A composite object is made of two disc-shaped sections A and B, as shown in the figure, and is rotating about an axis through the center of disc A. The masses and the radii of discs A and B are respectively 2 kg and 0.2 kg, and 25 cm and 2.5 cm. What is the moment of inertia of the composite object?

The moment of inertia of an object is a measure of its resistance to rotational motion about a particular axis. It depends on both the mass of the object and its distribution of mass around the axis of rotation.

For a disc rotating about an axis through its center, the moment of inertia is given by the formula I = 0.5mr^2, where m is the mass of the disc and r is its radius.

Let's calculate the moment of inertia for each disc separately:

For disc A: I_A = 0.5m_Ar_A^2 I_A = 0.52 kg(25 cm)^2 I_A = 0.52 kg(0.25 m)^2 I_A = 0.52 kg0.0625 m^2 I_A = 0.0625 kg*m^2

For disc B: I_B = 0.5m_Br_B^2 I_B = 0.50.2 kg(2.5 cm)^2 I_B = 0.50.2 kg(0.025 m)^2 I_B = 0.50.2 kg0.000625 m^2 I_B = 0.0000625 kg*m^2

However, disc B is not rotating about its own center, but about the center of disc A. Therefore, we need to use the parallel axis theorem, which states that the moment of inertia of a body about an axis parallel to and a distance d away from an axis through its center of mass is given by I = I_cm + m*d^2, where I_cm is the moment of inertia about the center of mass axis, m is the mass of the body, and d is the distance between the two axes.

In this case, d is the radius of disc A, so:

I_B = I_cm_B + m_Bd^2 I_B = 0.0000625 kgm^2 + 0.2 kg*(0.25 m)^2 I_B = 0.0000625 kgm^2 + 0.0125 kgm^2 I_B = 0.0125625 kg*m^2

Finally, the total moment of inertia of the composite object is the sum of the moments of inertia of the two discs:

I_total = I_A + I_B I_total = 0.0625 kgm^2 + 0.0125625 kgm^2 I_total = 0.0750625 kg*m^2

So, the moment of inertia of the composite object is 0.0750625 kg*m^2.