# Ball and Bowl and Newton's Second Law

RIT_Rich
Hello everyone.

I'm having a little trouble understanding the whole "ball and bowl" problem using Newton's Second Law. What I'm talking about involves a small shpere of mass m and radius r, rolling down a "bowl" (more like a hemisphere of radius R). How would one go about finding the KE of the sphere at the bottom of the hemisphere using Newton's Second Law?

Thanks a lot

HallsofIvy
Homework Helper
Do you really have to use Newton's second law? If I were doing a problem like this, I would use "conservation of energy".

Newton's law says "force is the rate of change of momentum" (which can actually be taken as a definition of force). In the simple case of constant mass, this is "force= mass times acceleration" or
m dv/dt= F.

Model the cross section of the bowl that includes the ball, as it is rolling down, as a circle of radius R. Let theta be the angle formed by the line from the ball to the center of the circle and then to the bottom of the circle. The gravitational force on the ball is -mg, straight down. The ball does not move straight down because the bowl is applying a force toward its center. The net force is tangent to the bowl. Dividing the gravitational force into components parallel and perpendicular to the tangent we see that the angle between perpendicular and vertical lines is also theta. The net force, tangent to the circle, is -mg sin(theta).

Newton's second law now says m dv/dt= -mg sin(theta) (v is the speed along the circumference of the circle). If we measure theta in radians, then the arclength is given by s= Rtheta so
v= ds/dt= R d theta/dt. Newton's second law is now:
R d2theta/dt20= -g sin(theta).
If we let w= dtheta/dt (the angular velocity), then d2theta/dt2= dw/dt. Using the chain rule, dw/dt= dtheta/dt dw/dtheta= w dw/dtheta. Now the equation is R w dw/dtheta= -g sin(theta). We can rewrite this as w dw= -(g/R) sin(theta) dtheta and integrate. That gives (1/2)w2= (g/R) cos(theta)+ C.

At the top of the bowl theta= pi/2 and cos(pi/2)= 0. Also, since the ball started from rest, w= 0 so we have 0= 0+ C or C= 0.

Our equation is (1/2)w2= (g/R) cos(theta).

At the bottom of the bowl, theta= 0 and cos(0)= 1. Therefore,
(1/2)w2= g/R.

We were asked for kinetic energy at the bottom of the bowl. That is (1/2)m v2= (1/2)(Rw)2= R2((1/2)w2)= R2(g/R)= gR.