
  • #17
Red_CCF said:
I actually have an additional question regarding the article, on the diagram, the article mentions that the beam hanging off the wire would rotate about the center of gravity? Why would it not pivot around the connection point between the wire and the beam? Is there a definition that would help us identify the pivot in a rotating object?
In the diagram in the article, the beam is "suspended from a wire by a U-bolt so that it can move freely along the wire."
Bob S
 
  • #18
Bob S said:
In the diagram in the article, the beam is "suspended from a wire by a U-bolt so that it can move freely along the wire."
Bob S

Sorry but I still don't see why the beam would rotate around its center of gravity instead of the connection point.
 
  • #19
Red_CCF said:
Sorry but I still don't see why the beam would rotate around its center of gravity instead of the connection point.

Because the "connection point" in this case can move. It isn't an anchor to one position on the wire. When you push your clothing aside in your closet, any rotation is caused by friction between the hangar and the dowel it hangs on, causing the hangar to stick a bit until you push hard enough to overcome the stickiness. In the absence of friction, the movement would just be translational. Well, since clothing isn't rigid, some of the movement would just distort the clothing, but you get the point.
 
  • #20
loseyourname said:
Because the "connection point" in this case can move. It isn't an anchor to one position on the wire. When you push your clothing aside in your closet, any rotation is caused by friction between the hangar and the dowel it hangs on, causing the hangar to stick a bit until you push hard enough to overcome the stickiness. In the absence of friction, the movement would just be translational. Well, since clothing isn't rigid, some of the movement would just distort the clothing, but you get the point.

But if this connection point isn't a pivot, then that means that point have to be able to follow a rotational/circular path but in this case I don't see how it can move in such a fashion around the center of gravity because the point only able to travel on the pre-defined line.
 

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