The magnitude of the angular momentum (L) of the bar is .[tex]\frac{1}{6}mbv[/tex].
From the information given:
The rigid uniform bar has a mass = m
Length of the bar = b, and;
Linear speed of the bar endpoints = v
If we consider taking the angular momentum of the rotation, we have:
L = I × ω
where;
I = moment of the inertia
ω = angular velocity
The moment of the inertia for the given rigid uniform bar can be expressed as:
I = [tex]\frac{1}{2}[/tex] X M X L²
where;
M = mass = m
L = bar's length = b
∴ I = [tex]\frac{1}{2}[/tex] x m x b²
Also, the angular velocity ω can be expressed as:
ω = v/r
where;
radius r = half of the length = b/2
ω = [tex]\frac{v}{\frac{b}{2} }[/tex] = 2v/b
Replacing the value of angular velocity and moment of inertia into the above equation for angular momentum; we have:
L = ([tex]\frac{1}{12}[/tex] x m x b²) x ([tex]\frac{2v}{b}[/tex])
L = [tex]\frac{1}{6}[/tex] mbv
Therefore, we can conclude that the magnitude of the angular momentum of the bar (L) is 1/6 mbv
To know more about angular momentum, refer: brainly.com/question/15104254
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