# Conservation of Energy, Simple Harmonic Motion

1. Dec 3, 2009

### jacksonpeeble

1. The problem statement, all variables and given/known data
A 0.200-m uniform bar has a mass of 0.790 kg and is released from rest in the vertical position, as the drawing indicates. The spring is initially unstrained and has a spring constant of k = 20.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.

2. Relevant equations
Eup=mgh=Edown=.5Iw2+.5ky2

$$\sqrt{3(mgL-ky^{2}}$$

3. The attempt at a solution
$$\sqrt{.1^{2}+.2^{2}}-.1=.124$$

$$\sqrt{\frac{3(.79 kg)(9.8 m/s/s)(.2 m)-(20 N/m)(.124 m)^{2}}{.790}}=2.343 m/s$$

I think that I went about this the right way and simply made an arithmetic error or plugged in something incorrect. Any help will be greatly appreciated, as I must have this assignment completed by midnight.

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Last edited: Dec 3, 2009
2. Dec 3, 2009

### jacksonpeeble

Any suggestions?

3. Dec 4, 2009

### Redbelly98

Staff Emeritus
Just saw this, looks like it's too late.

In the energy equation, you omitted the kinetic energy of the bar's center-of-mass, (1/2)mv2. (v is the velocity of the center of the bar.)

4. Dec 4, 2009

### jacksonpeeble

Thank you, I appreciate the help anyway so that I can do the problems in the future. I was able to turn in that problem later for partial credit - a friend helped me.

:]