String Tension

1. Mar 25, 2012

mr.physics

Hi there,

I have a very basic question about string tension. Say I have a mass at the end of some string whirled around in a vertical circle. At the top of the swing why is there still tension? I get that if the speed of the mass is large enough some tension would be needed to supply the centripetal force, but I don't understand why the tension would act in response to the "needs" of the centripetal force. It seems to me there would only be tension if the mass were trying to move farther away from the center of the circle than allowed by the string. If I were to release the mass in this same configuration from the top of the swing with some velocity, why isn't the string just slack and why doesn't the mass just fall down in response to gravity?

Thanks for the help

2. Mar 26, 2012

Khashishi

It depends how fast you are twirling it. If you twirl it with the minimum rotation necessary to make a circle, then the tension at the top is 0. But if you twirl any faster, then the string is pulling on the mass to accelerate the mass downward faster than free-fall. In the absence of forces, the mass will just travel in free-fall. Not sure what is confusing you.

If you release the mass in the same configuration, you get the same behavior, of course. You are launching the mass horizontally, but your string is attached vertically, so the mass isn't free to travel horizontally, but is pulled by the string downward.

3. Mar 26, 2012

sophiecentaur

Try not to think anthropomorphically. There is a constant acceleration downwards due to gravity, in addition to anything caused by the string. So, at the top, in the limiting case, there is just enough centripetal acceleration (g) to keep the motion in a circle and the tension would be zero. If the mass were going any slower then g would be more centripetal acceleration than needed to keep the circular motion and the mass would follow a parabola - not a circle.