Rotational Motion

Tension between and is
centripetal force on

Tension between and is
centripetal force on centripetal force on


Ratio of tension is

Moment of inertia of complete disc about its center is .
Hole radius .
Moment of inertia of the removed part about center of disc: Final moment of inertia =

Given: Angular momentum So: Now, average power is:

Step 1: Initial kinetic energy: Step 2: Angular momentum conserved: Step 3: Final energy: So, fraction of energy lost:

is a scalar, and cross product requires both operands to be vectors. So is invalid — not a vector, but undefined.

For a solid cylinder rolling without slipping, total mechanical energy is conserved. Potential energy at height : At the bottom, total energy is translational + rotational: For a solid cylinder, and , so: Equating energies:

For rolling motion without slipping, the condition is that the linear velocity must match the angular velocity condition. Using the relation between sliding and rolling velocity, we find that the velocity at which the sphere starts rolling is 2.5 m/s. Thus, the correct answer is option (1).

The speed of the satellite is given by the formula: Substituting the given values: Hence, the speed of the satellite is .

Using the equations of rotational motion with uniform acceleration, we calculate the total number of rotations in the next 3 seconds. The result is 42 rotations. Thus, the correct answer is option (3).

The initial angular velocity of the flywheel is rev/minute.
Convert this to rad/s: The constant angular retardation (angular acceleration) is rads (negative because it's retardation).
The final angular velocity is rad/s (since the wheel comes to rest).
We use the first equation of rotational motion: , where is the time taken.
Substitute the given values: s The time required for the wheel to come to rest is 5 seconds.

Let kg, kg, and the distance between them m.
The center of mass (COM) lies closer to the heavier mass. Let the distance of from COM be and from COM be , then:\ , and
So, ⇒ ⇒ ⇒ m, m
Now, kg m
Converting to standard form: g cm = kg m

Moment of inertia of solid sphere:
Let mass of large sphere be , then small sphere has mass since volume ratio is
Moment of removed part:
Remaining moment:
Original moment:

Use: , where

For a disc rolling without slipping, the total kinetic energy is the sum of the translational and rotational kinetic energy. The rotational kinetic energy is half of the translational kinetic energy, so: Substitute the given value of :
Hence, the total kinetic energy of the disc is 36 J.

Given: initial angular velocity , final rad/s, s
Angular displacement
First find angular acceleration: rad/s
Now compute displacement: rad

The angular momentum of a rotating object is given by , where is the moment of inertia and is the angular speed.
For a solid cylinder rotating about its geometric axis, the moment of inertia is given by , where is the mass of the cylinder and is its radius.
Given: Angular momentum kg m s Angular speed rad s Radius cm m We can find the moment of inertia using the formula : Now, we can use the formula for the moment of inertia of a solid cylinder to find the mass : kg The mass of the cylinder is 10 kg.

We are given that the angular speed rad/s and that the body eventually comes to rest. When the body comes to rest, . Next, we need to calculate the total angular displacement (the angle through which the body rotates) before it comes to rest. The angle is given by the integral of the angular velocity: Evaluating this integral: Thus, the total angle the body rotates before coming to rest is radians.

The energy of a rotating cylinder is , where and . The energy used in heating the cylinder is related to the temperature change, and after simplifying the formula, we obtain:

Using energy conservation, the potential energy lost by the sphere is converted into kinetic energy. The total kinetic energy at the bottom is the sum of the translational and rotational kinetic energy: where .
Using , we find:
Thus, the maximum velocity is 5 m/s.

The torque applied to a rotating object is related to its angular acceleration by: where is the moment of inertia. The angular speed after time is: Since the torques are equal and applied for the same time , the ratio of angular speeds is: - Moment of inertia of a solid sphere about its central axis: - Moment of inertia of a solid cylinder about its axis of symmetry: Now, the ratio of the moments of inertia: Thus, the ratio of the angular speeds is: So, the ratio of the angular speeds is 5:8.

First, calculate the radius of the ring. The wire of length m is bent into a ring, so the circumference of the ring is: Next, calculate the mass of the ring. The linear density is kg m , so: The moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is: Substituting the values: So, the moment of inertia of the ring is 0.25 kg m .

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

where is the mass of the sphere and is the radius (half the diameter of 20 cm).

Substituting the values:

To find the moment of inertia about a tangent, we use the parallel axis theorem:

Here, because the tangent is parallel and at a distance equal to the radius from the center.

Thus,

The correct moment of inertia about the tangent to the sphere is .

Radius
For a solid sphere about diameter:

Step 1: Set up the problem
Ball's initial velocity: , angle . Horizontal component: . Person's speed: . The person catches the ball when they reach the landing point at the same time the ball lands. Step 2: Time of flight and range
Time of flight: . Range: . Person's distance: . Set : , , , , , . Step 3: Match with options
The angle 60 matches option (1).

