A pendulum is made of two discs, one with mass M and radius R and the other with mass 2M and radius R, which are separated by a massless rod. The lighter disc is pivoted through its centre by a small pin. The discs lie on the same plane and their centres are a distance L apart. Find the period for small oscillations.

A uniform lamina consists of a disc with radius R with a disc of radius r cut off from its centre, with r < R. The total mass of the object is M. Find the moment of inertia when the lamina rotates about an axis perpendicular to the lamina, thought its centre.

An object consists of two thin rods, AB with mass M and length 2L, and CD, of mass 2M and length L, joined together at right angles at their midpoints to form a cross. Let E be the point where the two rods intersect. Find the moment of inertia of the object when it rotates about the following axes of rotation:

(a) An axis along rod AB.
(b) An axis along rod CD.
(c) An axis through point E, perpendicular to the cross.
(d) An axis through point A, perpendicular to the cross.

Find the moment of inertia of an A × A × A cube with mass M, for the following axes of rotation:

(a) An axis through the midpoints of two opposing sides.
(b) An axis along one of the edges of the cube.

To find the period for small oscillations of the pendulum, we can use the concept of rotational motion and the formula of the period of a simple pendulum.

The rotational motion of the pendulum is governed by the torque acting on it. The torque due to the gravitational force tries to bring the pendulum to its equilibrium position. For small oscillations, the torque is directly proportional to the angular displacement.

Let's consider the torque due to the gravitational force acting on the pendulum. The torque τ is given by the formula:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

The moment of inertia for each disc can be calculated using the formula:

I = (1/2)MR^2

where M is the mass and R is the radius of the disc.

Since the lighter disc is pivoted through its center, the moment of inertia for that disc is:

I₁ = (1/2)MR^2

For the heavier disc, we need to consider the parallel axis theorem, which states that the moment of inertia for a body rotating about an axis parallel to an axis through its center of mass is given by:

I = I_cm + M*d^2

where I is the moment of inertia, I_cm is the moment of inertia about the center of mass, M is the mass of the body, and d is the distance between the parallel axes.

The moment of inertia for the heavier disc about its center of mass is:

I_cm = (1/2)(2M)R^2 = 2MR^2

The distance d between the two parallel axes is the distance L between the centers of the two discs.

Therefore, the moment of inertia for the heavier disc is:

I₂ = I_cm + M*d^2 = 2MR^2 + M*L^2

Now, let's consider the torque acting on the pendulum due to the gravitational force. The torque is given by:

τ = -mg*(L/2)*sinθ

where m is the mass of each disc and θ is the angular displacement of the pendulum.

For small oscillations, we can assume that sinθ ≈ θ (using the small angle approximation).

Therefore, the torque can be simplified to:

τ = -mg*(L/2)*θ

Using the formula τ = Iα and the small angle approximation, we have:

-I₂*(d²θ/dt²) = -mg*(L/2)*θ

Dividing both sides by -I₂, we get:

(d²θ/dt²) = (mg*(L/2) / I₂) * θ

Comparing this equation with the equation of motion for a simple harmonic oscillator:

(d²θ/dt²) + (ω²)θ = 0

where ω is the angular frequency, we can see that:

ω² = mg*(L/2) / I₂

The period T of a simple harmonic oscillator is related to the angular frequency ω by the formula:

T = 2π / ω

Substituting ω² = mg*(L/2) / I₂ into the formula for T, we get:

T = 2π * sqrt(I₂ / (mg*(L/2)))

Substituting the values of the moment of inertia I₁ and I₂, we have:

T = 2π * sqrt((1/2)MR^2 / (mg*(L/2))) for the lighter disc
T = 2π * sqrt((2MR^2 + ML^2) / (mg*(L/2))) for the heavier disc

Simplifying these equations further will give you the period for small oscillations of the pendulum for each disc.