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

Respuesta :

The acceleration formula is
[tex]a_{c}=\frac{v^{2}}{r}[/tex]

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

The formula for force is
[tex]F=ma[/tex]

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

So the centripetal force is 9.75 Newtons.