The disk has mass 10kg and radius 0.25m. The rod has mass 5kg, length 0.5m and radius 0.05m. The rod is split down the middle and is hinged so that its two halves can lay flat against the disk. Suppose that the disk and rod are initially rotating at 3.0rad/s around the axis that is shown in the first diagram. What is the angular speed of the disk and rod after the rod flies open?

Please do show how you do it. TQ

To solve this problem, we can use the principle of conservation of angular momentum. According to this principle, the total angular momentum before the rod flies open is equal to the total angular momentum after the rod flies open.

The angular momentum of an object is given by the product of its moment of inertia and its angular velocity.

First, let's find the moment of inertia of the disk and rod system before the rod flies open.

The moment of inertia of a disk about its axis of rotation is given by the formula:

I_disk = (1/2) * m_disk * r_disk^2

where m_disk is the mass of the disk and r_disk is the radius of the disk.

Substituting the given values, we have:

I_disk = (1/2) * 10kg * (0.25m)^2
= 0.3125 kg*m^2

The moment of inertia of a rod about its center and perpendicular to its length is given by the formula:

I_rod = (1/12) * m_rod * L_rod^2

where m_rod is the mass of the rod and L_rod is the length of the rod.

Substituting the given values, we have:

I_rod = (1/12) * 5kg * (0.5m)^2
= 0.0521 kg*m^2

The initial total moment of inertia, I_initial, is the sum of the moments of inertia of the disk and the rod:

I_initial = I_disk + I_rod
= 0.3125 kg*m^2 + 0.0521 kg*m^2
= 0.3646 kg*m^2

The initial angular momentum, L_initial, is given by the product of the initial total moment of inertia and the initial angular velocity:

L_initial = I_initial * ω_initial

where ω_initial is the initial angular velocity.

Given ω_initial = 3.0 rad/s, we have:

L_initial = 0.3646 kg*m^2 * 3.0 rad/s
= 1.0938 kg*m^2/s

Now, let's find the moment of inertia of the disk and rod system after the rod flies open.

When the rod flies open, it becomes a separate rotating object from the disk. Since the rod is hinged at its center, it can rotate freely. Hence, the moment of inertia of the rod after it flies open will be the same as its initial moment of inertia: I_rod = 0.0521 kg*m^2.

The angular velocity of the disk after the rod flies open is the same as the initial angular velocity: ω_disk = 3.0 rad/s.

The moment of inertia of the disk after the rod flies open, I'_disk, is given by:

I'_disk = (1/2) * m_disk * r_disk^2

Substituting the given values, we have:

I'_disk = (1/2) * 10kg * (0.25m)^2
= 0.3125 kg*m^2

The final total moment of inertia, I'_final, is the sum of the moments of inertia of the disk and the rod:

I'_final = I'_disk + I_rod
= 0.3125 kg*m^2 + 0.0521 kg*m^2
= 0.3646 kg*m^2

According to the conservation of angular momentum principle, the final angular momentum, L'_final, is equal to the initial angular momentum, L_initial:

L'_final = L_initial

Now, we can find the final angular velocity of the system after the rod flies open, ω'_final, by rearranging the equation:

L'_final = I'_final * ω'_final

ω'_final = L'_final / I'_final

Substituting the values, we have:

ω'_final = 1.0938 kg*m^2/s / 0.3646 kg*m^2
≈ 3.0 rad/s

Therefore, the angular speed of the disk and rod after the rod flies open is approximately 3.0 rad/s.