What does a torque do? A torque causes an angular acceleration. To see why, Figure 12.26 shows a rigid body undergoing pure rotational motion about a fixed and unmoving axis. This might be a rotation about the object’s center of mass, such as we considered in Section 12.2. Or it might be an object, such as a turbine, rotating on an axle.

FIGURE 12.26 The external forces exert a torque about the rotation axis.

The forces ${\vec{F}}_1\text{,}{\vec{F}}_2\text{,}{\vec{F}}_3\text{,}\text{.}\text{.}$. in FIGURE 12.26 are external forces acting on particles of masses $m_1\text{,}{m}_2\text{,}{m}_3\text{,}\text{.}\text{.}$. that are part of the rigid body. These forces exert torques ${\tau }_1\text{,}{\tau }_2\text{,}{\tau }_3\text{,}\text{.}\text{.}$. about the rotation axis. The net torque on the object is the sum of the torques:

Focus on particle $i$, which is acted on by force $\vec{F}{}_i$ and undergoes circular motion with radius $r_i$. In Chapter 8, we found that the radial component of $\vec{F}{}_i$ is responsible for the centripetal acceleration of circular motion, while the tangential component ${\left(F_i\right)}_t$ causes the particle to speed up or slow down with a tangential acceleration ${\left(a_i\right)}_t$. Newton’s second law is

where in the last step we used the relationship between tangential and angular acceleration: $a_t=r\alpha$. The angular acceleration $\alpha$ does not have a subscript because all particles in the object have the same angular acceleration. That is, $\alpha$ is the angular acceleration of the entire object.

Multiplying both sides by $r_i$ gives

But $r_i{\left(F_i\right)}_t$ is the torque ${\tau }_i$ about the axis on particle $i;$ hence Newton’s second law for a single particle in the object is

Returning now to Equation 12.27, we see that the net torque on the object is

In the last step, we factored out $\alpha$ by using the key idea that every particle in a rotating rigid body has the same angular acceleration.

You’ll recognize the quantity in parentheses as the moment of inertia. Substituting I into Equation 12.31 puts the final piece of the puzzle into place. An object that experiences a net torque ${\tau }_{\text{net}}$ about the axis of rotation undergoes an angular acceleration

where I is the object’s moment of inertia about the rotation axis. This result, Newton’s second law for rotation, is the fundamental equation of rigid-body dynamics.

In practice we often write ${\tau }_{\text{net}}=I\alpha$, but Equation 12.32 better conveys the idea that torque is the cause of angular acceleration. In the absence of a net torque $({\tau }_{\text{net}}=0)$, the object either does not rotate $(\omega =0)$ or rotates with constant angular velocity $(\omega =\text{constant})$.

TABLE 12.3 summarizes the analogies between linear and rotational dynamics.

EXAMPLE 12.10 Rotating rockets

Far out in space, a 100,000 kg rocket and a 200,000 kg rocket are docked at opposite ends of a motionless 90-m-long connecting tunnel. The tunnel is rigid and its mass is much less than that of either rocket. The rockets start their engines simultaneously, each generating 50,000 N of thrust in opposite directions. What is the structure’s angular velocity after 30 s?

MODEL The entire structure can be modeled as two masses at the ends of a massless, rigid rod. There’s no net force, so the structure does not undergo translational motion, but the thrusts do create torques that will give the structure angular acceleration and cause it to rotate. We’ll assume the thrust forces are perpendicular to the connecting tunnel. This is an unconstrained rotation, so the structure will rotate about its center of mass.

VISUALIZE FIGURE 12.27 shows the rockets and defines distances $r_1$ and $r_2$ from the center of mass.

FIGURE 12.27 The thrusts exert a torque on the structure.

FIGURE 12.27 Full Alternative Text

SOLVE Our strategy will be to use Newton’s second law to find the angular acceleration, followed by rotational kinematics to find $\omega$. We’ll need to determine the moment of inertia, and that requires knowing the distances of the two rockets from the rotation axis. As we did in Example 12.1, we choose a coordinate system in which the masses are on the x-axis and in which $m_1$ is at the origin. Then

$$\begin{array}{ll}{x}_{\text{cm}}\hfill & ={\displaystyle \frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}}\hfill \\ \hfill & ={\displaystyle \frac{(100\text{,}000\text{ kg})\left(0\text{ m}\right)+\left(200\text{,}000\text{ kg}\right)\left(90\text{ m}\right)}{100\text{,}000\text{ kg}+200\text{,}000\text{ kg}}}=60\text{ m}\hfill \end{array}$$

The structure’s center of mass is $r_1=60\text{m}$ from the 100,000 kg rocket and $r_2=30\text{m}$ from the 200,000 kg rocket. The moment of inertia about the center of mass is

$$I=m_1{r}_1{}^2+m_2{r}_2{}^2=540\text{,}000\text{,}000{\text{ kg m}}^2$$

The two rocket thrusts exert net torque

$$\begin{array}{ll}{\tau }_{\text{net}}\hfill & =r_1{F}_1+r_2{F}_2=(60\text{m})(50\text{,}000\text{N})+(30\text{m})(50\text{,}000\text{N})\hfill \\ \hfill & =4\text{,}500\text{,}000\text{N}\text{m}\hfill \end{array}$$

With I and ${\tau }_{\text{net}}$ now known, we can use Newton’s second law to find the angular acceleration:

$$\alpha =\frac{\tau }{I}=\frac{4\text{,}500\text{,}000\text{ N m}}{540\text{,}000\text{,}000{\text{ kg m}}^2}=0.00833\text{ rad}\text{/}{\text{s}}^2$$

After 30 seconds, the structure’s angular velocity is

$$\omega =\alpha \text{Δ}t=0.25\text{rad}\text{/}\text{s}$$