A uniform rod of length L and mass M is held vertically with one end resting on the floor as shown below. When the rod is released, it rotates around its lower end until it hits the floor. Assuming the lower end of the rod does not slip, what is the linear velocity of the point one-seventh the length of the rod away from its lower end when it hits the floor? (Enter the magnitude. Use any variable or symbol stated above along with the following as necessary: g.)

To find the linear velocity of the point one-seventh the length of the rod away from its lower end when it hits the floor, we can use the principle of conservation of angular momentum and the principle of conservation of energy.

First, let's define some variables:
- L: length of the rod
- M: mass of the rod
- v: linear velocity of the point one-seventh the length of the rod away from its lower end when it hits the floor

Now, let's find the angular velocity of the rod when it hits the floor. We can use the conservation of angular momentum.

The angular momentum of the rod when it's initially held vertically is zero because there is no angular velocity.

When the rod rotates around its lower end, its moment of inertia changes, resulting in an angular velocity.

The moment of inertia of a uniform rod rotating about one end is (1/3) * M * L^2. Therefore, when the rod hits the floor, its angular momentum is given by:

(1/3) * M * L^2 * ω = M * L * v'

Where ω is the angular velocity and v' is the linear velocity at the end of the rod when it hits the floor.

Simplifying the equation, we have:

ω = 3v' / L

Now, let's find the velocity of the center of mass of the rod when it hits the floor. We can use the conservation of energy.

The potential energy of the rod when held vertically is M * L * g * 0 (height is zero).

When it hits the floor, the potential energy is zero. The kinetic energy can be calculated as the kinetic energy of the center of mass plus the rotational kinetic energy.

The kinetic energy of the center of mass is (1/2) * M * v^2, and the rotational kinetic energy is (1/2) * (1/3) * M * L^2 * ω^2.

Setting the initial potential energy equal to the final kinetic energy, we have:

M * L * g * 0 = (1/2) * M * v^2 + (1/2) * (1/3) * M * L^2 * (3v'^2 / L^2)

Simplifying the equation, we get:

0 = (1/2) * M * v^2 + (1/2) * M * v'^2

Now, let's solve for v':

(1/2) * M * v'^2 = - (1/2) * M * v^2
v'^2 = - v^2
v' = - v

Since velocity cannot be negative, we take the magnitude:

|v'| = |v| = |v_one_seventh|

Therefore, the linear velocity of the point one-seventh the length of the rod away from its lower end when it hits the floor is equal to the magnitude of the linear velocity at the end of the rod, |v|.