Question

A light, rigid rod of length $\mathit{l}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}$joins two particles, with masses ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}$and role="math" localid="1663651253607" ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{g}\mathbf{,}$at its ends. The combination rotates in the x y plane about a pivot through the center of the rod (Fig. P11.11). Determine the angular momentum of the system about the origin when the speed of each particle is$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{.}$

Short answer

$\overrightarrow{\mathbf{L}}\mathbf{=}\mathbf{17}\mathbf{.5}\widehat{\mathbf{k}}\mathit{k}\mathit{g}\mathbf{\cdot }{\mathit{m}}^{\mathbf{2}}\mathbf{/}\mathit{s}$

Step-by-step solution

Given Information

Given:

$\begin{array}{c}l=1\text{\hspace{0.17em}m}\\ m_1=4\text{kg}\\ m_2=3\text{kg}\\ v=5\text{m/s}\end{array}$

The angular momentum of the system

The magnitude of the angular momentum of the system of the two masses about the origin is defined as:

$L=(m_1+m_2)vr\sin \theta$

where the angle $\left(\theta \right)$ between the radius vector and the velocity vector is $\left({90}^{°}\right)$ because they are perpendicular to each other. And (r) is the distance between each of the mass and the origin which is equal to half the length of the $rod(l/2)$:

$L=(m_1+m_2)v\left(\frac{l}{2}\right)\sin {90}^{°}$

Substitute numerical values:

$\begin{array}{c}L=(4+3)\left(5\right)\left(\frac{1}{2}\right)\\ =17.5\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

By the right-hand rule, the direction of the angular momentum is perpendicular to the plane containing the radius and velocity vectors. Therefore, it is along the z-axis which has a unit vector $\widehat{k}:$

$\overrightarrow{L}=17.5\widehat{k}kg\cdot m^2/s$