Question

A particle is located at (3m , 4m) and moving with v=(4hati-3hatj)m//s. Find its angular velocity about origin at this instant.

Answer 1

The angular velocity of particle is ω =$-9\hat k -16\hat k = -25\hat k$ , and magnitude is 25 $\frac{m^2}{s}$

Step by step explanation:

Given details:

Location of point P = (3m , 4m)

Velocity vector v = $(4\hat i - 3\hat j) \frac{m}{s}$

To find:

Angular velocity ω

So, The displacement vector r  (OP)

= $3\hat i + 4\hat j$

The angle between r and v    

 $\theta = \cos^-^1 \frac{r\cdot v}{|r||v|}$

as magnitude of r and v ,

| r | = $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$

| v | =$\sqrt{4^2 + 3^2} = \sqrt{25} = 5$

So  angle between  r and v =

$\theta = \cos^-^1 \frac{(3\hat i + 4\hat j)\cdot (4\hat i - 3\hat j)}{|5||5|} = \cos^-^1 (0)$ =90°

so, v⊥r

Then angular velocity is the cross product between displacement and velocity vectors.

ω = $r\times v$

ω = $(3\hat i + 4\hat j)\times (4\hat i - 3\hat j)$

 ω = $-9\hat k -16\hat k = -25\hat k$

and

$\textbf{\Large magnitude of angular velocity = rvsin90 }\\= 5\times 5\times 1 =\textbf{\Large 25 } \frac{m^2}{s}$

Answer 2

Answer:

A particle is located at (3m , 4m) and moving with v=(4î-3j^) m/s. Find its angular velocity about origin at this instant.

Answer : 25 $m^{2} /s$

Explanation:

Given that

Position of vector P = (3m , 4m)

So, The displacement vector r  (OP) = (3î + 4j^)

Velocity vector v = (4î - 3j^) m/s

angular velocity about origin = r x v = rvsin∅

angle between r and v

cos∅ = (3î + 4j^).(4î - 3j^) m/s / 5 x 5

cos∅ = 0

∅ = $90^{0}$

Hence v is perpendicular to r i.e. v ⊥ r

ω = rv sin∅ = 5x 5 sin$90^{0}$ = 25 $m^{2} /s$

Therefore, angular velocity about origin will be 25 $m^{2} /s$