Rotational Motion

The correct answer is 120




⇒ s = 0.3×4
= 120cm

To solve the problem, we first determine the angular acceleration α of the pulley using the relationship between torque τ and moment of inertia — τ = Iα. The tangential force F is given as a function of time t: F = (12t – 3t²) N. Since torque τ is related to the force and radius r by τ = Fr, we have:
τ = (12t - 3t2) × 1.5 = 18t - 4.5t2 Nm.
Given that I = 4.5 kg m², the angular acceleration is computed as:
α = τ/I = (18t - 4.5t2)/4.5 = 4t - t2 rad/s2.
To find when the direction reverses, set the angular velocity ω to zero and integrate the expression for α:
ω = ∫(4t - t2)dt = 2t2 - (t3/3).
The initial angular velocity is zero, and we solve for t when ω again becomes zero:
2t2 - (t3/3) = 0.
Factor the equation:
t2(2 - t/3) = 0.
Thus, solutions are t=0 or t=6. For t=0, it’s the initial condition. We use t=6. Calculate the angular displacement θ by integrating:
θ = ∫ω dt = ∫(2t2 - t3/3) dt = (2t3/3) - (t4/12) |60.
Evaluate the definite integral:
θ = [(2×63/3) - (64/12)] - [0] = (2×216/3) - (1296/12).
This simplifies to:
θ = 144 - 108 = 36 radians.
The number of full rotations N is:
N = θ/2π = 36/2π = 18/π.
This gives N = K/π, and thus K = 18. The computed value falls within the provided range of 18,18, confirming the correctness of the value.

So,

The correct option is (B):

To find the expression for the angular position of the ball, we need to integrate the given angular acceleration to obtain angular velocity and then integrate again to get the angular position. Let's go through the steps one by one:

1. Given: Angular acceleration .
2. Integration to find angular velocity:

Integrate the angular acceleration to find velocity: Given that at , , we find : Thus,

1. Integration to find angular position:

Integrate the angular velocity to find the position: Simplifying, we get: Given that at , , we find : Thus,

Therefore, the expression for the angular position of the ball is .

The correct answer is: .

The correct answer is 91




Therefore The magnitude of its angular momentum is Nm where x is

The correct answer is 10

20 – T = 2a
and 0.1 × T = 0.02 α

T = 2a
⇒ a = 5 m/sec²
So T = 10 N
tension in the cord is 10 N.

Where cm

The correct answer is (C) : 300 N

Given:

• Initial angular velocity ( ) =
• Initial moment of inertia ( ) =
• Final moment of inertia ( ) =
• Initial kinetic energy ( ) =

Step 1: Apply Conservation of Angular Momentum

Since there is no external torque, angular momentum is conserved:

Substitute , , and :

Cancel :

Step 2: Calculate the Initial Kinetic Energy

The initial kinetic energy is given by:

Substitute and :

Step 3: Calculate the Final Kinetic Energy

The final kinetic energy is given by:

Substitute and :

Simplify:

Compare to the initial energy :

Step 4: Find the Value of

The final kinetic energy is given as . Comparing this with , we find:

Final Answer:

The value of is 3.

The moment of inertia of a hollow sphere about its diameter axis is given by:

where is the radius of gyration of the hollow sphere.

The moment of inertia of a solid cylinder about its diameter axis is:

Calculating the ratio of the radii of gyration:

Hence, the value of is 67.

The problem asks for the energy dissipated when a stationary disc is placed on top of an identical rotating disc, and they eventually rotate together with a common angular velocity.

Concept Used:

This problem is an example of an inelastic rotational collision. The key principles to use are:

1. Conservation of Angular Momentum: In the absence of external torques, the total angular momentum of a system is conserved. When the second disc is gently placed on the first one, the only horizontal forces are internal (friction between the discs), so there is no net external torque about the axis of rotation. The total angular momentum before the collision is equal to the total angular momentum after the collision.

2. Rotational Kinetic Energy: The rotational kinetic energy of a body with moment of inertia rotating with angular velocity is given by:

3. Energy Dissipation: The energy dissipated in an inelastic collision is the difference between the initial kinetic energy and the final kinetic energy of the system.

4. Moment of Inertia of a Disc: The moment of inertia of a solid disc of mass and radius about an axis perpendicular to its plane and passing through its center is:

Step-by-Step Solution:

Step 1: Calculate the moment of inertia of a single disc.

Given values:

• Mass of a disc,
• Radius of a disc,

The moment of inertia of one disc is:

Since the two discs are identical, they both have the same moment of inertia, .

Step 2: Calculate the initial angular momentum and initial kinetic energy.

