Question

A point 'p' is moving with constant velocity V along a line AB. O is a point on the line
perpendicular to AB at A and at a distance "p" from A. The angular velocity of P about o
​

Answer 1

Answer:

The angular velocity of P about O is $\frac{pv}{OP^2}$

Step-by-step explanation:

As shown in the figure

The rate of change of x is velocity v

Therefore,

$\frac{dx}{dt}=v$

Again

$\tan\theta=\frac{x}{p}$

Differentiating w.r.t. t

$\sec^2\theta\frac{d\theta}{dt}=\frac{1}{p}\frac{dx}{dt}$

But $\frac{d\theta}{dt}=\omega$ where $\omega$ is the angular velocity

And $\sec\theta=\frac{OP}{p}$

Therefore,

$(\frac{OP}{p})^2\times \omega=\frac{1}{p}\times v$

$\implies \omega=\frac{pv}{OP^2}$

Hope this helps.