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      $L = \frac{mvb}{6}$

Answer 2

The magnitude of the angular momentum (L) of the bar is $\mathbf{\dfrac{1}{6}mbv}$

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:

$\mathbf{I = \dfrac{1}{12}\times M \times L ^2}$

where;

• M = mass = m
• L = bar's length = b

∴

$\mathbf{I = \dfrac{1}{12}\times m \times b ^2}$

Also, the angular velocity ω can be expressed as:

$\mathbf{\omega = \dfrac{v}{r}}$

where;

• radius r = half of the length = b/2

$\mathbf{\omega = \dfrac{v}{\dfrac{b}{2}}}$

$\mathbf{\omega = \dfrac{2v}{b}}}$

replacing the value of angular velocity and moment of inertia into the above equation for angular momentum; we have:

$\mathbf{L = \Big (\dfrac{1}{12}\times m \times b^2 \Big) \times \Big ( \dfrac{2v}{b} \Big) }$

$\mathbf{L = \dfrac{1}{6} m b v }$

Therefore, we can conclude that the magnitude of the angular momentum of the bar (L) is $\mathbf{ \dfrac{1}{6} m b v }$

Learn more about angular momentum here:

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