Tension in string holding a ball which is swung in a circle.?

Question

What will be the tension in the string holding a ball which is being swung in a circle of raidus .7 m if the ball is going 2.5 rev/s. Assuming the sytem is in free fall?

What is the tension if the system was on a sationary with respect to surface of earth, w/the ball being swung in a vertical circle, when the ball is at the top of its arc.

What if the ball is at the bottom of the arc?
What if the ball is at 1/2 height?

Answer 1

Centripetal acceleration Ac = v^2/r = w^2*r, where w = angular rate in rad/sec. (1st answer assumes it's w^2/r, and uses w units of revolutions/sec, both erroneous.)
w = 2*pi*2.5
Ac = 172.72 m/s^2
Tension = centripetal force Fc = Ac * M = 172.72 * M
Mass M is not given, so this is the basic equation.
When rotated in a vertical circle under grav. acceleration g (= 9.81 m/s^2),
tension at top = (Ac - g) * m = 162.91 * M
tension at bottom = (Ac + g) * m = 182.53 * M
tension at half height = Ac * m = 172.72 * M

Answer 2

Stand by for a tiny bit of help: There is one formula which solves for the force needed to accelerate an object toward the center of a circle (which is what you are talking about.) It uses mass of the object, radius of the circle, and velocity of the object. I don't remember the formula and don't have access to my materials! You find the formula and you can solve for the velocity with the radius and the # rev./ sec., you have the radius. You need the mass. Then, in free fall the force (tension on the string) will be constant. On earth the tension at the top of the circle will be the tension found through the elusive formula MINUS the weight (mass x g) of the object, at the bottom of the circle it will be tension PLUS weight, at the 3- and 9-o'clock positions it will just be the tension.

Answer 3

The tension in the first system is going to be 8.9m(rev^2/s^2)/m
(sorry I don't know how to convert rev/s to m/s). The tension was equal to the centripetal force, and since you didn't specify a mass, I left it as a multiple, which I did for the rest.
For the second system (if the velocity is the same) the tension is equal to the centripetal force plus its weight, so the tension is 18.71m
The third system the tension is equal to the weight minus the centripetal force, so the tension is .91m
For the forth system, the tension will just be the centripetal force 8.9m