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Collisions and Combining Torques

  1. Aug 15, 2009 #1
    Torque in Physics Simulation

    Let's say there is flat floor at y=0 . There is a rectangle of which two vertices are at (0,0) and (w,h). There is a ball in mid-air. Someone let's it fall and the ball falls until it hits the (w,h) vertex of the rectangle. The push-back force is very small and can be ignored. I suppose the ball will be rotating around the vertex until it is able to move downwards at stop when it reaches the floor to the right of the rectangle.

    The question: I'm writing a physics simulation program. It each "step" of the simulation the collision can be recalculated and as long as there is a collision, it it obvious arounf which point the ball spins. But what happens when the collision stops? Does the ball just move according to linear mechanics only and spin around itself?

    And what happens in this case: there is a ball with the center at (0,h). (x,y+h) is a poijt on the circumference of the ball. A force F is applied to this point. How do I conbine r and F for this force with the existing linear and angular speed?

    Another question: ball1 and ball2 are identical. Same mass, same radius. Both have the same speed - same magnitude and same direction. The difference is that ball1 had torque few steps ago. Could this torque affect the ball1's route (besides making the ball spin) even though at the current step the velocities are equal?

    I never learnt these things in school (I would if they taught them!) and now I need them for the physics simulation program...
     
    Last edited: Aug 15, 2009
  2. jcsd
  3. Aug 16, 2009 #2
    Re: Torque in Physics Simulation

    If you know the F vector, r vector and the duration for which it acts, t seconds, you can calculate the final angular and linear speed using the equations of motion.

    No difference in the motion of COM.
     
  4. Aug 17, 2009 #3
    So a force [tex]\vec{F}[/tex] acting on a ball, its effect [tex]\vec{a}[/tex]=[tex]\vec{F}[/tex]/m on the ball's speed doesn't depend on the angle between [tex]\vec{r}[/tex] and [tex]\vec{F}[/tex]? The acceleration will be the same even if [tex]\vec{r}[/tex] x [tex]\vec{F}[/tex] is very close to zero and only the angular acceleration of the ball around itself will be tiny?
     
  5. Aug 17, 2009 #4
    Exactly..
     
  6. Aug 18, 2009 #5
    Two more questions:
    1. So only a collision can make the ball rotate around a point on its circumference (otherwise it rotates around its center)?
    2. A little question about I. Momentum of inertia. If a ball if in mid-air, not colliding, and a force applied to a point on the circumference makes the ball spin around itself, which I do I use in the equation [tex]\vec{\alpha}[/tex]= [tex]\vec{\tau}[/tex][tex]/[/tex]I , the centroid's I or I+mR[tex]^{2}[/tex] (which corresponds to a point on the circumference)? For a ball spinning around the point to which the force is applied (like in a collision) it's obvious, but I'm wondering which I is the correct one for a ball spinning around itself.
     
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