Solving for Tangential Speed of a Uniform Bar

[wlvanbesien]

A 0.200 m uniform bar has a mass of 0.770 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 = 28.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.

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I've been struggling with this problems for quite a while now.
I know that:
x = \theta r

v _{f} = \frac{x}{t}

Beyond these two simple steps I can't figure anything out... i tried experimenting by integrating with respect to theta and x, but it still didnt work out... any help would be great. Thanks.

 

[jdavel]

wlvanbiesen,

The first way I thought of was to consider the torques (there are two of them) that act on the rod. The rod has a moment of inertia about its pivot point, so you'd integrate angular acceleration over the time it takes for the rod to go from vertical to horizontal...That looked like a mess!

But then I thought of energy conservation! Give that a try; I think it will work.

 

[krab]

I would use energy. You know the potential gravitational energy before and after and the potential spring energy after. The left over energy is the kineti energy (1/2)I omega^2, where I is the moment of inertia of the bar.

edit: Sorry, I didn't see that jdavel had already suggested this.

 

[wlvanbesien]

Nobody in the class can figure this problem out? Can anyone show me how to solve this exactly?

 

[jdavel]

Did you try what krab suggested?

How much total energy is there at the beginnning? What are all the components of that energy? What do you know about the total energy at the end? What are the components of THAT energy?

If you give us answers to all those questions, and you're still stuck, we'll give you more help.