System Of Particles Rotational Motion

Mean speed of a moving body,

Mean velocity vector for a moving body,

A ball moving in a circular arc is shown in the figure below,

Here,
Similarly,
Hence,

Velocity of a rocket at any time ,

where, speed of exhausted gases,
initial mass of the rocket
and mass of the rocket at time
Given, fuel burned rate,
Then,
As, it is given that the gravitational force is ignored and mass of rocket is reduced to th of it's initial mass i.e, .
or
So,

As we know, the rotational kinetic energy of a body is given by,

where, radius of gyration
Given, and
Hence,

Step 1: Understanding the Concept:
The centre of mass of a system of particles is a weighted average of the positions of the individual particles, where the weighting factor is their mass. For continuous bodies, we can treat them as point masses located at their individual centres of mass.
Step 2: Setting up the Coordinate System:
To simplify the problem, let's place the origin of our coordinate system (x=0) at the centre of the sphere. The problem asks for the distance from this point.
Let be the mass of the sphere and be the mass of the cylinder.
Step 3: Locating the Centres of Mass of Individual Bodies:
Sphere:
Mass kg.
Radius cm.
The centre of mass of the sphere is at its geometric centre. So, its position is .
Cylinder:
Mass kg.
Length cm.
Radius cm.
The cylinder is in contact with the sphere. The centre of the cylinder is at the midpoint of its length. The distance from the origin (centre of the sphere) to the centre of the cylinder is the radius of the sphere plus half the length of the cylinder.
Position of the cylinder's centre of mass: cm.
Step 4: Calculating the Centre of Mass of the System:
We use the formula for the centre of mass of a two-body system:
Since the origin was chosen at the centre of the sphere, this value directly gives the distance of the system's centre of mass from the centre of the sphere.
Final Answer: The distance is 24 cm, so option (C) is correct.

Step 1: State the formula for the moment of inertia of a rod about its end.
The moment of inertia ( ) of a thin uniform rod of mass and length about an axis perpendicular to the rod and passing through one of its ends is given by:

Step 2: State the definition of the radius of gyration.
The radius of gyration ( ) is defined as the distance from the axis of rotation to a point where the entire mass of the body could be concentrated without changing its moment of inertia. The relationship is:

Step 3: Equate the two expressions for the moment of inertia to find K.
The mass cancels out. Taking the square root, we get:

Step 4: Find the required ratio K:L.
Therefore, the ratio is .

Step 1: Write the formulas for translational and rotational kinetic energy.
The translational kinetic energy ( ) of an object with mass and center-of-mass velocity is: The rotational kinetic energy ( ) of an object with moment of inertia and angular velocity is:

Step 2: Use the properties of a solid sphere and rolling without slipping.
For a solid sphere, the moment of inertia about its center is . The condition for rolling without slipping is , which implies .

Step 3: Express the rotational kinetic energy in terms of m and v.
Substitute the expressions for and into the formula for :

Step 4: Calculate the required ratio.
The ratio of rotational to translational kinetic energy is . The terms cancel out. The ratio is 2:5. Note that the mass (2 kg) and radius (0.5 m) are not needed for this calculation as the ratio is independent of these values.

Step 1: Relate power, work, and kinetic energy.
The Work-Energy Theorem states that the work done ( ) on an object is equal to the change in its kinetic energy ( ). Power ( ) is the rate at which work is done, . Therefore, the total work done over a time interval is the integral of power.

Step 2: Calculate the work done on the body from t=0 to t=3s.
We integrate the given power function .

Step 3: Apply the Work-Energy Theorem to find the final velocity.
We are given that the initial velocity is zero, so the initial kinetic energy . The final kinetic energy is , where kg.

Step 4: Solve for the final velocity .

Let be the radius of gyration and be the moment of inertia. The relationship is .
We need to find the moment of inertia for both a ring and a disc about a tangential axis in their plane.
Step 1: Moment of Inertia of a thin circular ring.
The moment of inertia of a ring about a diameter is .
Using the parallel axis theorem, the moment of inertia about a tangential axis in its plane (which is parallel to a diameter and at a distance R) is:
.
The radius of gyration of the ring is .
Step 2: Moment of Inertia of a circular disc.
The moment of inertia of a disc about a diameter is .
Using the parallel axis theorem, the moment of inertia about a tangential axis in its plane is:
.
The radius of gyration of the disc is .
Step 3: Find the ratio of the radii of gyration.
.
We are given the ratio as , which is .
Equating the two ratios: .
Squaring both sides: .
Solving for K: .