In simple harmonic motion, total energy , where is the amplitude and is the spring constant.
At displacement , potential energy , and kinetic energy .
Given , we have .
Thus, .
The ratio of kinetic energy to total energy is .

Step 1: Moment of inertia of a disc about a tangent in its plane.
Using the parallel axis theorem, moment of inertia Step 2: Radius of gyration is defined by: Step 3: Ratio of radius to radius of gyration: So, the required ratio is

Let the initial speed be , and the angle of projection be .
The height of the ceiling is .
The ball just moves very near to the ceiling, so the maximum height of the projectile is .
The formula for maximum height is .
Given .
(since is acute for projectile motion to reach maximum height).
Then .
So, .
The length of the auditorium is the horizontal range of the projectile.
The formula for range is .
.
The length of the auditorium is .
This matches option (1).

Step 1: Kinetic Energy in Rolling Motion The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energies: Since , we substitute . Step 2: Comparing a Disc and a Ring For a solid disc: Given J, For a thin ring: Using , solving for , Conclusion Thus, the correct answer is:

Step 1: Recall the formula for apparent weight in a lift moving upward. Step 2: Substitute the values. % Final Answer

Moment of inertia of a solid cylinder about its central axis is given by: Given: Calculate:

The velocity at the maximum height is the horizontal component of the velocity. Since the angle is , we know that the vertical component at the maximum height is zero. The formula for maximum height is: where . Thus,

Let the initial speed be , and the angle of projection be .
The height of the ceiling is .
The ball just moves very near to the ceiling, so the maximum height of the projectile is .
The formula for maximum height is .
Given .
(since is acute for projectile motion to reach maximum height).
Then .
So, .
The length of the auditorium is the horizontal range of the projectile.
The formula for range is .
.
The length of the auditorium is .
This matches option (1).

The given equation of motion: Step 1: Finding Maximum Height At the peak of the trajectory, . Differentiating: Setting , Solving for , Substituting into the equation of motion: Step 2: Finding Range The projectile reaches the ground when . Setting : Solving for , Thus, the range is: Step 3: Computing the Ratio Thus, the correct answer is:

Initial velocity vector:
Final velocity vector:
Displacement vector (assuming uniform change):

Average velocity
Magnitude

- For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energy.
- Translational kinetic energy:
- Rotational kinetic energy: The moment of inertia of a solid sphere about its axis is . The angular velocity (since it rolls without slipping).
- Total kinetic energy:
- This matches option (B).

Given: Initial angular velocity \omega ₀ = 20 rad s^-1, final angular velocity \omega = 50 rad s^-1, time t = 10 s.
First, calculate the angular acceleration \alpha using \omega = \omega ₀ + \alpha t:
50 = 20 + \alpha (10), so \alpha = 3 rad s^-2.
Next, find the angular displacement \theta using \theta = \omega ₀ t + ½ \alpha t²:
\theta = (20)(10) + ½(3)(10)² = 200 + 150 = 350 radians.
Finally, convert \theta to rotations:
Number of rotations = \theta / (2\pi ) = 350 / (2\times 3.14) ≈ 55.73 rotations.

Total mass
x-momentum \quad y-momentum
, \quad
Magnitude:

The time period of a spring-mass system is given by , where is the mass and is the spring constant.
Given: N m , g = 0.65 kg.
So, seconds.
Using , s = 628 ms.

The given equation of motion: Step 1: Finding Maximum Height At the peak of the trajectory, . Differentiating: Setting , Solving for , Substituting into the equation of motion: Step 2: Finding Range The projectile reaches the ground when . Setting : Solving for , Thus, the range is: Step 3: Computing the Ratio Thus, the correct answer is:

Case 1: Rolling down without sliding.
Let the angle of inclination be .
The acceleration of a body rolling down an inclined plane without slipping is given by: For a solid sphere, moment of inertia .
So, .
Given .
We can find : Case 2: Sliding down without rolling.
When the sphere slides down without rolling, friction is either absent or not sufficient to cause rolling (e.
g.
, a perfectly smooth plane, or kinetic friction if it's just sliding).
If it's "sliding without rolling", it typically means friction is negligible or we consider the case where only gravity component along the incline acts.
The force causing the sliding is the component of gravity along the incline, which is .
The acceleration when sliding down is .
From Case 1, we found .
So, .
The mass (4 kg) and radius (28 cm) are not directly needed to find from , and only depends on .
This matches option (4).

Step 1: Compare with standard SHM equation: Step 2: Use the relation between angular frequency and time period: Step 3: Calculate: % Final Answer

The moment of inertia for a solid sphere about its diameter is: The moment of inertia for a solid cylinder about its axis is: The ratio of the moment of inertia of the sphere to the cylinder is: