Let's break down each of these physics problems step by step, ensuring we understand the concepts involved and how to arrive at the answers logically.

Finding the Center of Mass of a Circular Plate

In the first problem, we have a circular plate with a diameter of 56 cm, and a circular part with a diameter of 42 cm is removed from one edge. To find the center of mass of the remaining part, we can use the concept of the center of mass of composite shapes.

Step-by-Step Calculation

• The radius of the original plate is 28 cm (56 cm / 2).
• The radius of the removed part is 21 cm (42 cm / 2).
• We can consider the original plate as a whole and the removed part as a negative mass.

To find the center of mass, we can use the formula:

x_{cm} = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2}

Here, m_1 is the mass of the original plate, x_1 is its center (0 cm), m_2 is the mass of the removed part, and x_2 is the distance from the center of the original plate to the center of the removed part (which is 28 cm - 21 cm = 7 cm).

After calculating, we find that the center of mass of the remaining part is located 9 cm from the center of the original plate.

Rotational Dynamics of a Ring and a Disc

Next, we have a ring and a disc rotating with the same kinetic energy. When a retarding torque is applied to the ring, it stops after making 'n' revolutions. The question asks how many revolutions the disc will make under the same conditions.

Understanding Kinetic Energy

The kinetic energy (KE) for a rotating object is given by:

KE = \frac{1}{2} I \omega^2

Where I is the moment of inertia and ω is the angular velocity. Since both objects have the same kinetic energy, we can set up the equation:

I_{ring} \omega_{ring}^2 = I_{disc} \omega_{disc}^2

When the torque is applied, the angular deceleration will be the same for both objects, leading to the conclusion that both will stop after the same number of revolutions, which is 'n'.

Rotational Inertia of Two Wheels

In this scenario, we have two wheels connected by a belt, where the radius of the larger wheel is three times that of the smaller one. We need to find the ratio of their rotational inertia when both have the same angular momentum.

Calculating Rotational Inertia

The moment of inertia for a solid disc is given by:

I = \frac{1}{2} m r^2

For the larger wheel (radius = 3r), its moment of inertia becomes:

I_{large} = \frac{1}{2} m (3r)^2 = \frac{9}{2} m r^2

For the smaller wheel (radius = r), its moment of inertia is:

I_{small} = \frac{1}{2} m r^2

The ratio of their moments of inertia is:

\frac{I_{large}}{I_{small}} = \frac{9/2 \, m r^2}{1/2 \, m r^2} = 9

Thus, the ratio of the rotational inertia of the larger wheel to the smaller wheel is 9.

Coins on a Gramophone Record

In the next problem, we have a coin placed on a gramophone record that rotates at 45 rpm and flies off at 50 rpm. If two coins are stacked on top of each other, we need to determine the speed at which both will fly off.

Understanding Centripetal Force

The force keeping the coins on the record is centripetal force, which depends on the rotational speed. Since both coins experience the same conditions, they will fly off at the same rotational speed of 50 rpm, regardless of their stacking.

Moment of Inertia of Two Discs

Now, we have two discs: disc 'X' with radius 'R' and thickness 't', and disc 'Y' with radius '4R' and thickness 't/4'. We need to find the relationship between their moments of inertia.

Calculating Moments of Inertia

The moment of inertia for a disc is given by:

I = \frac{1}{2} m r^2

For disc 'X', the moment of inertia is:

I_x = \frac{1}{2} m_x R^2

For disc 'Y', we need to consider its mass:

m_y = \text{density} \times \text{volume} = \text{density} \times \pi (4R)^2 \frac{t}{4}

Thus, the moment of inertia for disc 'Y' becomes:

I_y = \frac{1}{2} m_y (4R)^2 = \frac{1}{2} \left(\text{density} \times \pi (4R)^2 \frac{t}{4}\right) (16R^2) = 64 I_x

Therefore, we find that I_y = 64 I_x.

Force Exerted by a Liquid in a Rotating Tube

Lastly, we have a tube filled with an incompressible liquid of mass 'M', rotated at one end with uniform angular speed 'ω'. We need to find the force exerted by the liquid at the other end.

Applying Centrifugal Force Concepts

When the tube rotates, the liquid experiences a centrifugal force. The force exerted at the other end can be derived from the formula:

F = M \cdot a =