A ball of mass m sits on an inclined plane, with its center of mass at a height h above the ground. It is
released from rest and allowed to roll without slipping down the plane. What is its velocity when itreaches the ground? I (^) ball = ( )mr 2 .
This is a situation you’ve seen before, except there’s a twist: this time, when the object moves down the
inclined plane, it gains both linear and rotational kinetic energy. However, it’s still just a conservation of
energy problem at heart. Initially, the ball just has gravitational potential energy, and when it reaches the
ground, it has both linear kinetic and rotational kinetic energy.
A bit of algebra, and we find that
If the ball in this problem hadn’t rolled down the plane—if it had just slid—its final velocity would have
been . (Don’t believe us? Try the calculation yourself for practice!) So it makes sense that the final
velocity of the ball when it does roll down the plane is less than ; only a fraction of the initial
potential energy is converted to linear kinetic energy.
Angular Momentum and Its Conservation
It probably won’t surprise you by this point that momentum, too, has a rotational form. It’s called angular
momentum (abbreviated, oddly, as L ), and it is found by this formula:
This formula makes intuitive sense. If you think of angular momentum as, roughly, the amount of effort it
would take to make a rotating object stop spinning, then it should seem logical that an object with a large
rotational inertia or with a high angular velocity (or both) would be pretty tough to bring to rest.
For a point particle, this formula can be rewritten as