The dimension of angular momentum is-
A). $\left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
B). $\left[ {{M^1}{L^2}{T^{ - 2}}} \right]$
C). $\left[ {{M^1}{L^2}{T^{ - 1}}} \right]$
D). $\left[ {{M^2}{L^1}{T^{ - 2}}} \right]$
Answer
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Hint: The angular momentum is the rotary equivalent to the linear momentum (rarely, momentum or rotating momentum). It is an essential parameter because it is a conserved quantity – a closed system's total angular momentum remains constant.
Complete step-by-step answer:
For three dimensions, the pseudovector $r \times p$, the cross product of the vector location (for reference to any origin) of a particle, and its vector momentum, the current $p = mv$ in Newtonian mechanics are the twisted momentum for a point particle. The description may be used in continuous solids, fluids, or physical fields at each level. In relation to momentum, angular momentum relies on where the source is chosen, since it measures the particle's location. There are two different angular momentum types: spin-angular momentum and orbital-angular momentum. The angular momentum of the rotation of an object is described by its angular momentum in its centre of mass. The angular momentum of an object on a chosen source is known as the angular momentum of the mass centre on the origin. The overall angular momentum of an object is the sum of the rotation and the angular orbital momentum.
The general formula for angular momentum is-
Angular momentum = $mvr$
Here the dimension formula of mass(m) is $[M]$, dimension formula of velocity (v) is $\left[{L}{T^{-1}}\right]$ and dimension formula of radius(r) is $[L]$.
So, The angular momentum’s dimensional formula is given by-
$ \Rightarrow \left[ {{M^1}{L^2}{T^{ - 1}}} \right]$
Hence, it is clear that option C is the correct option.
Note: Angular momentum is a massive amount; i.e. every hybrid system's overall angular momentum is a percentage of the angular momenta of its component pieces. The total angular momentum for a continuous rigid body is the integral volume of the angular momentum distribution over the whole body ( i.e. angular momentum per unit of the region in the boundary as volume falls to nil) over the whole body.
Complete step-by-step answer:
For three dimensions, the pseudovector $r \times p$, the cross product of the vector location (for reference to any origin) of a particle, and its vector momentum, the current $p = mv$ in Newtonian mechanics are the twisted momentum for a point particle. The description may be used in continuous solids, fluids, or physical fields at each level. In relation to momentum, angular momentum relies on where the source is chosen, since it measures the particle's location. There are two different angular momentum types: spin-angular momentum and orbital-angular momentum. The angular momentum of the rotation of an object is described by its angular momentum in its centre of mass. The angular momentum of an object on a chosen source is known as the angular momentum of the mass centre on the origin. The overall angular momentum of an object is the sum of the rotation and the angular orbital momentum.
The general formula for angular momentum is-
Angular momentum = $mvr$
Here the dimension formula of mass(m) is $[M]$, dimension formula of velocity (v) is $\left[{L}{T^{-1}}\right]$ and dimension formula of radius(r) is $[L]$.
So, The angular momentum’s dimensional formula is given by-
$ \Rightarrow \left[ {{M^1}{L^2}{T^{ - 1}}} \right]$
Hence, it is clear that option C is the correct option.
Note: Angular momentum is a massive amount; i.e. every hybrid system's overall angular momentum is a percentage of the angular momenta of its component pieces. The total angular momentum for a continuous rigid body is the integral volume of the angular momentum distribution over the whole body ( i.e. angular momentum per unit of the region in the boundary as volume falls to nil) over the whole body.
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