1. Homework Statement
When the 3.2-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force F on the cord is increased, the bob rises and then rotates around the horizontal circular path B. (picture included)
L = I ω v = rω
The Attempt at a Solution
The solution from the book goes into a long procedure with summation of forces to find the angle etc. I tried this
Since there is no moment about the z axis angular momentum is conserved about the z axis. I1ω1 = I2ω2
For a particle I= mr2 so initially, I = (3.2)(0.6sinθ)2 and after I = (3.2)(0.3sinθ)2. For ω, initially it is (0.6)(sinθ)(1.5m/s) at the end is is (0.3sinθ)(v2)
Setting these equal to each other and solving gives v=12m/s But the answer is 2m/s Not sure where im going wrong. Thank you!