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I Translational + Angular Acceleration of Free Body (not fixed)

  1. Dec 28, 2017 #1
    Imagine a long brick in outer space. You apply a force tangential to the center of mass. The brick accelerates in a transitional and angular fashion. There are no constraints or fixed axis. How would I calculate the translational and angular acceleration? I would like to run some simulations with Mathematica. I understand rotational problems with fixed axis and translation problems if the force vector passes through the center of mass. This problem is different. One force will cause rotation and the center of mass to move. I would appreciate any kind of help. Maybe I just don't know the correct google search terms to figure this out.
    358AtoL.jpg
     
  2. jcsd
  3. Dec 28, 2017 #2

    kuruman

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    Hi Julian and welcome to PF.

    It is important to specify if the force is applied for a short duration, like a sharp kick, or if it acts continuously. If the latter is the case, then you need to specify its direction as the brick moves. Your drawing suggests that it's a sharp kick (impulse). Is it?
     
  4. Dec 28, 2017 #3
    Thank you for pointing that out. For now, let's say it is an impulse or high force for a short interval. If the force is closer to the center of the brick, the brick will move more but rotate less. If the force is far from the center of mass, the brick will rotate a lot and not move much. Correct? Am I making this too complicated? Can I somehow break this into two problems (rotation and movement)? That doesn't seem right. Do I need to know about Euler's equations of motion? I really appreciate the help.
     
  5. Dec 28, 2017 #4

    kuruman

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    Correct.
    Yes you can. There is translational motion of the center of mass (CM) and rotational motion about the center of mass. In other words, the CM will move with velocity VCM and at the same time the brick will rotate about the CM with angular speed ω. An observer moving along the brick with speed VCM will see the brick just spin.
    You do not. You need to conserve linear and angular momentum before and after the impulse is delivered to the brick. Hint: Call the impulse delivered to the brick ##J## and proceed from there. The linear momentum before the kick is ##P_{before}=J## and the angular momentum before the kick is ##L_{before}=r~J## where ##r## is the distance from the CM to the point on the brick where it is kicked. You should be able to derive expressions for ##V_{CM}## and ##\omega## in terms of ##J## and the brick's dimensions and mass.
     
  6. Dec 28, 2017 #5

    A.T.

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    Yes, you can avoid rotation completely, by applying the force through the CM.

    No, the linear velocity will be the same as in the previous case, if you apply the same high force for the same short time interval.
     
    Last edited: Dec 28, 2017
  7. Dec 28, 2017 #6

    FactChecker

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    Essentially, you can calculate the acceleration of the CG and the rotational acceleration separately. F = ma is true for the center of mass, regardless of where the force is applied. So you know what the acceleration of the center is. The force at a particular location also creates a torque, which gives you the rotational acceleration when the moments of inertia are considered. Combining the rotational acceleration and the linear acceleration of the center gives the entire answer.
     
  8. Dec 29, 2017 #7

    jbriggs444

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    Not necessarily correct. Much depends on the details of the impulse that is applied.

    If one is applying a fixed force for a fixed time (i.e. a fixed amount of momentum) then the delivered linear impulse and the resulting linear momentum is the same regardless of where the force is applied. The delivered rotational torque [about an axis located at the center of mass] will then be directly proportional to the offset of the force from the center of mass.

    This may seem counter-intuitive at first. How can the same impulse applies at different places deliver result in obviously different energies? Does that not violate energy conservation? It does not. If one applies the same force for a fixed time and if the object is able to rotate under that impulse then the result is a different distance moved by the point of application under the applied force. The delivered work is greater when the force is delivered near the end of the block than when it is delivered near the center.

    If one applies a fixed force over a fixed distance (i.e. a fixed amount of work) then all is exactly as you say. If delivered closer to the center, the brick will move more but rotate less. If delivered closer to the end, the brick will rotate a lot and not move much.
     
  9. Dec 29, 2017 #8
    Thank you for the wonderful replies. I have a better handle on the problem now.

    My simulation will be split into fixed time intervals. For each interval, an object may have a force vector applied to the normal of one of its faces. That force vector is an impulse when the time interval is taken into account. Time is constant. Force is a variable. I think I know how to simulate the linear and rotational motion based on impulse vectors now.

    I formulated this understanding based on the replies:
    I would like to examine this as a perfectly elastic collision.
    If I have an impulse on one object, I must have an equal and opposite impulse on another (Newton).
    If we have a ball and a brick of equal mass hit each other dead center (Case A), they would both flip the direction of their velocities but keep the same magnitudes.
    If the ball hits off center from the brick’s center of mass (Case B), we have a different situation. The ball velocity will flip but the magnitude will go decrease. This means that the impulse is smaller for case B compared to case A. The ball loses kinetic energy from the impact.

    From the brick’s perspective, the impulse is also less for case B. However, the impact causes the brick’s kinetic energy to increase. The rotational energy of the brick makes up for the loss of translation energy of the ball and brick. So the total kinetic energy of the entire system remains unchanged for this perfectly elastic collision. But the ball lost energy and the brick gained energy in Case B.

    So I could build this system of equations:
    Ball impulse = brick impulse
    (Ball energy + brick linear energy + brick rotation energy) of time 0 = (Ball energy + brick linear energy + brick rotation energy) of time x
    F = ma stuff split into two problems based on the same impulse:
    Relate impulse, mass, and velocity to new linear velocity.
    Relate impulse, mass, and angular velocity to new angular velocity.

    And the system of equations should at least be solvable numerically. And my resulting answers would be impulse vectors. Now I can use those vectors to calculate rotational and linear motion separately.

