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stiks02 [169]
3 years ago
10

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

o 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?
Physics
1 answer:
Pie3 years ago
7 0

Answer:

I = \frac{mvb}{6}

Explanation:

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

       L = Iω ....                        (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}

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