Moment Of Inertia

To solve this problem, we need to ensure that the rod is in rotational equilibrium. Therefore, the sum of the torques around the pivot point (the wedge at the 40 cm mark) must be zero.

Let's analyze the situation:

• The length of the rod is , and its center of mass is at the mark.
• The rod's mass is . Thus, its weight is .
• A mass of is suspended at the mark, creating a force of .
• An unknown mass is suspended at the mark.

We will calculate the torques around the pivot point ( mark). The torque caused by a force at a distance from the pivot is given by:

Let's calculate the torque for each force:

1. Torque due to the 2 kg mass at 20 cm mark:
   • Distance from pivot =
   • Torque = (Clockwise)
2. Torque due to the rod's weight:
   • Distance from pivot =
   • Torque = (Anticlockwise)
3. Torque due to the unknown mass at 160 cm mark:
   • Distance from pivot =
   • Torque = (Anticlockwise)

For rotational equilibrium, the sum of the clockwise torques must equal the sum of anticlockwise torques:

Solving for :

Therefore, the mass required for the rod to be in equilibrium is .

The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is given by the formula:

$ is the mass of the rod, and is the length of the rod.

Given:

• g cm
• g

Substituting the given values into the formula:

\) :

\) :

\) $

Thus, the length of the rod is approximately 8.5 cm.

To find the ratio of the moment of inertia of the smaller sphere to that of the rest of the larger sphere about the Y-axis, we begin by calculating the moment of inertia for both parts.

The moment of inertia for a solid sphere about an axis passing through its center is given by:

Formula:

where is the mass and is the radius of the sphere.

Step 1: Calculate the moment of inertia of the smaller sphere.

The radius of the smaller sphere is . Let the density of the material be , then the mass of the smaller sphere is:

The moment of inertia of the smaller sphere about its own diameter is:

Step 2: Calculate the moment of inertia of the larger sphere and the rest of the sphere.

The radius of the larger sphere is . Hence, the mass of the larger sphere is:

The moment of inertia of the larger sphere about its diameter is:

The rest of the sphere has a mass of and its moment of inertia .

Step 3: Calculate the ratio of the moments of inertia.

Therefore, the ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere is:

Correction: There seems to be discrepancies in above calculation, re-checking will show correct ratio: