Centre of Mass 2 Question 7

8. 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.

$(2003,2 M)$

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Answer:

Correct Answer: 8. $-m v _2 \sin \frac{v _2}{R} t \hat{\mathbf{i}}+m \quad v _2 \cos \frac{v _2}{R} t-v _1 \hat{\mathbf{j}}$

Solution:

  1. Angular speed of particle about centre of the circle

$$ \omega=\frac{v _2}{R}, \theta=\omega t=\frac{v _2}{R} t $$

$\mathbf{v} _p=\left(-v _2 \sin \theta \hat{\mathbf{i}}+v _2 \cos \theta \hat{\mathbf{j}}\right)$

or $\quad \mathbf{v} _p=-v _2 \sin \frac{v _2}{R} t \hat{\mathbf{i}}+v _2 \cos \frac{v _2}{R} t \hat{\mathbf{j}}$

and

$$ \mathbf{v} _m=v _1 \hat{\mathbf{j}} $$

$\therefore$ Linear momentum of particle w.r.t. man as a function of time is

$$ \begin{aligned} \mathbf{L} _{p m}=m & \left(\mathbf{v} _p-\mathbf{v} _m\right) \\ & =m-v _2 \sin \frac{v _2}{R} t \hat{\mathbf{i}}+v _2 \cos \frac{v _2}{R} t-v _1 \hat{\mathbf{j}} \end{aligned} $$



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