### Video Transcript

A solid sphere of mass 48 kilograms
is rolling in a straight line at a speed of 6.0 meters per second across a
horizontal surface. What is the magnitude of work
required to bring the sphere to rest?

We’ll label this work magnitude
capital 𝑊. And we’ll rely on the work-energy
theorem to help us solve for it. This theorem tells us that the work
done on an object is equal to that object’s change in kinetic energy. So in our case, if we can calculate
the change in this solid sphere’s kinetic energy as it goes from moving at 6.0
meters per second to a stop, then we’ll have solved for the work done on it.

If we make a sketch of this solid
sphere as it rolls along flat ground, we know that not only is it moving linearly or
translationally with speed 𝑣, but it must also be rotating about its own
center. The total kinetic energy of this
rolling sphere will involve adding its rotational kinetic energy along to its linear
or translational kinetic energy.

In order to solve then for the
sphere’s overall change in kinetic energy, we’ll recall the relationships for
rotational and translational kinetic energy. Rotational kinetic energy is equal
to an object’s moment of inertia, 𝐼, times its angular speed squared. And translational kinetic energy is
equal to one-half its mass times its linear speed squared. Writing these into our equation, we
know because we’ve been given 𝑚 and 𝑣, we can solve for those values. But we don’t yet know 𝐼 or 𝜔.

If we consider first the moment of
inertia 𝐼, we know that because we’re working with a solid sphere and it’s rotating
about its center, when we look up the moment of inertia of such a shape rotating in
such a way, we find it’s equal to two-fifths the mass of the sphere times its radius
squared. Plugging that expression in for 𝐼,
now we focus on the angular speed 𝜔.

We’ll recall that there’s a
relationship between linear speed 𝑣 and angular speed 𝜔. They’re related by the radius 𝑟,
𝑣 equals 𝑟𝜔. This means we can replace 𝜔 in our
expression with 𝑣 over 𝑟. And we see that when we do, the
factors of 𝑟 squared in this term cancel out, which is good because we don’t know
the radius of the sphere.

Our equation for Δ𝐾𝐸 then
simplifies to one-fifth 𝑚𝑣 squared plus one-half 𝑚𝑣 squared. This is equal to seven-tenths 𝑚𝑣
squared or plugging in for 𝑚 and 𝑣 seven-tenths times 48 kilograms times 6.0
meters per second squared. When we multiply out this
expression, to two significant figures, we find a result of 1.2 kilojoules. That’s the change in kinetic energy
of the solid sphere as it comes to a stop. And by the work energy theorem,
that’s also the work done on the sphere to stop it.