Finding the Speed of a Wedge and Rod System

In summary: CM of the rod since the CM is not perpendicular to the face of the wedge. If I consider the upper end of the rod and the Center then the only useful information we have is the component of velocity of upper end perpendicular to the face of the wedge .The velocities of wedge and rod perpendicular to the contact surface should be same.So Vcos60=vcos30.
  • #71
Hey Guys,
I'm a little confused. Why would some of the P.E from the center of mass of the rod be converted to Rotational Kinetic energy? The the rod is not rotating. I see the idea that as it slips down its "rotating" about the center of mass, but is it correct to imply that in our reference frame its actually rotating. If this was the case, it seems like all problems of this sort would involve a rotational kinetic energy. If a ladder was perched up on a wall and fell over, you wouldn't assume that some of the P.E is converted into rotational K.E would you (assuming the end on the ground was stationary)? Or a rod was sliding down a parabola shaped ramp? I just have a weird feeling about it. Would any of you guys care to explain?
 
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  • #72
You would have to take rotation into account in all of the examples you described. Also if you consider a ball rolling down a slope you would have to take rotation into account. The point is that you cannot just describe things based on how the CoM moves, this will in general not give you the full kinetic energy.

Of course you could consider the movement of every single mass point in an object and how it moves (including any motion from rotation), but you can just as well split the energy cleverly into rotation and translation and end up with a simpler problem.
 

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