Initially, only the first disc is rotating with angular velocity . The second disc is at rest ( ).

Initial angular momentum:

Initial kinetic energy:

Step 3: Apply conservation of angular momentum to find the final common angular velocity.

After the collision, both discs rotate together with a common final angular velocity, . The total moment of inertia of the combined system is .

The final angular momentum is:

Equating initial and final angular momentum:

Step 4: Calculate the final kinetic energy of the system.

Final Computation & Result:

Step 5: Calculate the energy dissipated.

The energy dissipated is the loss in kinetic energy:

The energy dissipated so that both the discs continue to rotate together is 250 J.

To solve this problem, we need to determine the ratio of the kinetic energies of a ring and a solid sphere rolling down an inclined plane without slipping. Both start from rest, and their radii are identical. We can start by expressing the kinetic energy for both objects.
Kinetic energy ( ) for an object rolling without slipping consists of translational kinetic energy ( ) and rotational kinetic energy ( ).
1. Kinetic Energy of the Ring:
The total kinetic energy is given by:

For a ring, and :

2. Kinetic Energy of the Solid Sphere:
For a solid sphere, :


3. Ratio of Kinetic Energies:
The ratio of the kinetic energies of the ring to the sphere is determined as follows:

Therefore, the value of in the ratio where the kinetic energies are compared as is confirmed to be 7.

To solve this problem, we need to apply the principle of conservation of angular momentum. When no external torques are acting on a system, the angular momentum before and after an event must be the same.

Let's break down the problem step-by-step:

1. Initially, only the first disc with mass and radius is rotating with an angular velocity . The moment of inertia for a disc rotating about its center is given by .
2. The initial angular momentum is:
3. The second disc with mass and radius is placed gently on the first disc, co-axially, causing a new system to form.
4. The moment of inertia for the second disc is:
5. The total moment of inertia for the new system is:
6. By conservation of angular momentum, the total angular momentum before and after placing the second disc must be equal:
7. Thus,
8. Solving for , we get:

Therefore, the new angular velocity of the system after the second disc is placed gently is .

The correct answer is:

Conservation of Energy:
When the solid sphere and hollow cylinder roll up the incline, their initial kinetic energy (both translational and rotational) is converted into gravitational potential energy at the maximum height.

Total Initial Kinetic Energy for Each Object:
For the solid sphere:

where and .

For the hollow cylinder:

where .

Potential Energy at Maximum Height:
At maximum height, all kinetic energy is converted to potential energy:

For the solid sphere:

For the hollow cylinder:

Calculate the Ratio :

Conclusion:
The ratio is , so the value of is .

To find the additional distance traveled by the tip of the second hand compared to the minute hand in 30 minutes, we need to calculate the respective circumferences traveled by each hand and then find their difference.

1. First, calculate the circumference traced by each hand:

• The second hand has a length of 75 cm. The circumference traveled by the tip of the second hand in one complete rotation is given by: . Using , we get: .
• The minute hand has a length of 60 cm. The circumference traveled by the tip of the minute hand in one complete rotation is: . Similarly: .

1. Now, calculate the distance each hand travels in 30 minutes:

• The second hand completes one full rotation every minute, so in 30 minutes, it completes 30 rotations. The total distance traveled is: .
• The minute hand completes one full rotation every 60 minutes. In 30 minutes, it completes half a rotation. The total distance traveled is: .

1. Calculate the additional distance the second hand travels over the minute hand:

• The extra distance is given by: .
• Convert this value to meters by dividing by 100: .

Thus, the value of in meters is nearly 139.4 meters.

To solve this problem, we need to find the time taken by the rod to turn through a right angle when an impulse is applied. We'll calculate this by finding the angular velocity first and then determining the time for the rod to rotate by radians.

Step 1: Determine the Moment of Inertia

The rod is uniform, and its moment of inertia about an axis through its center perpendicular to its length is given by . For rotation about an end, we use the parallel axis theorem: where . Here, and .

First, calculate :

Then, the moment of inertia about end B:

Step 2: Find Angular Velocity

The angular impulse is given by . The impulse , leading to:

Step 3: Calculate Time to Rotate

Since the angular velocity is constant (no external torque after the impulse), the time is:

Step 4: Compare with Given Range

The time given in the question is . Thus, equate and solve for :

Implying

Upon rounding, .

Conclusion:
.

Given:

The object is rolling with velocity and angular velocity . The reference line is taken such that the potential energy at the surface of point B/C, which is smooth. Also, the relationship between velocity and angular velocity for rolling motion is given by:

Step 1: Conservation of Energy:

The principle of conservation of energy is applied: where: - and are the initial kinetic and potential energies, - and are the final kinetic and potential energies.

