The Angular Momentum MCQ Quiz in मल्याळम - Objective Question with Answer for The Angular Momentum - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

option(2)

CONCEPT:

• Angular momentum (L): It is a vector quantity that requires both a magnitude and a direction.
  • The magnitude of the angular momentum is equal to its linear momentum and perpendicular distance r from the center of rotation to a line.
    • The unit of Angular Momentum is Kg m²/s.

​L = p × r

Where p is linear momentum and r is the radius vector 

EXPLANATION:

Angular momentum: The vector product of the distance r and linear momentum (mv).

​L = p × r

L = m v r

Since p = mass (m) × velocity (v)

 L = Kg ms^-1 m = Kg m2/s

• Hence the unit of Angular Momentum is Kg m2/s.

Additional Information

• Vector Quantity: That quantity that contains both magnitude and direction is called a vector quantity.
  • Examples: Velocity, Force, Angular momentum, Displacement, etc.
• Linear Momentum: That physical quantity which the vector product of mass and velocity.

p = m × v (where m is the mass and v is the velocity)

CONCEPT:

• The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.
• It is measured as the product of linear momentum and the perpendicular distance of its line of action from the axis of rotation.
• The relation between the angular momentum and moment of inertia is given by

L = Iω

Where

I = moment of inertia,

L = angular momentum,

ω = angular velocity.

Explanation:

From the above explanation, we can see that angular momentum is the product of its moment of inertia and angular velocity

Hence option 1 is correct among all

The correct answer is –3.2 ms–1.

Key Points

• Given, 
  • Mass of bullet, m₁ = 40g (= 0.04 kg)
  • Mass of pistol, m₂ = 2 kg
  • The initial velocity of the bullet (u₁) and pistol (u₂) = 0
  • Final velocity of the bullet, v₁ = +160m s^-1
  • Let, v₂ be the recoil velocity of the pistol.
  • The total momentum of the pistol and bullet is zero before the fire because both are at rest.
  • The total momentum of the pistol and bullet after it is fired is

                     = (0.0 kg 4x 160 m s^-1) + (2 kg x v₂ m s^-1)

                     = (6.4 + 2v₂) kg m s^-1

• Total momentum after the fire = Total momentum before the fire
•  6.4 + 2v₂ = 0
• →v₂ = 3.2 m/s  
• Thus, the recoil velocity of the pistol is 3.2 m/s.

CONCEPT:

Angular Momentum:

• It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object. It is a vector quantity

​
⇒ L = Iω

where I = moment of inertia and ω = angular velocity

EXPLANATION:

The angular momentum of a body is given as,

​⇒ L = Iω

If the angular momentum is constant, then,

$\Rightarrow \omega \propto \frac{1}{I}$     ---(1)

• Since there is no external torque is acting on the dancer, so his angular momentum will remain to conserve when he folds his arms.
• When the dancer folds his arms his moment of inertia decreases so the angular velocity increases. Hence, option 1 is correct.

CONCEPT

Angular momentum (L): It is defined as the momentum of the rotating body or angular momentum of a rigid body is the product of moment of inertia and the angular velocity.

• It is analogous to linear momentum.
• It follows the angular momentum principle if no external torque on the object.

Torque (τ): The physical quantity, similar as force that causes the rotational motion. It is the cross product of the force with the perpendicular distance between the axis of rotation and the point of application of the force with the force.

• Torque (τ) = r × F = r F sin θ

[Where, r = distance from point of application of force (in metre), f = force (in Newton) and θ = angle between r→ and F^→]

• Also, Torque (τ) = r_⊥ f = r F_⊥

Angular momentum for the particle about the origin (O)

$Angularmomentum\left(\overrightarrow{L}\right)=\overrightarrow{r}\times \overrightarrow{p}=\overrightarrow{r}\left(m\overrightarrow{v}\right)$

Where, r^→ = position vector of a particle, v^→ = velocity

Also, L = mv (r sin θ) = mv r_⊥

• Angular momentum for a particle moving in a circle of radius r with the speed v, its linear momentum is mv, its angular momentum L

$\overrightarrow{L}=mv{r}_{\mathrm{\perp }}=mvr$

• Angular momentum for a rigid body

$\overrightarrow{L}=I\overrightarrow{\omega }$

$\frac{d\overrightarrow{L}}{dt}=I\frac{d\overrightarrow{\omega }}{dt}=I\overrightarrow{\alpha }={\overrightarrow{\tau }}_{net}$

Where, I = moment of inertia, ω = angular velocity, α = angular acceleration, τ_net = total external torque

EXPLANATION

Conservation of Angular Momentum: If the total external torque on a system is zero, its angular momentum remains constant.

• This is known as the principle of conservation of angular momentum.

We know that L = I × α = τ_net

Where, L = angular momentum, α = angular acceleration, τ_net = Net external torque.

If τ_net = 0

⇒ dL/dt = 0

⇒ L = constant

⇒ L_final = L_initial

Here, both conditions satisfied for conservation of momentum.

