# Angular Momentum of a rotating mass 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!

#### Attachments

Merlin3189
Homework Helper
Gold Member
You don't seem to mark theta on your diagram. I assume it is the angle the chord makes with the z axis. If so, it will not be the same in both positions.
As the tension increases, the "orbit" becomes more horizontal and theta changes.

Ah that would explain it. I think i need to find theta using newton's laws. Then I can apply it to the conservation of angular momentum.

Thank You.

Merlin3189
Homework Helper
Gold Member
Remember that the reason the ball is not just spinning with the chord horizontal, is gravity. The tension in the chord is providing the opposition to gravity as well as centripetal force.