# Calculating work done on a rigid body

How in general does one calculate the work done by some force acting by a rigid body? Do you have to take into account the torque and the translation? For example:

Suppose we have a ball rolling down an incline (not slipping, there is rolling friction), starting at rest at the top. We have only two forces acting on the ball: gravity and friction.

We have $W_{grav}+W_{fric}=K$

How do you calculate the work done by a force on a rigid body? The work done by gravity is $mgh$ but how can we cacluate the work done by friction? Naively I think it is $Fd$ where $F$ is the friction force, but friction also induces the rotational motion of the ball.

• Friction does zero work on an object that is rolling without slipping. – David Hammen Jan 3 '15 at 21:02
• @DavidHammen : are you sure? The friction produces a torque that opposes the rolling of the object. The object has a rotational energy $\frac {I\omega ^2}{2}$ and since the friction force is parallel to the velocity at the contact between the ball and the ground, there is a mechanical work consuming energy. – Sofia Jan 3 '15 at 22:30
• @Sophia. Absolutely positive. This is basic stuff. See, for example, books.google.com/… . – David Hammen Jan 4 '15 at 4:09

There is more than one approach one can use to calculate the work done by a contact force on a rigid object. All approaches are valid so long as you're careful about how you actually use your calculated work.

One way is to use the infinitesimal displacement of the point of application $d\vec l_\text{app}$:

$$W = \int \vec F \cdot d\vec l_\text{app} \tag{1}$$

In the absence of other forces (and forms of potential energy), this will correctly give you the quantity $1/2 \int v^2 dm$, i.e., the change in total kinetic energy.

In your object-down-the-ramp example, the static frictional force would do zero work.

Another way to approach the same problem is to use the center of mass displacement $d\vec l_\text{CM}$:

$$W = \int \vec F \cdot d\vec l_\text{CM} \tag{2}$$

In the absence of other forces (and forms of potential energy), this will give you the change in center-of-mass kinetic energy $\Delta K_\text{CM} \equiv \Delta\left(\frac{1}{2}mv_\text{CM}^2\right)$.

Yet another way is an extension of method #2, in which one considers the rotational aspects of the motion:

$$W = W_\text{CM} + W_\text{rot} = \int \vec F \cdot d\vec l_\text{CM} + \int \vec\tau \cdot d\vec\theta \tag{3}$$

This latter approach lends itself well to separating out work that causes changes in the center-of-mass kinetic energy and changes in the rotational energy of a rigid body:

$$\Delta K_\text{tot} = \Delta K_\text{CM} + \Delta K_\text{rot}$$

For a rigid body, we have two different types of kinetic energy:

Translational KE

This is the usual $\frac{1}{2}mv^2$

Rotational KE

This is new, and takes the form $\frac{1}{2}J\omega^2$, where $J$ is the polar moment of inertia taken about the axis of rotation that passes through the centre of mass.

Work

To cause a change in translational KE, you need to do work by a resultant force, and the work done is the displacement of the centre of mass multiplied by the component of force that lies in the direction of the force.

However, we need to explain changes in rotational KE. This is caused by doing work by a resultant torque about the centre of mass. This work is equal to the rotation multiplied by the component of the resultant torque that lies in the plane of rotation.

Therefore, as an equation, the total work is given by:

$$W = \int \vec F \cdot d \vec r + \int \vec \tau \cdot d \vec \theta$$

Where $\vec F$ is the result force, $\vec \tau$ is the resultant torque, $d \vec r$ is an infinitesimal change in displacement of the centre of mass, and $d \vec \theta$ is an infinitesimal rotation about the axis of rotation going through the centre of mass.

• we meet again. You told me that when a force is perpendicular to the velocity is does no mechanical work. And I agree. But here, the friction is parallel with the velocity at the contact between the wheel and the floor. So, yes it does work, as the result of which the tangential velocity of the wheel is slowed down. All of you say that the friction doesn't do work. Then, why is it relevant at all if the floor opposes or doesn't oppose friction? There is a rolling friction, it is said explicitly. How do you stop your car? By imposing friction on the wheels. – Sofia Jan 4 '15 at 0:11
• Hello again. Actually, friction does indeed do work. The frictional force is contained inside the "resultant force" expression. You can either find the total work by adding up all the works done by individual forces and torques, or you can group all the forces and torques into a resultant force and a resultant torque, and find the work done by both to get the same total work. It is entirely your choice, but I personally prefer grouping stuff into resultants. If you want to work with, say, individual forces, then let $\vec F = \vec F_1 + \vec F_2 + ...$ – Involutius Jan 4 '15 at 0:19
• EternalCode : thanks, I see now that you speak in terms of resultant force and torque. – Sofia Jan 4 '15 at 0:29

The frictional force that keeps an object rolling without slipping does no work on the object. How can it? The point at which it operates is not moving with respect to the surface. That the velocity of the contact point is identically zero is the quintessential nature of rolling without slipping.

If the frictional force doesn't do work, what does it do? It is instead is a constraint force that acts to keep the object rolling without slipping.

The kinetic energy of an object that is translating and rotating is $KE = \frac12 mv^2 + \frac12 I\omega^2$ where $m$ is the mass of the object, $v$ is the velocity of the center of mass, $I$ is the moment of inertia about the center of mass, and $\omega$ is the angular velocity. In the case of an object that is rolling without slipping, the relation between angular velocity and center of mass velocity is given by $\omega = \frac v r$. The kinetic energy for such an object can thus be re-expressed as $KE = \frac12 m\bigl(1 + \frac I{mr^2}\bigr) v^2$.

In the case of an object rolling down a ramp starting from at rest and rolling down a vertical height $h$, the gravitational force does work $W=mgh$. The frictional force does no work, so the work-energy principle dictates that $mgh = \frac12 m\bigl(1 + \frac I{mr^2}\bigr) v^2$, or $$v = \sqrt{\frac 1 {1+\frac I{mr^2}} g h}$$

In the case of a solid sphere, $I=\frac 2 5 mr^2$ and thus the velocity is given by $v = \sqrt{\frac{10}7 gh}$. (See http://hyperphysics.phy-astr.gsu.edu/hbase/sphinc.html for an alternate derivation.) For a hollow sphere, $I=\frac 2 3 mr^2$, yielding $v = \sqrt{\frac65 gh}$.