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Suppose there is a lever arm fixed at one end, and it is parallel to the ground. There is an object resting somewhere on top of the lever arm (the object is not attached to the lever). At the moment when the lever arm is released, does the object have any angular velocity? According to my physics teacher, the answer is yes. I know the approximation $\sin y \approx y$ for $y \approx 0$ but I fail to see how this situation (from the object's perspective) is different than just a regular free fall situation.

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  • $\begingroup$ Well since the block will start falling, it has some velocty and therefore some angular velocity about the pivot $\endgroup$ – Rijul Gupta Jan 13 '14 at 6:56
  • $\begingroup$ @rijulgupta but the block is not attached to the lever $\endgroup$ – Ovi Jan 13 '14 at 6:59
  • $\begingroup$ @rijulgupta, angular velocity is not $r \times v$! $\endgroup$ – richard Jan 13 '14 at 7:10
  • $\begingroup$ @rijulgupta But why? What difference does the lever make? If I removed the lever from under the object and just let it fall, you wouldn't say that the object has angular acceleration, right? So why does the lever being there make a difference if it is not connected in any way? $\endgroup$ – Ovi Jan 13 '14 at 7:19
  • $\begingroup$ Thank you richard I dont know what possessed me @OVI: angular velocity is $v/r$ and angular acceleration is $a/r$ in your question the lever has no role as we deal with the situation after it has been removed, yes we would say that it has angular acceleration about that point, you can calculate it $\endgroup$ – Rijul Gupta Jan 13 '14 at 7:23
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Angular velocity is $ v/ r$ while angular acceleration is $ a/r$.
Since the block has a vertical velocity(assumption :it was at the end of the rod) after the rod/lever is released, it also has an angular velocity about the point where the lever has been pivoted. If the block is near to the pivot then for some time the block will tend to slide down the lever as well as go down, but even in this case you can find out angular velocity with respect to pivot.

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  • $\begingroup$ If it is at the end of the rod, I don't think the object will slide down the lever, at least initially. From the two equations of torque, we have $Fr=I \alpha$ . Let the lever have mass $M$ and length $L$ . Then, $MgL/2=\dfrac 13 ML^2 \alpha$ . Solving for $\alpha L$ we get $3g/2$ as the initial tangential acceleration, which is greater than $g$ . However, I guess the object maybe could catch up later, but at least initially it doesn't look like it would roll on the lever if it starts at the end. $\endgroup$ – Ovi Jan 13 '14 at 7:54
  • $\begingroup$ I have written tend to slide down initially, since I did not do the calculations I did not assert whether it will actually slide down or not $\endgroup$ – Rijul Gupta Jan 13 '14 at 7:55
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I would say, from your description, that at the instant of release the answer is yes. The Instant before release the answer is no. If you think of a delta t over which you measure the positions, shrink delta t to an infinitesimal dt. Can you tell me at what point you would go from moving to not moving?

But, what level of a problem is this? It is actually much more complicated. Since the end of the arm is fixed and the arm rotates about that fixed point, the gravitational force on the portions closer to the pivot is exerting a torque on the arm. It is restrained from falling as fast as it could if unrestrained. (This is why the top of a falling tree bends so far upward. It may look like wind or air resistance, but it is the end being "whipped" downward by the torque and moving much faster than if it had fallen freely).

This means along some portion of the arm closer to the pivot your object will stay in contact with the arm since it is also restrained from free fall and it will also rotate as it lies on the arm.

If the object is near the free end of the arm, it will fall free because the arm will be torqued out from under it at a rate faster than free fall. Is your teacher addressing all this? Or is this a concept physics question about when things start to happen?

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  • $\begingroup$ Thanks for the answer. We didn't really talk about this in class but I just remembered I have done a similar problem like this in the past, and I just calculated the initial tangential acceleration to be $3g/2$ at the end of the lever. $\endgroup$ – Ovi Jan 13 '14 at 7:51

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