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svetlana [45]
3 years ago
8

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 poi

nt 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?
Physics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

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

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

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

\overrightarrow{L}=\overrightarrow{r}\times \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 )\times\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 )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )

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

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