Saturday, November 19, 2011
HW9 #4
I'm really stuck on this problem... I know how to find the instantaneous angular velocity and the average angular acceleration, but I don't know how to calculate the instantaneous angular acceleration. I tried a few different methods, but none of them are right. For example, I tried alpha = (v^2)/(r^2), since a = alpha*r and in this case, the acceleration is centripetal acceleration. Did anyone figure out how to do this one?
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Interesting question. Actually, you can't use the usual kinematics equations (even if you put angular quantities in for the linear ones) because the angular acceleration is going to change with the angle. You know that the angular acceleration is going to change because the torque changes, and the torque causes an angular acceleration (τ=Iα. just like F=ma).
So to find the angular acceleration you need to use τ=Iα (just like in linear problems if you can't use kinematics you use F=ma).
So first find the rotational inertia I. It is the rotational inertia of a rod, but you have to then use the parallel axis theorem I=I(cm)+mh^2 because the rod is rotating about a point some distance h from its center of mass.
Then find the torque. In this case the torque is caused by the weight (force due to gravity) of the rod itself, and that acts at the rod's center of mass. In the horizontal position, the angle between r and F is 90 degrees (the fact that the angle changes the torque tells you that the torque is changing so the angular acceleration is changing), so τ=hmg, where I used h for the distance from the pivot point to the center of mass.
Then you put it all together in τ=Iα and solve for the angular acceleration α.
Hope that helps. Good luck.
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