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Ball and Bowl and Newton's Second Law

  1. Sep 24, 2003 #1
    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
  2. jcsd
  3. Sep 25, 2003 #2


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    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.
    That's your answer.

    Using "conservation of energy", we would have argued that at the top of the bowl, v is 0 and the height is R so the potential energy (relative to the bottom of the bowl) is mgR and the total energy is 0+ mgR= mgR. At the bottom of the bowl, the potential energy is 0 so the kinetic energy must be mgR in order to keep total energy the same. It's that easy!

    Actually, if you look at the integral of the equation resulting form Newton's second law, you will see that that IS conservation of energy. If the force, F, depends only on position: F(x), it is a "conservative" force. Newton's second law is m dv/dt= F(x). Again, we can use the chain rule to write this as m v dv/dx= F(x) and rewrite that as m v dv= F(x) dx. Integrating, (1/2)mv2= integral(F(x)dx)+ C or (1/2)mv2- integral(F(x)dx)= C.
    The first part of that is kinetic energy. The (negative) integral is the work done against the force: the potential energy. That just says that the total energy is a constant: conservation of energy.
  4. Sep 25, 2003 #3
    Thanks. Thats the answer I got as well. I just needed to know because it got marked wrong on a webassign HW. I probably just didn't write it the right way...but it had to be right.

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