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A particle is moving with constant speed v in xy plane as shown in the figure . The magnitude of its angular velocity about the point O is1) v/(√a^2 + b^2)2) v/b3) vb/(a^2 + b^2)4) v/a

```
2 years ago

```							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.
```
2 years ago
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