Step 2: Applying the energy equation:

The initial and final energies are expressed as:

Step 3: Simplifying the equation:

Simplifying the above equation:

Step 4: Finding the maximum height:

Solving for :

The given problem involves the analysis of motion on a rotating table. We need to determine the value of in the expression for the radial velocity of a steel ball leaving the table.

Consider the forces acting on the ball. As the table rotates with angular velocity , the centrifugal force acts outward on the ball. This force is given by , where is the radial distance from the center of the table.

The ball is initially placed at a distance from the center. As it moves outward, it experiences no tangential force (due to the smooth groove) and gains only radial kinetic energy due to centrifugal force. As the ball reaches the edge of the table at , its radial kinetic energy can be expressed as:

Solving the integral, we find:

Equating the kinetic energy:

Simplifying,

Thus, the final radial velocity of the ball is , and comparing with , we determine that .

This value fits perfectly within the specified range of .

To solve the problem, we need to compare the time taken by the disc to slip versus roll down an inclined plane. We start by analyzing both cases:

1. Slipping Down:
When the disc slips without rotational motion, it only accelerates due to gravity along the incline. For an incline of length and angle , the acceleration along the incline is given by:
Where is the acceleration due to gravity.
The time taken to slip down the plane is given by the equation of motion:


2. Rolling Down:
When the disc rolls, both translational and rotational motions occur. The effective acceleration is given by considering both motion dynamics, calculated as:
Where is the radius of gyration, and for a disc, , thus:

The time taken to roll, , relates as follows:

We are given that , so equating equations for , we have:


Squaring both sides and simplifying:


Thus, the value of is 3.
This calculated value correctly fits within the provided range [3,3]. Therefore, the solution is verified as accurate.

To find the maximum angular velocity that can be given to a disc without a coin slipping off, we need to consider the forces acting on the coin due to the rotation of the disc. The key force here is the centripetal force required to keep the coin in circular motion.

The centripetal force needed to keep the coin moving in a circle of radius with angular velocity is given by:

where:

• is the mass of the coin,
• is the angular velocity,
• is the distance from the center of the disc to the coin.

The frictional force between the coin and the disc must be equal to or greater than the centripetal force to prevent slipping. This frictional force is given by:

where:

• is the coefficient of friction,
• is the acceleration due to gravity.

For the coin not to slip, the frictional force must be equal to or greater than the centripetal force:

Dividing throughout by (assuming ) and solving for , we get:

Thus, the maximum angular velocity that can be given to the disc without the coin slipping is:

Therefore, the correct answer is: .

To determine the height by which the outer rail should be raised, we use the concept of railway banking, which involves balancing the centripetal force needed to negotiate a curve with the gravitational force on the inclined plane. The formula for the height of the outer rail is given by:

where:

• is the velocity of the train ( ),
• is the distance between the rails ( ),
• is the acceleration due to gravity ( ),
• is the radius of the curve ( ).

Plugging these values into the formula, we have:

Simplifying:

Converting this into centimeters, we find:

Thus, the height by which the outer rail should be raised is 5.4 cm.

Therefore, the correct answer is 5.4 cm.

The problem asks for the acceleration of the center of a solid sphere that is rolling without slipping on a rough horizontal plane. A tangential force is applied to the highest point of the sphere.

Concept Used:

To solve this problem, we will use the equations for linear and rotational motion of a rigid body, along with the condition for rolling without slipping.

1. Newton's Second Law for Linear Motion: The net external force on the body equals its mass times the acceleration of its center of mass.

2. Newton's Second Law for Rotational Motion: The net external torque about the center of mass equals the moment of inertia about the center of mass times the angular acceleration.

3. Moment of Inertia of a Solid Sphere: For a solid sphere of mass m and radius r, the moment of inertia about its center is:

4. Condition for Rolling without Slipping: The linear acceleration of the center of mass is related to the angular acceleration by:

Step-by-Step Solution:

Step 1: Identify the forces and set up the equations of motion.

Let be the acceleration of the center of the sphere, be its angular acceleration, and be the force of static friction acting at the point of contact. The applied force is and the mass is .

The applied force tends to cause the sphere to accelerate to the right. It also creates a clockwise torque that would cause the point of contact to slip backward. To prevent this, the static friction force must act in the forward (right) direction.

The equation for linear motion in the horizontal direction is:

Step 2: Set up the equation for rotational motion.

We take torques about the center of mass. The applied force creates a clockwise torque of magnitude . The friction force creates a counter-clockwise torque of magnitude . Taking the direction of angular acceleration (clockwise) as positive, the net torque is:

Using the rotational form of Newton's second law, :

Step 3: Use the condition for rolling without slipping.

