a small sphere of 10 kg is attached to the of a chord f length 2 meter and set into motion in a vertical circle about a fixed point 0.determine the tension in the chord at any time when speed of the sphere is 20 m/s the chord makes an angle of 69 degrees with the vertical?
There are two forces that contribute to the tension in the chord:
Originally Posted by alderon
---one due to the weight of the sphere
---another due to the centripetal force because of the circular mtion of the ball
Due to the weight, it is tension or pulling away away from point O.
(F1)cos(69deg) = 10(9.8)
F1 = 98 /cos(69deg) = 273.46 newtons
The centripetal force is going towards the point O. The reaction on the sphere is an equivalent force, called "centrifugal" force that pulls the sphere away from point O.
F2 = m(v^2)/r
F2 = 10(20^2)/2 = 2000 newtons
Since both F1 and F2 are pulling the sphere away frpom point O, then the tension in the chord is
T = F1 +F2 = 273.46 +2000 = 2273.46 newtons. -----answer.
I'm super rusty on Physics. No practice.
Just a quick note on the English here. The poster meant to say "cord" not "chord." (Music)