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acceleration at the bar tip

 
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kimjh0



Joined: 04 Dec 2010
Posts: 1
Location: Portland

PostPosted: Sat Dec 04, 2010 10:22 pm    Post subject: acceleration at the bar tip Reply with quote

I have a bar that has length L. The center of the bar is travelling along a 1/2 circle about a point O horizontally at an angular velocity Omega. The circle has a radius of R.

At the center of the bar, things are straight forward. Treating as a point;

The normal acceleration (An) is R * Omega^2 or Vt^2 / R. The tangential acceleration (At) is R * Alpha, Alpha being the angular acceleration.

Now, how do you compute the acceleration at the bar tip (at L)? Say the bar is a rigid body. I know that both normal/tangential acceleration is larger than the center of the bar just from the increase in radius at the bar tip. However, I think there is more than this.

If I have an object at the tip of the bar, the object will be thrown out. Two points on the bar are fixed. They don't change. If the bar is very long, the vector will almost be normal to the bar. The object could be launched at 90 deg to the bar, depending on the L value. The bar doesn't rotate about the center of the bar although it seems. The rotation is only caused by the entire bar rotating about the point O.

I would like to know how much greater the acceleration is at the tip of the bar. If I put an object at the two ends of the bar and the center, the object at the center stays. However, the objects at both ends move a lot more. It must be due to the increase in the acceleration, but also due to the throwing effect. I would like to come up with an equation that explains this. It must be pretty simple, but not sure how to approach.

I thought I could treat this as the "relative acceleration" problem. But, there is no relative motion between the center of the bar and the end of the bar. They are turning together. It's strictly because of the geometry.

Any help would be greatly appreciated.
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