Video: Work–Energy Theorem in Rotational Motion

A solid sphere of mass 48 kg is rolling in a straight line at a speed of 6.0 m/s across a horizontal surface. What is the magnitude of work required to bring the sphere to rest?

02:56

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.

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