Mechanical Energy:Potential, kinect, etc.

[Patolord]

Homework Statement

A sphere of mass 2kg is release from point A off the edge of a hole that is represented of as a semi-circle of R=0,2 m.
consider g=10m/s²
What is the force that the ground apply on the sphere when it passes trough point B(bottom of the hole?

Homework Equations

Ep=m*g*h
Ec=m*v²/2
Q=m*v
F=m*a

The Attempt at a Solution

I tried finding the Potential energy of the sphere at point A that is
Ep=2kg*0,2m*10m/s² = 4J
and the weight of the sphere that is 2*10= 20 N
I don't know what to do anymore the velocity at point B i considered 2m/s because of
Ep=Ec=m*v²/2 then V=2m/s
Help?
Image not from actual problem point A would be the left point B where it is on the pic

 

[ShayanJ]

You can simply find the speed of the sphere when it reaches point B using conservation of energy.Call it $v_B$. When the sphere is moving in the semi-circle,it needs a centripetal force which at point B is equal to $\frac{mv_B^2}{R}$. But this centripetal force can be provided only by the circular track and nothing else so your answer is $\frac{mv_B^2}{R}$.

 

[Patolord]

AH thanks the answer is 60N i guess you have to sum that with the weight that is 20N
40 +20 =60 N
but i don't get where mv2/r comes .. i don't have that on my theory
they just gave mv²/2 and m*g*h

 

[PhanthomJay]

You need to understand the centripetal force concepts before solving this problem. You should know that the centripetal force mv^2/r is the net force acting inward toward the center of the circle. So the centripetal force is 40 N inward and is comprised of the net force of the ground normal force acting up
and the weight acting down, the net force

 

[Patolord]

Thank you, now i see why couldn't soulve any problems envolving circles xD