Conservation of Energy Question?

Question

A 2.8 m rod standing on its end is allowed to fall. The falling tip traces an ARC. Find the speed of the tip when it hits the floor.
I would like to see how it is done. The answer is supposed to be 6.4m/s
I am asking to find the velocity of the tip that is falling, so the top one.

Answer 1

The rod’s mass is concentrated at the center of the rod.

Its potential energy at the erect position is mg R where R is half the length of the rod

When it rotates to the horizontal position, the decrease of potential energy is mg R

This is converted into kinetic energy of rotation of the rod = 1/3 m V^2 where V is the velocity of the end point of the rod at the horizontal position.

{Using I = 1/3 m L^2 and K.E = ½ I V^2/ L where L is the full length of the rod}

1/3 m V^2 = mg R

V^2 = 3 g R

V = √ [3 *9.8 * 1.4}

V = 6.4156 m/s

Answer 2

Hints:

At the instant that the rod strikes the floor, the tip is moving perpendicularly wrt the floor

The rod has acquired both rotational and linear momentum.

The center of gravity of the rod (the middle) is moving downwards at exactly half of the velocity as the tip

The rate of spin of the rod can be determined by the relationship between the downward velocity of the tip compared to the middle.

Answer 3

The object goes from a state where its center of mass is 1.4 m off the ground, to 0 m off the ground, so the amount of potential energy it loses is
U = m*g*h
U = m*(9.81 m/s^2)*(1.4 m)
U = m*(13.734 (m/s)^2)

The object is effectively rotating. The moment of inertia of a rod of length L, rotating about its tip, is
I = 1/3*m*L^2
I = 1/3*m*(2.8 m)^2
I = m*(2.6133 m^2)

The kinetic energy of a rotating object is
E = 1/2*I*omega^2
where I is the moment of inertia and omega is the angular velocity.

Setting the kinetic energy equal to the initial potential energy, we have
E = U
1/2*I*omega^2 = m*g*h
1/2*m*(2.6133 m^2)*omega^2 = m*(13.734 (m/s)^2)
1/2*(2.6133 m^2)*omega^2 = 13.734 (m/s)^2
omega = 3.2302 rad/s

So, the object has a rotational velocity of 3.2302 rad/s at the moment it hits the ground. The linear velocity of the tip is then
v = omega*L
v = (3.2302 rad/s)*(2.8 m)
v = 9.045 m/s

I don't know where you got an answer of 6.4 m/s from, but I believe it is wrong. I checked my calculations several times, but it's possible I could have missed something. If anyone spots an error, let me know.