Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

## Comments

Why is kinetic energy not accounted just before it hits the ground? Also, is the velocity of the center of mass the same with the end of the pole?

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

All the best,

Mr. Dychko