Question

A particle of mass $m$, moving in a circular path of radius $R$ with a constant speed $v_{2}$ is located at point $(2 R, 0)$ at time $t=0$ and a man starts moving with a velocity $v_{1}$ along the positive $y$-axis from origin at time $t=0$. Calculate the linear momentum of the particle w.r.t. man as a function of time.

Answer 1

Angular speed of particle about centre of the circle

$ \omega =\frac{v_{2}}{R}, \theta=\omega t=\frac{v_{2}}{R} t $
$v _{p} =\left(-v_{2} \sin \theta \hat{ i }+v_{2} \cos \theta \hat{ j }\right) $
or $v _{p} =\left(-v_{2} \sin \frac{v_{2}}{R} t \hat{ i }+v_{2} \cos \frac{v_{2}}{R} t \hat{ j }\right) $
and $ v_{m} =v_{1} \hat{ j } $
$\therefore$ Linear momentum of particle w.r.t. man as a function of time is
$L _{p m} =m\left( v _{p}- v _{m}\right) $
$=m\left[\left(-v_{2} \sin \frac{v_{2}}{R} t\right) \hat{ i }+\left(v_{2} \cos \frac{v_{2}}{R} t-v_{1}\right) \hat{ j }\right]$