a particle is moving with constant speed v along x axis in positive direction. find the angular velocity of the particle about the point (0,b),when position of the parricle is (a, 0).
Answers
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
https://s.yimg.com/tr/i/a29415b5f8e14a3395caffb5448a64aa_A.jpeg
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.