Question

A particle is moving with constant speed v along x axis in positive direction find the angle of the particle about the points (0, b) when the position the particle is (a,0)

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Answer 1

Answer:

W=v(perpendicular)/r ( where r is the distance between the position of the particle and the point about  which the angular velocity is to be calculated)  

thus,W=vcos@/r {m using '@' for theta}  

From the triangle ABC,sin@=b/root of (a^2+b^2)  

where @= 90-@,  

Thus sin(90-@)= cos@,  

Therefore, cos@=b/root of (a^2+b^2).  

So,W={v*b/root of (a^2+b^2)}/root of(a^2+b^2).  

which is equal to vb/a^2+b^2.  

Thus angular velocity = vb/a^2+b^2.  

[NOTE: Angular velocity is always equal to the perpendiculat velocity divided by radiusi.e radius and velocity should be perpendicular.]  

I am attaching a picture to clear the vector components.Thanks

Explanation:

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Answer 2

Question :-

A particle is moving with constant speed v along x axis in positive direction find the angle of the particle about the points (0, b) when the position the particle is (a,0) .

Answer :-

$\implies \:(1) \: \: \: \theta \: = { \sin }^{ - 1} \bigg( \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg) \: \\ \\ \implies \: (2) \: \: \: \: \: \omega = \frac{vb}{ \sqrt{ {a}^{2} + {b}^{2} } } \:$

Formula used :-

$\star \: \sin( \theta) = \frac{perpendicular}{hypotenuse} \: \\ \\ \star \: \omega \: = \frac{v}{r}$

To find :-

Find the angle .

Step - by - step explanation :-

Solution :-

Distance from origin in x axis is "a",

and distance from origin at point in y axis is "b" ,

Let distance from point (0,b) to (0,a) is r ,

Then Angle is ,

We know that,

$\implies \: \sin( \theta) = \frac{perpendicular}{hypotenuse} \\ \\ \implies \: \sin( \theta) = \frac{b}{r} \\ \\ \implies \: \boxed{ \theta \: = { \sin }^{ - 1} \big( \frac{b}{r} \big)} \\ \\ \: or \\ \implies \: \theta \: = { \sin }^{ - 1} \bigg( \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg)$

To finding angular velocity ,

$\implies \: \omega \: = \frac{v \: \sin( \theta) }{r} \\ \\ \implies \: \omega \: = \frac{v}{r} \times \frac{b}{r} \\ \\ \implies \: \boxed{ \omega \: = \frac{vb}{ {r}^{2} } } \\ \\ or \\ \\ \implies \: \boxed{ \omega = \frac{vb}{ \sqrt{ {a}^{2} + {b}^{2} } } }$

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