Smooth particle with mass of 80 grams is attached by an elastic cord from O to P. O is at the base of a circle with radius .4m. Angle theta is 60deg from horizontal. The cords unstretched length is .25m. K of the cord is 30N/m
The path the particle makes is r=0.8sin(theta)m. The constant angular velocity is 5 rad/s. An arm moves this particle.
What I am looking for is the force that the guide has on the particle. This particular one is in the horizontal plane so gravity is out. How do you do this?
The Attempt at a Solution
Force that the cord produces is (0.69m-.25m)30N/m or 6.64 newtons
The listed answer is 7.67N. I know this has something to do with the tangential and normal forces due to the particle itself.