One end of a thin uniform rod, 1.00m in length, is attached to a pivot. The rod is free to rotate about the pivot without friction or air resistance. Initially, it is hanging straight down (the "6 o' clock position). Then the lower tip is given an initial horizontal speed of Vti, so that the rod rotates upward, about its pivot. Find the value of Vti so that the rod just reaches the "12 o' clock" position (straight up), coming to a momentary halt there.

The initial kinetic energy must be enough to increase the potential energy enough to let it reach the 12 O'clock position.

The initial kinetic energy is:
KE = (1/2) I w^2 = (1/2)(1/3)M L^2*(Vti/L)^2
= (1/6)M Vti^2
(I is the moment of inertia and w is angular velocity)

The potential energy change (with the CM in the middle of the rod) is
deltaPE = M g [L/2 - (-L/2)] = M g L

Therefore the requirement is
(1/6)Vti^2 = gL
Vti = sqrt(6 g L)

To find the value of Vti so that the rod just reaches the "12 o'clock" position, we need to consider the conservation of angular momentum.

The initial angular momentum (L_i) of the rod can be calculated as the product of its moment of inertia (I) and angular velocity (ω), which is given by:

L_i = I * ω

The moment of inertia of a uniform rod rotating about its pivot can be calculated as (1/3) * m * L^2, where m is the mass of the rod and L is its length.

The mass of the rod is not given in the problem statement, but we can cancel out the mass by dividing the moment of inertia on both sides of the equation.

L_i / (1/3 * m * L^2) = ω

At the "12 o'clock" position, the rod comes to a momentary halt, meaning its final angular velocity (ω_f) is zero.

So, we can rewrite the equation as:

L_f / (1/3 * m * L^2) = ω_f

Since the rod starts from rest at the "6 o'clock" position, its initial angular momentum is zero.

Thus, we have:

0 = L_f / (1/3 * m * L^2)

Rearranging the equation, we get:

L_f = 0

Now, let's consider the angular velocity at the "6 o'clock" position.

The angular velocity can be calculated as the ratio of the linear velocity (Vt) at the end of the rod to the distance (L/2) from the pivot to the end of the rod.

So, ω_i = Vt / (L/2)

Since the rod just reaches the "12 o'clock" position, its linear velocity at that point is zero.

Hence, we have:

0 = Vti / (L/2)

Solving for Vti, we get:

Vti = 0

Therefore, the value of Vti that makes the rod just reach the "12 o'clock" position while momentarily coming to a halt is zero.

To find the initial velocity required for the rod to reach the "12 o'clock" position (straight up) and momentarily stop, we can use the principle of conservation of mechanical energy. At the "6 o'clock" position, the rod is at its lowest point, with potential energy equal to zero. As the rod swings up to the "12 o'clock" position, it gains potential energy in exchange for a loss of kinetic energy.

Let's go through the steps to find the required initial velocity:

1. Determine the change in height (vertical displacement) of the rod.
Since the rod goes from hanging straight down to reaching the "12 o'clock" position, its height changes by its full length, which is 1.00m.

2. Calculate the change in potential energy.
The change in potential energy is given by the equation ΔPE = m * g * Δh, where m is the mass of the rod, g is the acceleration due to gravity (9.8 m/s^2), and Δh is the change in height.

In this case, since the rod is uniform and thin, its mass can be considered to be concentrated at its center of mass. The center of mass of a uniform rod is located at its geometrical center, which is at a distance of half its length from the pivot point.

The change in potential energy is therefore ΔPE = m * g * Δh = m * g * L, where L is the length of the rod.

Substituting the given values, ΔPE = (1.00kg) * (9.8m/s^2) * (1.00m) = 9.8 Joules.

3. Calculate the change in kinetic energy.
At the "6 o'clock" position, the rod has no kinetic energy since it is momentarily at rest. At the "12 o'clock" position, the rod also has no kinetic energy because it momentarily stops. Therefore, the change in kinetic energy is zero.

4. Apply the conservation of mechanical energy principle.
According to the principle of conservation of mechanical energy, the total mechanical energy E of the system remains constant. The equation for mechanical energy is given by E = KE + PE, where KE is the kinetic energy and PE is the potential energy.

Since the change in kinetic energy is zero, the initial kinetic energy must be equal to the final potential energy: KE_initial = -PE_final.

Substituting the values, KE_initial = -ΔPE = -9.8 Joules.

5. Calculate the initial velocity.
The initial kinetic energy can be written as KE_initial = (1/2) * m * Vti^2, where m is the mass of the rod and Vti is the initial velocity.

Solving for Vti, we have Vti^2 = (2 * KE_initial) / m.

Substituting the values, Vti^2 = (2 * -9.8 J) / 1.00 kg.

Taking the square root of both sides, we find Vti ≈ -9.9 m/s.

The negative sign indicates that the initial velocity is in the opposite direction of the gravitational force (upward).

Therefore, the required initial velocity for the rod to reach the "12 o'clock" position and momentarily stop is approximately 9.9 m/s upward.