Question

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar is vvv. What is the magnitude of the angular momentum LLL of the bar?

Answer 1

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       $L = I$ω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      $I = \frac{ mb^2}{12}$  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 $L = \frac{mb^2}{12}$×$\frac{2v}{b}$

              so final answer will be      $L = \frac{mvb}{6}$