Does the potential energy change if we take a different point on a rod? [high school physics]

[Kevin248]

Hi all, I'm doing rotational motion in school right now, and I had a question on potential energy.

If we have a vertical rod tipping over in place (so no translational motion, only rotational motion), and the point of motion we're looking for the velocity of is the tip of the rod (at the very top), then the potential energy before it tips would be (mg)(L/2), with L being the height of the rod.

If we were looking for the velocity of the point on the center of the rod instead of the tip, then would the potential energy then become (mg)(L/4)? And the energy equation is (mg)(L/4) = (1/2)(1/3)(m)(L/2)2 (W)2 ?

Sorry if I'm not wording this right. I can try providing a drawing if someone needs

[panb]

Hi all, I'm doing rotational motion in school right now, and I had a question on potential energy.

If we have a vertical rod tipping over in place (so no translational motion, only rotational motion), and the point of motion we're looking for the velocity of is the tip of the rod (at the very top), then the potential energy before it tips would be (mg)(L/2), with L being the height of the rod.

If we were looking for the velocity of the point on the center of the rod instead of the tip, then would C then become (mg)(L/4)? And the energy equation is (mg)(L/4) = (1/2)(1/3)(m)(L/2)2 (W)2 ?

Sorry if I'm not wording this right. I can try providing a drawing if someone needs

The potential energy will be the same \( \frac{1}{2}mgl \) an it will be equal to energy of rotation (around the bottom of the rod) \( \frac{1}{2} \frac{ml^2}{3} \omega^2\) where \(\omega\) is the angular velocity.
The velovity (linear) of the midpoint \(v=\omega \frac{l}{2} \). Putting all these together we get
\(\displaystyle 0,5mgl= \frac{1}{2} \cdot\frac{ml^2}{3}\cdot \omega^2= \frac{ml^2}{6}\cdot \frac{4v^2}{l^2} \So v^2= \frac{3}{4}gl \So v= \frac{ \sqrt{3gl} }{2} \)