Center of Gravity: Rod Kicked Quickly - What Happens?

[bsmith2000]

Homework Statement

My friend's little brother took some physics test a week or so ago, and he was asking about some of the questions that were on it. This was one of them:
There is a rod (uniform density, mass) which is balancing on the ground on its end (It basically looks like a flag pole). Someone comes along and kicks the bottom of it really quickly - describe what happens to the center of gravity (also the geometric center) of the rod.

Homework Equations

$$F_g = m_c*g$$
$$\tau = F \times d$$

The Attempt at a Solution

I figured that kicking the bottom of the rod would cause a torque around the center of gravity, so the rod would rotate. However, you also have gravity acting on the center of gravity, so the rod would fall downwards. Therefore, the center of gravity just moves straight down until it hits the floor.

Any thoughts/critiques/suggestions?

 

[Doc Al]

To find out how the center of mass would accelerate, consider all the forces acting on the rod.

 

[bsmith2000]

To find out how the center of mass would accelerate, consider all the forces acting on the rod.

I think I know what you mean. If this was the set-up,

O ||
\|/ ||
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and the man kicked it, then the force from the kick going to the right plus the downward force of gravity would cause the center of mass to move in a "southeast" direction, correct?

 

[bsmith2000]

Does anyone have any suggestions? I am curious to know the answer.

 

[Dick]

The rod would have an initial impulse from the kick that would give the center of mass a non-zero initial velocity in addition to the angular momentum. The center of mass would follow the same parabola as a point mass with the same initial velocity (until some part of the rod hit the ground).