with its Derivation<\/strong><\/span><\/p>\n\n\n\n In this article, we will find the dimension of Angular Momentum<\/p>\n\n\n\n Dimensional formula for Angular Momentum is<\/p>\n\n\n\n $[M^1L^2T^{-1}]$<\/p>\n\n\n\n Where L<\/strong> -> Length<\/p>\n\n\n\n T<\/strong> -> Time Angular Momentum is defined<\/p>\n\n\n\n $L = I \\omega$ $\\omega $-> Angular velocity<\/p>\n\n\n\n Now of Angular velocity is defined as<\/p>\n\n\n\n $\\omega =\\frac {d\\theta }{dt} = \\frac {\\text {change in angular displacement}}{time}$<\/p>\n\n\n\n Now Angular displacement is Dimension less and Dimension of time is $[T^1]$<\/p>\n\n\n\n So, Dimension of Angular velocity $\\omega$ is $= \\frac {[M^0L^0T^0}{[T^1]} = [M^0L^0T^{-1}]$<\/p>\n\n\n\n Now Moment of inertia is defined as<\/p>\n\n\n\n $I= mr^2$<\/p>\n\n\n\n Dimension of Mass =$[M^1]$<\/p>\n\n\n\n Dimension of distance(r) = $[L^1]$<\/p>\n\n\n\n So,Dimension of Moment of inertia =$[M^1] \\times [L^1] \\times [L^1]= [M^1L^2]$<\/p>\n\n\n\n Now we know both the Moment of Inertia and angular velocity dimension , we can calculate the dimension of Angular momentum easily as<\/p>\n\n\n\n $\\text {dimension of Angular momentum} = \\text {dimension of moment of inertia} \\times \\text {dimension of Angular velocity}$<\/p>\n\n\n\n $= [M^1L^2] \\times [T^{-1}] = [M^1L^2T^{-1}]$<\/p>\n\n\n\n Unit of Angular Momentum is $Kg-m^2\/sec$ and It is generally denoted by letter $L$<\/p>\n\n\n\n Try the free Quiz given below to check your knowledge of Dimension Analysis:-<\/p>\n\n\n\n
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\nM<\/strong> -> Mass<\/p>\n\n\n\n
\nWe would now derive this dimensional formula.<\/p>\n\n\n\nDerivation for expression of Dimension of Angular Momentum<\/h3>\n\n\n\n
\nWhere L -> Angular Momentum
\nI -> Moment of Inertia<\/p>\n\n\n\n
\nQuiz on Dimensional Analysis<\/strong><\/span><\/h4>\n
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