    If I’m wrong about something, please let me know. Thank you for taking the time to help me.
     
  10. Dec 29, 2017 #9

    kuruman

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    You are not wrong, but I would recommend using conservation of mechanical energy and angular and linear momentum since you have chosen to simulate an elastic collision. You don't (and can't) know the details of the forces developed between the colliding objects, but you do know that the kinetic energy and the total linear and angular momentum after the collision are the same as before the collision. This is enough to enable you to find the final linear and rotational speeds of the brick and the ball that collides with it. These are your resulting answers and you can find them analytically. There are three conservation equations and three unknowns, the final linear velocity of the brick, the final linear velocity of the ball and the angular speed of the brick about its center of mass.
     
  11. Dec 29, 2017 #10
    Conservation of kinetic energy makes sense to me. But I don't understand how the conservation of linear momentum is true. And I do not understand how the conservation of angular momentum is true. In Case B of the ball-brick problem, some linear momentum is lost. Some angular momentum is gained (nothing was rotating but the brick rotates after the impact). If you sum linear and angular momentum to be total momentum, I can see how total momentum is conserved. But that would leave us with only two conservation equations. However, I believe you when you say there are three equations. I must have my brain twisted into a knot. Please stay with me here. I almost understand.

    EDIT:
    If I consider the center of mass of the entire system, the system does have initial angular momentum. That initial angular momentum of the entire system never changes. Does that sound right?
     
    Last edited: Dec 29, 2017
  12. Dec 29, 2017 #11

    kuruman

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    You don't see it quite right but you're almost there. Before the collision, the only thing that is moving is the ball. It has linear momentum, ##\vec{p}_{before}=m\vec{v}##, kinetic energy, ##K_{before}=\frac{1}{2}mv^2## and angular momentum about the CM of the ball-brick system ##\vec{L}_{before}=\vec{r}\times \vec{p}## where ##\vec{r}## is the position vector from the system's CM to the ball. Linear momentum is conserved because there are no external forces acting on the ball-brick system, angular momentum is conserved because there are no external torques acting on the ball-brick system and kinetic energy is conserved because you specified tht the collision is elastic. There is no physical mechanism here that will convert linear momentum to angular momentum. "Conserved" means that what you have before the collision (three separate things) is what you have after the collision. That's where the three equations come from.
     
  13. Dec 29, 2017 #12

    jbriggs444

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    That would be dimensionally incorrect. Linear momentum has dimensions of mass times velocity. Angular momentum has dimensions of mass times velocity times displacement.

    In fact, linear momentum and angular momentum are each conserved independently of the other.
     
  14. Dec 30, 2017 #13
    This really cleared things up for me. I never thought to see the angular momentum that way. Here is my current standing:
    Mechanical energy is collectively conserved. I can make equation number one that includes linear and rotational energy. Am I wrong on this as well? Can linear and rotation energy be conserved separately like we do with momentum?
    Linear momentum is conserved independently. I can make equation number two that only includes linear momentum.
    Angular momentum is conserved independently. I can make equation number three that only includes angular momentum.

    However, I understand that one object can have two forms of angular momentum. It can have angular momentum about its own CM and angular momentum about the system CM. This means that there are two additional variables to worry about: There is the ball angular momentum around the system CM. There is the brick angular momentum around the system CM (after the impact). So I have five variables: ball linear, ball system-angular, brick linear, brick self-angular, and brick system-angular. But I have only three equations/constraints. No worries!
    The fancy system-CM angular momentum can be related to the cross product of linear momentum and position from the system CM. I can apply that constraint to the ball. And I can apply that constraint to the brick.
    Now my degrees of freedom equals my number of constraints! That is a good deal. (5 = 5)

    So I believe everything I just said is just another way to look everything I've been taught on this thread.

    Now, I'll throw a wrench into the mix. I want the ball to be a small block. Let's call it the block-brick system. The small-block makes the impact off-center from its own CM. This means we have yet another variable: The small-block gains self-angular momentum after the impact.
    Here is the only way I can think to solve this:
    Take angular energy to be conserved independently.
    Take linear energy to be conserved independently.
    That way, I can split equation number one into two energy equations. I don't know if that is physically accurate though. I'll have six equations and six variables for the block-brick system. If you haven't already guessed, I'm trying to understand how to write an algorithm for general simulation of elastic collisions.

    Again, I really applicate all the help. This kind of discussion is invaluable to me.

    I attempted to quote that equation from kuruman exactly but it was giving me trouble with the special characters.
     
  15. Dec 30, 2017 #14

    kuruman

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    No. The linear and rotational energy kinetic energies are conserved as a sum. There are many ways to express the kinetic energy of an object that translates and rotates simultaneously. The preferred description is kinetic energy of the center of mass and rotational energy about the center of mass.
    Don't do that, at least not yet. I recommend that you first solve the problem without this complication to make sure you understand all the facets of the calculation, and then add the refinement. I strongly urge you to solve the simpler problem (which is not all that simple) analytically first and then program it in Mathematica. This will enable you to check the accuracy of your programming independently. Remember, garbage in → garbage out. Also, before you throw in the monkey wrench, you need to understand that, for the small block to exchange spin angular momentum with the brick, there must be a torque acting on the block and an equal and opposite torque acting on the brick during the short time the two are in contact. The geometry of the collision becomes important and you will probably need to do the collision in two dimensions which opens a new can of worms.
     
  16. Dec 30, 2017 #15
    I'll take your advice and solve "simple" problem analytically on my own. At least I have have a better understanding on the rules of conservation now. I'll have to practice applying them. Thank you.
     
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