Gravity exerts a torque on many objects. If the object in FIGURE 12.23 is released, a torque due to gravity will cause it to rotate around the axle. To calculate the torque about the axle, we start with the fact that gravity acts on every particle in the object, exerting a downward force of magnitude $F_i=m_{i}g$ on particle i. The magnitude of the gravitational torque on particle i is $\left|{\tau }_i\right|=d_i{m}_{i}g$, where $d_i$ is the moment arm. But we need to be careful with signs.

FIGURE 12.23 Gravitational torque.

A moment arm must be a positive number because it’s a distance. If we establish a coordinate system with the origin at the axle, then you can see from Figure 12.23a that the moment arm $d_i$ of particle i is $\left|x_i\right|$. A particle to the right of the axle (positive $x_i$ experiences a negative torque because gravity tries to rotate this particle in a clockwise direction. Similarly, a particle to the left of the axle (negative $x_i)$ has a positive torque. The torque is opposite in sign to $x_i$, so we can get the sign right by writing

The net torque due to gravity is found by summing Equation 12.24 over all particles:

But according to the definition of center of mass, Equations 12.4, $\sum m_i{x}_i=M{x}_{\text{cm}}$. Thus the torque due to gravity is

where $x_{\text{cm}}$ is the position of the center of mass relative to the axis of rotation.

Equation 12.26 has the simple interpretation shown in Figure 12.23b. Mg is the net gravitational force on the entire object, and $x_{\text{cm}}$ is the moment arm between the rotation axis and the center of mass. The gravitational torque on an extended object of mass M is equivalent to the torque of a single force vector $\vec{F}{}_{\text{G}}=-Mgĵ$ acting at the object’s center of mass.

In other words, the gravitational torque is found by treating the object as if all its mass were concentrated at the center of mass. This is the basis for the well-known technique of finding an object’s center of mass by balancing it. An object will balance on a pivot, as shown in FIGURE 12.24, only if the center of mass is directly above the pivot point. If the pivot is not under the center of mass, the gravitational torque will cause the object to rotate.

FIGURE 12.24 An object balances on a pivot that is directly under the center of mass.

EXAMPLE 12.9 The gravitational torque on a beam

The 4.00-m-long, 500 kg steel beam shown in FIGURE 12.25 is supported 1.20 m from the right end. What is the gravitational torque about the support?

FIGURE 12.25 A steel beam supported at one point.

FIGURE 12.25 Full Alternative Text

MODEL The center of mass of the beam is at the midpoint. $x_{\text{cm}}=-0.80\text{m}$ is measured from the pivot point.

SOLVE This is a straightforward application of Equation 12.26. The gravitational torque is

$${\tau }_{\text{grav}}=-Mg{x}_{\text{cm}}=-(500\text{ kg})(9.80{\text{ m/s}}^2)(-0.80\text{ m})=3920\text{ N m}$$

REVIEW The torque is positive because gravity tries to rotate the beam ccw. Notice that the beam in Figure 12.25 is not in equilibrium. It will fall over unless other forces, not shown, are supporting it.