Question

a 0.65 kg ball is moving in a circular path which has a radius of 1.35 meters with a linear speed of 4.5m/s. what is the centriperal force acting on this ball

Answer 1

The acceleration formula is
$a_{c}=\frac{v^{2}}{r}$

We are given
$v=4.5\ m/s$
$r=1.35\ m$
Plug in:
$a_{c}=\frac{(4.5\ m/s)^{2}}{1.35\ m}=15\ m/s^{2}$

The formula for force is
$F=ma$

We now have
$a=15\ m/s^{2}$
$m=0.65\ kg$
Plug in:
$F=15\ m/s^{2}*0.65\ kg=9.75\ N$

So the centripetal force is 9.75 Newtons.