Question

Let r=coswti + sinwtj. Show that acceleration a is directed towards origin and has magnitude proportional to distance from origin

Answer 1

A particle moves so that its position vector is given by r = cosWt i+ sin wt j where cv is a constant. Show that (a) the velocity v of the particle is perpendicular to r, (b) the acceleration a is directed toward the origin and has mag- nitude proportional to the distance from the origin, (c) rxv = a constant vector.

Answer 2

Full explanation is given below .

Given :

Displacement of the body is :

$r = cos wt i + sin wt j$

We know , acceleration is differentiation of velocity and velocity is differentiation of displacement .

$v=\dfrac{dr}{dt}=\dfrac{d(cos \ wt)}{dt}i +\dfrac{d(sin \ wt)}{dt}j\\\\v=w(-sin\ wt)\i+w(cos\ wt)\ j\\\\a=\dfrac{dv}{dt}=w\dfrac{d(-sin\ wt)}{dt}i+w\dfrac{d(cos\ wt)}{dt}j\\\\a=-w^2cos\ wti-w^2sin\ wtj\\\\a=-w^2[cos\ wt\ i+sin\ wt\ j]\\a=-w^2r$

From above we can see that the acceleration is directly proportional to displacement .

Hence , this is the required solution .

Learn More :

Kinematics

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