Question

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?

Answer 1

(a)
`\overrightarrow{L}=885.5\widehat{k}`

(b)
`\overrightarrow{L}=1046.5\widehat{k}`

Step-by-step explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

`\overrightarrow{L}=\overrightarrow{r}* \overrightarrow{p}`

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

Here,
`\overrightarrow{r}=3\widehat{i}-4\widehat{j}`

`\overrightarrow{p}=m\overrightarrow{v}`

`\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )`

`\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}`

So, the angular momentum

`\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )*\left ( 92\widehat{i}+172.5\widehat{j} \right )`

`\overrightarrow{L}=885.5\widehat{k}`

(b) Here,
`\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}`

`\overrightarrow{r}=5\widehat{i}-2\widehat{j}`

`\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )*\left ( 92\widehat{i}+172.5\widehat{j} \right )`

`\overrightarrow{L}=1046.5\widehat{k}`