• τ_net = 0
• initial angular momentum = final angular momentum

∴ Option 3 is correct

CONCEPT:

Angular Momentum: It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object. It is a vector quantity

​⇒ L = Iω

where I = moment of inertia and ω = angular velocity​

CALCULATION:

Given I = 20 kg-m2 and α = 5 rad/sec2

We know that the angular momentum is given as,

​⇒ L = Iω     -----(1)

So the rate of change of angular momentum is given as,

$\Rightarrow \frac{dL}{dt}=\frac{d(I\omega )}{dt}$

∵ The moment of inertia is constant,

∴ $\frac{dL}{dt}=I\frac{d\omega }{dt}$     -----(2)

We know that rate of change of angular velocity is equal to the angular acceleration, therefore,

$\Rightarrow \alpha =\frac{d\omega }{dt}$        -----(3)

By equation 2 and equation 3,

$\Rightarrow \frac{dL}{dt}=I\alpha$

$\Rightarrow \frac{dL}{dt}=20\times 5$

$\Rightarrow \frac{dL}{dt}=100$ kg-m2/sec2

Hence, option 2 is correct.

Concept:

Kepler’s laws of planetary motion:

• The law of Orbits: Every planet moves around the sun in an elliptical orbit with the sun at one of the foci.
• The law of Area: The line joining the sun to the planet sweeps out equal areas in equal interval of time. i.e. areal velocity is constant.

This diagram shows that the areal velocity of satellite is constant for area A and B

• According to this law, the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. It is similar to the law of conservation of angular momentum.
• The law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit i.e. T² ∝ r³

Explanation:

• From above it is clear that the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun
• According to this law, the planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. This is because of law of conservation of angular momentum.
• Hence angular momentum remains constant throughout its motion

CONCEPT:

• Rotational motion: When a block is moving about a fixed axis on a circular path then this type of motion is called rotational motion.
• Torque (τ): It is the twisting force that tends to cause rotation.
  • The point where the object rotates is known as the axis of rotation. 

Mathematically it is written as,

τ = r F sin θ 

Where r is the distance between the axis of rotation and point of application of force, F is force and θ is the angle between r and F.

• Rotational power (P): The rate of work done by torque is called power.

The power associated with torque is given by the product of torque and angular velocity of the body about an axis of rotation i.e.,

⇒ P = τ ω 

Where τ = torque and ω = angular velocity

Angular momentum (L) = Rotational inertia (I) × Angular velocity (ω)

Explanation:

From the above explanation, we can see that the product of torque and angular velocity is known as Rotational power (P)

Important Points

In rotational dynamics

• Moment of inertia is the analogue of mass
• Angular velocity is analogue of linear velocity
• Angular acceleration is analogue of linear acceleration

Thus, in linear motion mass x velocity = momentum

The analogues of the moment of inertia x angular velocity = angular momentum

The correct answer is option 3) i.e. $\frac{\omega M}{M+2m}$.

CONCEPT:

• Angular momentum: The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity.
  • Angular momentum also obeys the law of conservation of momentum i.e. angular momentum before and after is conserved.
  • Angular momentum = I × ω = mr²ω 

Where I is the moment of inertia, ω is the angular velocity, m is the mass of the object, and r is the radius of curvature.

CALCULATION:

Let ω be the initial angular velocity of the ring and ω' be the new angular velocity after placement of two bodies on it.

From the conservation of angular momentum:

Momentum before = Momentum after

The momentum of the ring before = Momentum of the ring after placing the bodies + Momentum of the bodies

Mr2ω = Mr²ω' + 2 × (mr²ω')

Mω = ω' (M+2m)

⇒ ω' = $\frac{\omega M}{M+2m}$

Explanation:-

Conditions for a rigid body to be in equilibrium - 

For a rigid object which is not moving at all, we have the following conditions:

The (vector) sum of the external forces on the rigid object must equal zero:

∑F = 0

When this condition is satisfied we say that the object is in translational equilibrium. (It really only tells us that a_CM is zero, but of course, that includes the case where the object is motionless.) 

The sum of the external torques on the rigid object must equal zero:

∑τ = 0

When this condition is satisfied we say that the object is in rotational equilibrium. (It really only tells us that α  about the given axis is zero, but again that includes the case where the object is motionless.)

When these conditions are satisfied then we can say that the object is in static equilibrium.

So if the net moment of the body is zero that does not mean that the distance between the force and the rotational axis is zero, It simply means that the moment caused by different forces gets canceled out.

That's why only statement 2 is correct.

Additional Information 

Equilibrium

• The term equilibrium implies that either the body is at rest or it moves with a constant velocity. We shall deal with bodies at rest, or bodies in static equilibrium. A body is in static equilibrium when the force system acting on it tends to produce no net translation or rotation of the body.
• A body is said to be in static equilibrium when the resultant force on it must be zero and the body must have no tendency to rotate. This second condition of equilibrium requires that the net moment about any point be zero.