This problem involves motion with non-constant acceleration due to air resistance. However, we are not given the law for air resistance. The options suggest that we can find an answer using the basic kinematic equations, which implies we should use an average acceleration or relate the average velocity.
Let's use the kinematic equation that relates displacement, initial velocity, final velocity, and time, assuming a constant (average) acceleration:
.
Here, the displacement is m.
The initial velocity is m/s (since it's dropped).
The final velocity is m/s.
We need to find the time .
.
.
s.
s.
This value is approximately 2.2 s, which matches option (B). This approach assumes the acceleration is constant, which isn't true with air resistance, but it's the most reasonable interpretation given the limited information and options.

Let's analyze the torques acting on the metre scale when it is balanced. The pivot is at the 45 cm mark.
Forces creating clockwise torque (tending to rotate right): 1. Weight of the right plate: Mass g at the 100 cm mark. Lever arm = cm. 2. Weight of the vegetables: Mass (unknown) at the 100 cm mark. Lever arm = 55 cm. 3. Weight of the right part of the scale: The scale is uniform, so its center of mass is at 50 cm. The right part (from 45 cm to 100 cm) has length 55 cm and mass g. Its center of mass is at cm. Lever arm = cm.
Forces creating counter-clockwise torque (tending to rotate left): 1. Weight of the left plate: Mass g at the 0 cm mark. Lever arm = cm. 2. The added weight: Mass g at the 0 cm mark. Lever arm = 45 cm. 3. Weight of the left part of the scale: The left part (from 0 cm to 45 cm) has length 45 cm and mass g. Its center of mass is at cm. Lever arm = cm.
Alternatively, the center of mass of the entire 100g scale is at 50 cm. Its lever arm is cm, creating a clockwise torque.
Let's use the second method for the scale's torque. Principle of moments: Clockwise torque = Counter-clockwise torque.
Torque\textsubscript{cw} = (Torque from right plate) + (Torque from vegetables) + (Torque from scale's CoM) . .
Torque\textsubscript{ccw} = (Torque from left plate) + (Torque from added weight) .
Equating the torques: .
.
g.
The measured weight of the vegetables is 200 g. The weight used for balancing is 300 g.
The error is the difference between the true weight of the vegetables and the weight used to measure them.
Error = . However, error is typically defined as (Measured Value - True Value) or the discrepancy in what the scale reads. The scale reads 300 g because that's the weight used. The true weight is 200 g. Error = Reading - True Value = g.

\textit{Note: The mass in the image appears to be , but this leads to an answer with . Assuming a typo and using kg allows us to match the given options.}
Step 1: Understanding the Concept:
The work done on a rotating object is equal to the change in its rotational kinetic energy. This is the Work-Energy Theorem for rotation. We need to calculate the initial and final rotational kinetic energies.
Step 2: Key Formula or Approach:
Work Done,
Moment of inertia of a disc about its central axis:
Step 3: Calculations:
Convert Angular Speeds to SI units (rad/s):
The conversion factor is: 1 rev/min = rad/s.
Initial angular speed, rad/s.
Final angular speed, rad/s.
Calculate Moment of Inertia (I):
Given: kg and m.
Calculate Work Done (W):
Now, substitute the values of I, , and into the work-energy theorem.
The terms cancel out.
Step 4: Final Answer:
The work done to increase the angular speed is 70 J. Therefore, option (B) is correct.

Step 1: Understanding the Concept:
This problem involves a one-dimensional (head-on) collision. To solve it, we need to apply two fundamental principles: the conservation of linear momentum and the definition of the coefficient of restitution.
Step 2: Setting up the Equations:
Let the initial velocities be and .
Let the masses be and .
Let the final velocities be and .
Equation from Conservation of Momentum:
Dividing by :
Equation from Coefficient of Restitution (e):
Step 3: Solving the Equations:
We have a system of two linear equations with two unknowns, and . We need to find the ratio .
From equation (2), we can express as .
Substitute this expression for into equation (1):
Now, substitute this result for back into the expression for :
Step 4: Finding the Ratio:
Now we can find the ratio of the velocities, .
This matches option (D).