Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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
#SPJ4
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.