The property of any rotating body given by moment of inertia times angular velocity is defined as angular momentum. That is, it is the property of a rotating body given by the product of the rotating object's moment of inertia and angular velocity. It is obvious that this is a vector quantity; in addition to magnitude, the direction is taken into account.
Any object or body moving with mass has momentum, and angular momentum is the property that characterises the rotary inertia of an object or system of objects in motion around an axis that may or may not pass through the object or system. The Earth also has orbital angular momentum from its annual revolution around the Sun, as well as spin angular momentum from its daily rotation around its axis.
The magnitude of an orbiting object's angular momentum is equal to its linear momentum (times of the perpendicular distance r from the centre of rotation to the line drawn in the direction of the object's instantaneous motion and passing through the object's centre of gravity). Here you can find information about the conservation of angular momentum from various physics-related articles.
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Angular momentum and its conservation law are important topics in physics. Students who want to excel in physics must be well-versed in momentum in order to perform well on exams. The relationship between torque and angular momentum, conservation of angular momentum, and its applications are discussed here to help students understand the subject. Continue to visit our website for more physics assistance.
This is a cross product of r, which is the radius of the circle formed by the body in rotational motion, and p, which is the linear momentum of the body. The magnitude of a cross product of two vectors is always the product of their magnitude multiplied by the sine of the angle between them, so the magnitude of angular momentum is given by, Angular momentum is the rotational analogue of linear momentum and it is denoted by L.
Now, the angular momentum of a particle in rotational motion is defined as,
$${l=r\:x\:p}$$
The magnitude of a cross product of two vectors is always the product of their magnitude multiplied by the sine of the angle between them. It is a cross product of r, which is the radius of the circle formed by the body in rotational motion, p, which is the linear momentum of the body.
So, in the case of angular momentum, the magnitude is given by,
$${l=r\:x\:p \:sin \:\theta}$$
The angular momentum of a system is conserved as long as there is no net external torque acting on the system. Because of the law of conservation of angular momentum, the earth has been rotating on its axis since the formation of the solar system. There are two methods for calculating an object's angular momentum. If the object is a point in a rotation, our angular momentum is equal to the radius times the linear momentum of the object.
$$\mathrm{\vec{l}=\vec{r}\:\vec{p}}$$
Differentiate L. H. S. and R. H. S.
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\frac{d}{dt}(\vec{r}\:x\:\vec{p})}$$
The expression can be written as follows using the property of differentiation on cross products:
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\frac{dr}{dt}x\:\vec{p}+r \frac{\overrightarrow{dp}}{dt}}$$
Thus, it is linear velocity $\mathrm{\vec{\nu}}$.
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\vec{\nu}\:x\:\vec{p}+r\frac{\overrightarrow{dp}}{dt}}$$
Here, p is linear momentum that is, mass times velocity. Now,
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\vec{\nu}\:x\:m\vec{\nu}+\vec{r}\frac{\overrightarrow{dp}}{dt}}$$
When noticing the first term, there is $\mathrm{\vec{\nu}\:\vec{\nu}}$
The magnitude of the cross product is,
$$\mathrm{\vec{\nu}\:x\:\vec{\nu} sin \theta}$$
where the angle is 0.
Thus, the whole term becomes 0. From newton's second law, we know that $\mathrm{\frac{\overrightarrow{dp}}{dt}}$ is force so,
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\vec{r}\:\vec{F}}$$
As we know that $\mathrm{\vec{r}\:\vec{F}}$ is torque
$$\mathrm{\frac{\overrightarrow{dl}}{dt}=\vec{r}}$$
Hence we get, the rate of change of angular momentum is torque.
The angular momentum of any system is conserved as long as there is no net external torque acting on the system; due to the law of conservation of angular momentum, the earth has been rotating on its axis since the formation of the solar system.
There are currently two methods for calculating an object's angular momentum. If the object is a point in a rotation, the angular momentum is equal to the radius multiplied by the linear momentum. That is to say,
$$\mathrm{\vec{l}=\vec{r}\:\vec{p}}$$
If there is an extended object, such as the earth, the angular momentum is given by the moment of inertia, which is the amount of mass in motion in the object and how far it is from the centre, multiplied by the angular velocity.
$$\mathrm{\vec{l}=\vec{I}\:x\:\vec{\omega}}$$
However, in both cases, as long as there is no net force acting on it, the angular momentum before is equal to the angular momentum after some given time. For example, imagine rotating a ball tied to a long string; the angular momentum would be,
$$\mathrm{\vec{l}=\vec{r}\:\vec{p}=\vec{r}\:m\:\vec{\nu}}$$
Now, if we reduce the radius of the ball by shortening the string while it is rotating, the r will decrease. Then, according to the law of conservation of angular momentum, L should remain constant. There is no way for the masses to change. Thus $\mathrm{\vec{v}}$ should increase. In order to keep the angular momentum constant. Therefore, this is the proof for the conservation of angular momentum.
There are numerous applications for the law of conservation of angular momentum, including:
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Q1. What is the conservation of angular momentum?
Ans. The angular momentum is constant for a system with no external torque.
Q2. Applications of conversation of angular momentum?
Ans. Aircraft Engine, Electric Generators etc.
Q3. Where can we find the centre of mass of two particles having equal mass?
Ans. The centre of mass of two particles of equal mass is located in the middle of them.
Q4. When a particle moves in such a way that its angular position relative to its reference axis changes, it is said to have angular momentum.? (Say whether True or False)
Ans. If a particle moves in a way that its angular position changes relative to its reference axis, is said to have angular momentum.
Q5. Consider an ice skater who begins to spin so that his arms are as far apart and parallel to the ice as possible. What happens to the skater's angular velocity when he pulls his arms inwards and raises his arms vertically?
Ans. The angular velocity of the skater increases when he pulls his arms inwards since the moment of inertia is lowered. Angular velocity of the skater stays the same when he raises his arms vertically because the distribution of radius of mass does not change.
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