Stop a Pendulum: Calculate Stopping Force

  • Thread starter Thread starter samodelov.1
  • Start date Start date
  • Tags Tags
    Pendulum
AI Thread Summary
Calculating the stopping force for a pendulum involves understanding the relationship between momentum and time, where the stopping force is defined as F = -dp/dt. The energy lost during an impact can be approximated using the kinetic energy formula, where energy loss is (1/2) M v². If the pendulum hits a rigid stop and bounces back with a coefficient of restitution of 100%, it theoretically retains its kinetic energy, while practical scenarios may involve energy loss due to elasticity. Stopping the pendulum at its highest point theoretically results in zero force, but forces will be applied as it swings back. Overall, precise calculations are complicated by factors like mass shape and impact elasticity.
samodelov.1
Messages
5
Reaction score
0
I am looking for a way to calculate the stopping force for a pendulum but have two problems here. Number one is that I can not find any information on a (non-experimental) way of calculating the force a pendulum would have if you pulled against its direction of motion or put something in front of it. Second: I need to stop it after an initial impact and was wondering if anyone knows of a way to calculate the energy lost to a pendulum after impact. Thanks! Any and all help is appreciated!
 
Physics news on Phys.org
samodelov.1 said:
I am looking for a way to calculate the stopping force for a pendulum but have two problems here. Number one is that I can not find any information on a (non-experimental) way of calculating the force a pendulum would have if you pulled against its direction of motion or put something in front of it. Second: I need to stop it after an initial impact and was wondering if anyone knows of a way to calculate the energy lost to a pendulum after impact. Thanks! Any and all help is appreciated!

The stopping force depends directly on time interval within which you wand to stop it:
F=-dp/dt. Here dp is the current momentum minus the final momentum (=0 for stopped body), dt is the time interval during which you stop the body. For example, if you want to stop it instantly, the force should be inifinite.
If dt is much smaller than the oscillation period (dt<<T), you can speak of a constant force F=-dp/dt. If not, your force will time dependent as F(t)=-dp(t)/dt (ordinary derivative). Anyway, you have to spend the work=the body kinetic energy. At the turning point not force is necessary.

Bob.
 
Last edited:
If you stop a pendulum at the bottom of its cycle, so that all of its energy is kinetic energy of the (point) mass itself, then the equations are the same as for a mass sliding horizontally without friction. In short, if the mass M has a velocity v, then the impulse to stop it is the integral of force F(t) times time dt = M v. The energy lost is just (1/2) M v2.

If the mass hits a rigid (infinite mass) stop and bounces off the stop with 100% coefficient of restitution, the direction of pendulum moton is changed, no momentum is transferred to the stop (because the integral F(x,t) dx is zero), and the pendulum loses no kinetic energy. In this case, the integral F(x,t) dt is doubled.

If the pendulum mass is finite size and therefore has angular momentum, then the mass has to hit the stop at the center of its percussion so that both the linear kinetic energy and the rotational kinetic energy are completely absorbed.

For a simple pendulum, you could put a mass equal to the pendulum mass at the bottom of the swing, and if the collision between the pendulum and the mass is completely elastic, all the momentum (and energy) would be transferred to the mass, and the pendulum would stop.
 
Last edited:
Thanks for all the help! I think I am beginning to get a better picture of what is happening theoretically. (I was never good at circular motion.)

On a practical basis: My mass is not spherical, so it cannot impact at the center of percusion (or I can't ensure that it does, in any case) and it is impacting an elastic fixed body. I know now that there is no way of calculating exactly how much energy is lost at impact (the body changes and each has a different elasticity). I will just go with a worst case scenario and assume no energy is lost upon impact.

If I do try to stop the pendulum at it's highest point after it has impacted and changed direction, even though the force is theoretically zero, wouldn't there be a force applied to my stop once the pendulum begins to swing in the other direction again?

Thanks again!
 
samodelov.1 said:
If I do try to stop the pendulum at it's highest point after it has impacted and changed direction, even though the force is theoretically zero, wouldn't there be a force applied to my stop once the pendulum begins to swing in the other direction again?

Thanks again!

Why do you think the force is theoretically zero? Does it accelerate? Try drawing a graph of displacement/velocity/acceleration (use different colors) vs time. Or you can try this neat applet
http://www.walter-fendt.de/ph11e/pendulum.htm
 
Last edited by a moderator:
That applet is really helpful. Thanks!

What formula is it using to calculate the tangential force? I thought it was F=mg*sin(theta) but their maximum force always seems to be somewhat higher than that, especially at larger angles.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top