Centre Of Mass

Let the center of the larger circular plate B be the origin (0, 0).
The radius of plate B is .
The area of plate B is .
Let the mass per unit area of the uniform plate be .
The mass of plate B is .
The center of mass of plate B is at the origin (0, 0).
The circular plate A of radius is removed from one edge of plate B.
The center of plate A is at a distance from the center of plate B along the x-axis.
So, the center of mass of the removed portion A is at .
The area of plate A is .
The mass of plate A is .
Let the remaining portion be C.
The mass of the remaining portion is .
Let the center of mass of the remaining portion C be at .
Using the formula for the center of mass of a system of particles: (for the x-coordinate) Due to symmetry about the x-axis, the y-coordinate of the center of mass is .
The distance of the center of mass of the remaining portion from the center of the plate B is .

Since no external horizontal forces act on the system, the center of mass of the system remains stationary. Initially, the center of mass is at rest, and as the boys move towards each other, the boat moves in the opposite direction to conserve momentum. The mass of the boat and the boys results in the boat moving a small distance compared to the boys. The total distance moved by the boat and boys is 50 cm.

Let the initial positions of the blocks be: m, kg m, kg m, kg Initial centre of mass .
Total mass kg.
m.
New positions after displacement (before B is moved): Block A is displaced towards B by 2 m: m.
Block C is displaced towards B by 3 m: m.
Let block B be moved by a distance .
Its new position is .
The new centre of mass .
We want the centre of mass to not change, so m.
m.
Since is positive, block B should be moved 1 m in the positive x-direction.
In the setup, positive x-direction is from A towards C.
So, block B should be moved 1 m towards block C.
This matches option (4).

Step 1: Defining the Radius of Gyration The radius of gyration is related to the moment of inertia and mass by: For a thin uniform rod rotating about its centre, the moment of inertia is: Thus, Conclusion Thus, the correct answer is:

Let the initial positions of the blocks be: m, kg m, kg m, kg Initial centre of mass .
Total mass kg.
m.
New positions after displacement (before B is moved): Block A is displaced towards B by 2 m: m.
Block C is displaced towards B by 3 m: m.
Let block B be moved by a distance .
Its new position is .
The new centre of mass .
We want the centre of mass to not change, so m.
m.
Since is positive, block B should be moved 1 m in the positive x-direction.
In the setup, positive x-direction is from A towards C.
So, block B should be moved 1 m towards block C.
This matches option (4).

Step 1: Given: Step 2: Angular momentum Step 3: Rate of change of angular momentum: % Final Answer

Angular velocity
,

Step 1: Defining the Radius of Gyration The radius of gyration is related to the moment of inertia and mass by: For a thin uniform rod rotating about its centre, the moment of inertia is: Thus, Conclusion Thus, the correct answer is:

Step 1: Divide the L-shape into two rectangles.
Let’s denote:
Block A: vertical rectangle of dimension
Block B: horizontal rectangle of dimension
Assume uniform density. Total mass = 3 kg. So, each block has:
Area of A = units,
Area of B = unit,
Total area = 3 units.
So, mass of A = 2 kg, mass of B = 1 kg.
Step 2: Coordinates of individual centroids
Centroid of A is at
Centroid of B is at
Step 3: Use the center of mass formula:
So, the center of mass =

Given: Mass of body 1, , velocity (north)
Mass of body 2, , velocity (east)
Velocity of centre of mass components: Assigning coordinate axes: north as y-axis, east as x-axis, we get: Calculate components: Magnitude of :

Step 1: Relate radius of gyration to moment of inertia
The radius of gyration about an axis is given by , where is the moment of inertia and is the mass. For a thin circular ring of radius about an axis through its center and perpendicular to its plane, , so . Given cm, the radius cm. Step 2: Find the radius of gyration about the diameter
For a thin circular ring, the moment of inertia about a diameter (using the perpendicular axis theorem) is . The radius of gyration about the diameter is . Substitute cm: cm. Step 3: Match with options
The radius of gyration about the diameter is 10 cm, matching option (1).