Question

At a particular instant the location of an object relative to location \(A\) is given by the vector \({\overrightarrow r _A} = \left\langle {6,6,0} \right\rangle {\rm{m}}\). At this instant the momentum of the object is \(\overrightarrow p = \left\langle { - 11,13,0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}.\) What is the angular momentum of the object about location \(A\)?

Short answer

The angular momentum of the object about location \(A\) is \(\left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\)

Step-by-step solution

Definition of Angular Momentum

The rotating inertia of an object or system of objects in motion about an axis that may or may not pass through the object or system is described by angular momentum.

Given data

Position of an object to point \(A\) is \({\overrightarrow r _A} = \left\langle {6,6,0} \right\rangle m\)

Here\(x - \) component of the position of the object, \({r_x} = 6\)

\(y - \)component of the position of the object, \({r_y} = 6\)

\(z - \)component of the position of the object,\({r_z} = 0\)

Momentum of the object relative to the point \(A\),\({\overrightarrow p _A} = \left\langle { - 11,13,0} \right\rangle kg \cdot m/s\)

Here\(x - \) component of the momentum of the object, \({p_x} = - 11\)

\(y - \)component of the momentum of the object, \({p_y} = 13\)

\(z - \)component of the momentum of the object, \({p_z} = 0\)

Find the Angular Momentum of the object at point A 

Angular momentum of the object relative to point \(A\)is

\({\overrightarrow L _A} = {\overrightarrow r _A} \times \overrightarrow p \)

\(\begin{aligned}{}{\overrightarrow L _A} &= \left( {{r_y}{p_z} - {r_z}{p_x} - {r_x}{p_z},{r_x}{p_y} - {r_y}{p_x}} \right)\\ &= \left( {0 - 0,0 - 0,\left( 6 \right)\left( {13} \right) - \left( 6 \right)\left( { - 11} \right)} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\\ &= \left( {0,078 + 66} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\\ &= \left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\end{aligned}\)

Hence, the Angular momentum of the object about point \(A\)is\(\left( {0,1,144} \right){\rm{kg}} \cdot {\rm{m}}/{{\rm{s}}^2}\).