A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the point with (x, y) coordinates of (3.0, −4.0) m. (Express your answers in vector form.)

(a) What is its angular momentum relative to the origin at this moment?
(b) the point located at (-2.0, -2.0) m?

Question

contestada

Respuesta :

Vespertilio Vespertilio · Answer 1 · 2019-12-14T12:34:23+01:00

Answer:

(a) [tex]\overrightarrow{L}=885.5\widehat{k}[/tex]

(b) [tex]\overrightarrow{L}=1046.5\widehat{k}[/tex]

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

[tex]\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}[/tex]

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, [tex]\overrightarrow{r}=3\widehat{i}-4\widehat{j}[/tex]

[tex]\overrightarrow{p}=m\overrightarrow{v}[/tex]

[tex]\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )[/tex]

[tex]\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}[/tex]

So, the angular momentum

[tex]\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )[/tex]

[tex]\overrightarrow{L}=885.5\widehat{k}[/tex]

(b) Here, [tex]\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}[/tex]

[tex]\overrightarrow{r}=5\widehat{i}-2\widehat{j}[/tex]

[tex]\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )[/tex]

[tex]\overrightarrow{L}=1046.5\widehat{k}[/tex]