# Conservation of energy

1. Oct 28, 2013

### oneplusone

1. The problem statement, all variables and given/known data

A 2 kg ball is attached to the bottom end of a length of fishline with a breaking strength of 44.5 N. The top end of the fishline is held stationary. The ball is released from rest with the line taut and horizontal (theta = 90 degrees). At what angle theta (measured w/vertical) will the fishline break?

2. Relevant equations

none i can think of

3. The attempt at a solution

$$F=\dfrac{mv^2}{l}$$ where l is the length of the wire. then from conservation:

$$mgh = \dfrac{1}{2}mv^2$$

$$mg(l \sin \theta) = \dfrac{1}{2} ( F\cdot l)$$

$$F = 2mg\sin \theta = 44.5$$

Which is incorrect as sin theta must be < 1.

2. Oct 28, 2013

### rock.freak667

At the angle θ, what would be the components of the forces radially and tangentially and in what direction would the tension T act?

After that, how would you find the centripetal force in terms of these forces?

3. Oct 29, 2013

### oneplusone

Sorry I still can't get it with your information. Any other hints/solution?

4. Oct 29, 2013

### Hollumber

Account for the weight opposing tension force.

(At the bottom, Fnet = 2mg = Tension - mg(weight), so max tension would have to = 60N for the line to never break.)

5. Oct 29, 2013

### Staff: Mentor

- You have sine and cosine reversed; note that θ is measured from the vertical.
- What you call F is the radial component of the net force on the ball, not the tension in the string.

Draw a free body diagram of the forces on the ball.

6. Oct 29, 2013

### oneplusone

Oops, is this correct now:

$$F=T-mg\cos\theta \implies T-mg\cos\theta = \dfrac{mv^2}{l}$$
Where l is the length of the line.

$$mgh = \dfrac{1}{2}mv^2\implies mg(l\cos\theta) = \dfrac{1}{2}(T-mg\cos\theta)\cdot l$$

$$2\cdot 9.8\cos\theta = \dfrac{1}{2}(44.5-2\cdot 9.8\cdot\cos\theta)$$

$$\theta = 40.82^{\circ}$$