Chapter 12: Q.44 (page 332)

What is the angular momentum vector of the rotating bar in

Short Answer

Expert verified

Therefore the angular momentum of the rotating bar is $(2.1 kg \cdot \frac{m^{2}}{s}) \hat{k}$

Step by step solution

Step :1 Introduction

The rotation between angular momentum and angular velocity of the bar can be

$\vec{L} = I \vec{ω}$

Here, $l$ is the moment of inertia of the bar and $ω$ is the angular velocity of the bar, and the bar is rotating counter clockwise the angular velocity is in the $\hat{k}$ direction.

Step :2 Explanation

The rotating bar is as shown in the below figure:

Step : 3 Angular velocity

The mass of bar is $500 g$ convert the units of mass from $g t o k g$

$M = 500 g$

$= (500 g) (\frac{1 kg}{1000 g})$

$= 0.5 kg$

The angular velocity of the bar is $120 r p m .$ Convert the units of bar from $r p m a n d r a d s$

$\vec{ω} = (120 \frac{rev}{\min}) (\frac{2 π rad}{1 rev}) (\frac{1 \min}{60 s})$

$= 12.6 \hat{k} \frac{rad}{s}$

The moment of inertia $l$ can be found using the equation for a thin rod around the center:

$I = \frac{1}{12} M m^{2}$

Substitute $0.5 k g$ for $2.0 m$ for $l$ in the above frmula t solve the mooemnt of inertia of the bar

$I = \frac{1}{12} (0.5 kg) (2.0 m)^{2}$

$= 0.167 kg \cdot m^{2}$

Substitute $0.167 kg \cdot m^{2} for / and 12.6 \hat{k} \frac{rad}{s} for \vec{ω}$ in equation $\vec{L} = I \vec{ω}$ to solve the angular momentum of the rotating bar.