Question

A particle is moving with constant speed V in xy plane as shown in figure the magnitude of its angular velocity about o is

Answer 1

Remember that you should only consider magnitude of velocity which is tangential , in circular motion.

Answer 2

Given:

Speed of the particle = V

To Find:

We have to find the magnitude of the particle's angular velocity about the point O.

Solution:

Angular velocity, ω = $\frac{dθ}{dt} = \frac{V. Sin θ}{r} .$

In the above equation, "θ" is the angular displacement and "t" is the time.

From the figure,

Sin θ = $\frac{b}{\sqrt{a^{2}+b^{2} } }$

r = $\sqrt{a^{2}+b^{2} } }$

Substituting these values in equation for angular velocity, we get:

$ω = \frac{V. \frac{b}{\sqrt{a^{2}+b^{2} } } }{\sqrt{a^{2}+b^{2} }}$

$ω = \frac{V.b}{a^{2} + b^{2} }$.

Hence,  the magnitude of the particle's angular velocity about the point O is $\frac{V.b}{a^{2} + b^{2} }$.

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