Is angular momentum equal to inertia times angular velocity and why?
Answer
539.1k+ views
Hint: Inertia of a rigid body acts like mass when the body is rotating with respect to some fixed axis. Angular momentum of a rigid body is the momentum of the body in rotational motion with respect to some fixed axis. Angular velocity of the body is the rotational velocity of the body with respect to some fixed axis
Formula used:
Angular momentum of a particle is given by,
\[L = mv \times r\]
Where, \[m\] is the mass of the particle \[v\] is the linear velocity of the particle \[r\]is the distance from the axis of rotation.
Angular velocity of a body is related to linear velocity as,
\[v = \omega r\]
where \[\omega \] is the angular velocity of the body.
The moment of inertia of a body is given by,
\[I = \sum {{m_i}{r_i}^2} \]
where \[{m_i}\] is the mass of the \[{i^{th}}\] particle and \[{r_i}\] is the perpendicular distance of the \[{i^{th}}\] particle from axis of rotation.
Complete step by step answer:
We know that the angular momentum or moment of momentum is nothing but the cross product of linear momentum and the distance from the axis of rotation if the body is rotating. That can be written as,
\[L = mv \times r\]
\[\Rightarrow L = mvr\]
Now, we know that the angular velocity of any body is related to linear velocity as,
\[v = \omega r\] where \[\omega \] is the angular velocity of the body.
So, we can have,
\[L = mr.\omega r\]
\[\Rightarrow L = m\omega {r^2}\]
So, the total angular momentum of a rigid body made of \[N\] number of particle will be,
\[L = \omega \sum {{m_i}{r_i}^2} \]
now, we know that moment of inertia of a rigid body is given by,
\[I = \sum {{m_i}{r_i}^2} \]
So, we can write, \[L = I\omega \]
Hence, we can say that the angular momentum of a body is equal to the inertia times the angular velocity of the body.
Note:When a rigid body is subjected to rotation with respect to a fixed axis the angular velocity vector is a constant vector. When a rigid body is rotating since all the particles are connected with each other and the distance between them is not movable so the angular velocity for each of the particles is always the same for rigid bodies.
Formula used:
Angular momentum of a particle is given by,
\[L = mv \times r\]
Where, \[m\] is the mass of the particle \[v\] is the linear velocity of the particle \[r\]is the distance from the axis of rotation.
Angular velocity of a body is related to linear velocity as,
\[v = \omega r\]
where \[\omega \] is the angular velocity of the body.
The moment of inertia of a body is given by,
\[I = \sum {{m_i}{r_i}^2} \]
where \[{m_i}\] is the mass of the \[{i^{th}}\] particle and \[{r_i}\] is the perpendicular distance of the \[{i^{th}}\] particle from axis of rotation.
Complete step by step answer:
We know that the angular momentum or moment of momentum is nothing but the cross product of linear momentum and the distance from the axis of rotation if the body is rotating. That can be written as,
\[L = mv \times r\]
\[\Rightarrow L = mvr\]
Now, we know that the angular velocity of any body is related to linear velocity as,
\[v = \omega r\] where \[\omega \] is the angular velocity of the body.
So, we can have,
\[L = mr.\omega r\]
\[\Rightarrow L = m\omega {r^2}\]
So, the total angular momentum of a rigid body made of \[N\] number of particle will be,
\[L = \omega \sum {{m_i}{r_i}^2} \]
now, we know that moment of inertia of a rigid body is given by,
\[I = \sum {{m_i}{r_i}^2} \]
So, we can write, \[L = I\omega \]
Hence, we can say that the angular momentum of a body is equal to the inertia times the angular velocity of the body.
Note:When a rigid body is subjected to rotation with respect to a fixed axis the angular velocity vector is a constant vector. When a rigid body is rotating since all the particles are connected with each other and the distance between them is not movable so the angular velocity for each of the particles is always the same for rigid bodies.
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