Question

A particle of mass 3 kg is rotating in a circle of radius 1 m such that the angle rotated by its radius is given by q=3(t+sint). Find the net force acting on the particle when t=p/2.

Answer 1

We have,

$m=310g$

$R=1m$

$q=3(t+sint)$

So,

$v=\frac{dq}{dt}=(3+3tcost)$

$v==3(1+cost)$

Now,

Tangential acceleration is $a_t=\frac{dv}{dt}=-3sint$

Centripital acceleration is $a_c=\frac{v^2}{r}9(1+cost{)}^2$

Let the acceleration $a=\sqrt{a_t^2{a}_c^2}$

$=\sqrt{9si{n}^{2}t81(1+cost{)}^4}$

Since,

$f=ma$

$f_t=\frac{\pi }{2}=30\sqrt{9+81(1+0{)}^4}$

$=30\sqrt{9+81}$

$=90\sqrt{10}N$