2

Consider the following situation: "A thin stick of mass m and length l is positioned vertically, with its tip on a frictionless table. It is released and allowed to fall."

This question may also be asked: "What is the speed of the center of mass of the stick just before it hits the table?"

You can say initial potential energy is equal to final kinetic energy conservation of energy:

first

Then you relate translational and rotational velocity and solve for translational velocity. The final answer for impact velocity comes out to this:

this

All the solutions I can find, including my own, agree with this, however they do not all agree with on that relation between translational and rotational velocity. Most solutions use the first one below, but I believe it should be the second (theta is the angle off the horizontal):

this

that

This doesn't make a difference for the impact velocity since the two expressions become equal as this, but it's still a pretty nontrivial mistake. I thought thing only described circular motion of the center of mass. There is a similar problem in which one end of a stick is constrained to a hinge on a wall such that one of its ends can only slide up and down. In this problem the center of mass does undergo circular motion and I believe that relation would be accurate, but not in the problem of just falling straight down. Am I misunderstanding something?

Edit: I came up with my expression because the stick must fall straight down -- there is no initial horizontal velocity and there are no horizontal forces.

thing

partyphysics
  • 121

2 Answers2

1

$v_{CM} = \frac{l}{2} \omega$ gives the speed of the stick's center of mass relative to the point of contact between the stick and ground, which is not stationary with respect to the surface in this instance (since we are given a frictionless surface). In this case, since friction is not present, there are no horizontal forces on the stick, meaning its center of mass will move only in a vertical line. And if you consider the triangle formed by that vertical line, the lower half of the stick, and the surface, it can be seen that the ratio of the height of the center of mass $h$ to the lower half of the stick $\frac{l}{2}$ is equal to $\cos{\theta}$, that is $\frac{h}{\frac{l}{2}}=\cos{\theta}$. Differentiation wrt time gives, $\frac{\dot{h}}{\frac{l}{2}}=\cos{\theta} \, \dot{\theta}$, since $l$ is constant, and once rearranged, that is $v_{CM}=\frac{l}{2}\omega \cos{\theta}$.

JoDraX
  • 189
0

I would say your formula is correct, but, as @sammy gerbil notes, the second formula is good enough for the purpose of solving the problem, as it is correct at the moment when the stick hits the ground. His remark about the speed of your formula being negative is correct, but that does not make your formula incorrect: both the speed and the angular velocity are negative (both $y_{cm}$ and $\theta$ decrease).

akhmeteli
  • 27,962