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- Thread starter samodelov.1
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- #2

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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.

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- #3

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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 v^{2}.

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.

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.

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- #4

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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!

- #5

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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 [Broken]

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- #6

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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.

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