Rotational motion

1. Jun 26, 2012

Saitama

1. The problem statement, all variables and given/known data
A thin rod of length L and mass m is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed ω. In terms of these symbols and g, and neglecting friction and air resistance, find the rod's kinetic energy at it's lowest position.

2. Relevant equations

3. The attempt at a solution
I am a poor guy in rotational dynamics, please bare with me.
$$KE=\frac{1}{2}Iω^2+\frac{1}{2}Mv^2$$
If i use this formula i get the wrong answer. The answer is $\frac{ML^2ω^2}{6}$.
If i use $$KE=\frac{1}{2}Iω^2$$, i get the right answer but i don't understand why is this correct? Why i can't use the first formula? Any explanation would be very helpful.

Thanks!

2. Jun 26, 2012

Staff: Mentor

You can use the first formula. Just be sure you are taking I about the center of mass and that the other KE term is the KE of the center of mass.

Of course it's much easier to use the second formula, as long as you take I about the pivot point. Taken about that point, the rod can be considered as purely rotating.

3. Jun 26, 2012

Saitama

Thank you, i was considering the center of mass but did not calculate I about the center of mass.

Any explanation on this?

4. Jun 26, 2012

ehild

The rod rotates about a fix axis, and the KE of a body rotating around a fix axis is 1/2 Iω2, where I is the moment of inertia with respect to the axis.

At the same time, you can consider the motion as translation of the CM (along a circle of radius L/2) and rotation about the CM. In this case, you have to use the moment of inertia with respect to the CM in the expression for the rotational KE. Try to figure out the expression for the KE. It must be the same as the former one.

ehild

5. Jun 26, 2012

Saitama

Thank you for the explanation!

I have tried solving the question by both the ways and i get the same answer, i was just making sure why both of them are correct.