We substitute and into equation (2):

Dividing by r on both sides, we get:

Step 4: Solve the system of linear equations for the acceleration .

We now have two equations with two unknowns ( and ):

From (1):

From (3):

Adding these two equations together eliminates :

Solving for :

Final Computation & Result:

Substitute the given values and into the expression for acceleration:

The acceleration of the center of the sphere is 3.5 m/s². This corresponds to option (1).

Let the moment of inertia of the rod about the axis passing through its center and normal to its length be , where is the mass and is the length.
Now, the rod is cut into two equal parts, each having mass and length .
Each part has a moment of inertia .
For the cross shape, the total moment of inertia will be the sum of the moments of inertia of the two parts, considering the distance from the center of the rod.
After using the parallel axis theorem, we get: Thus, the correct option is .

The given equation is:

Substitute the given values:

Simplify to find :

Now, solving for :

Thus, the value of is:

And the distance is:

To solve the problem, we need to apply the principle of conservation of angular momentum. Initially, the sphere has mass , radius , and is rotating with angular velocity . The moment of inertia of a solid sphere is given by .

The initial angular momentum is:

.

As the sphere loses mass uniformly with no change in shape, its density remains constant. Thus, when the radius is , its new mass can be determined using the density ratio since :

Original density .

New density when radius becomes is also .

Setting the original and new densities equal, solve for :

.

Therefore, .

The moment of inertia at the new state is:

.

.

By conservation of angular momentum :

.

.

Simplifying, .

Thus, the computed value of is 32, which falls within the provided range of (32,32).

The work done by the force is given by: This is the change in kinetic energy of the wheel: Using where and : Now equating the work done to the change in kinetic energy:

To solve the problem of finding the moment of inertia of a ring formed from a rod, we start with understanding the distribution of mass and geometry:

1. The rod of linear mass density and length is bent to form a ring of radius .
2. The length of the rod is equal to the circumference of the ring: . This implies:
   
3. The total mass of the rod (thus the ring) is .
4. The moment of inertia of a ring about its diameter is given by: .
5. Substitute the values of and :

1. Simplify the expression:

Thus, the moment of inertia of the ring about any of its diameters is , which corresponds to the correct answer.

To determine the moment of inertia (I) of the wheel, we start by using the relationship between torque (τ), moment of inertia (I), and angular acceleration (α). The equation is:

Given the force applied (F) and the radius (r) of the wheel, the torque can be calculated as:

Substituting the given values, we have:

Next, use this torque in the relation to solve for the moment of inertia (I):

Thus, the moment of inertia of the wheel is . This result fits exactly within the given range (1, 1).

The moment of inertia of the original disc is:

Explanation: The first term, , represents the moment of inertia of a solid disc about an axis perpendicular to the disc and passing through its center. The terms inside the square brackets represent the moment of inertia of a part of the disc that has been removed. The expression relates to the mass and radius of a removed section. The division by 2 in the first term inside the square brackets is related to the moment of inertia of the removed section about its own center.

Step 1: Simplify the terms inside the square brackets

Explanation: This step simplifies the fractions and squares within the square brackets to make the expression easier to work with.

Step 2: Combine the terms inside the square brackets Explanation: This step combines the simplified terms within the square brackets into a single fraction.

Step 3: Substitute the combined term back into the original equation

Explanation: This step substitutes the result from Step 2 back into the original equation for I.

Step 4: Find a common denominator and simplify

Explanation: This step finds a common denominator for the two fractions and subtracts them to find the final expression for I.

Final Result:

We are asked to identify the correct expressions for the torque acting on a body from the given options.

Concept Used:

Torque is the rotational analogue of force. It can be defined in various equivalent forms as follows:

Also, since linear momentum and , we can express torque in different forms using these relations. For rotational motion, the torque can also be expressed as:

Step-by-Step Solution:

Step 1: Check Option A:

This is incorrect because torque is the time derivative of angular momentum, not its cross product with position vector.

Step 2: Check Option B:

We know . Therefore, taking derivative with respect to time:

Hence, Option B is correct.

Step 3: Check Option C:

Since , this becomes:

Therefore, Option C is correct.

Step 4: Check Option D:

This is the rotational form of Newton's second law, valid for rigid body rotation about a fixed axis. Hence, Option D is correct.

Step 5: Check Option E:

This is the fundamental definition of torque. Hence, Option E is correct.

Final Computation & Result:

The correct expressions for torque are Options B, C, D, and E.

Final Answer: B, C, D and E Only