12.7.2 The Constant-Torque Model

If all the torques exerted on an object are constant, the object rotates with constant angular acceleration. Even if the torques aren’t perfectly constant, there are many situations where it’s reasonable to model them as if they were. The constant-torque model, analogous to the constant-force model of Section 6.2, is the most important model of rotational dynamics.

MODEL 12.2 Constant torque

For objects on which the net torque is constant.

Model the object as a rigid body with constant angular acceleration.
Take into account constraints due to ropes and pulleys.
Mathematically:
- Newton’s second law is $τ_{net} = I α$ .
- Use the kinematics of constant angular acceleration.
Limitations: Model fails if the torque isn’t constant.

The figure illustrates a rolling circular object and explains the constant torque model.

Full Alternative Text

Example 12.12 Lowering a bucket

A 2.0 kg bucket is attached to a massless string that is wrapped around a 1.0 kg, 4.0-cm-diameter cylinder, as shown in FIGURE 12.30a. The cylinder rotates on an axle through the center. The bucket is released from rest 1.0 m above the floor. How long does it take to reach the floor?

The figure consists of two parts labeled ("a") and (b) that pictorially represent a falling bucket attached to a rope that turns a pulley in the form of a cylinder.

FIGURE 12.30 Full Alternative Text

MODEL Assume the string does not slip.

VISUALIZE FIGURE 12.30b shows the free-body diagram for the cylinder and the bucket. The string tension exerts an upward force on the bucket and a downward force on the outer edge of the cylinder. The string is massless, so these two tension forces act as if they are an action/reaction pair: $T_{b} = T_{c} = T$ .

SOLVE Newton’s second law applied to the linear motion of the bucket is

m a_{y} = T - m g

where, as usual, the y-axis points upward. What about the cylinder? The only torque comes from the string tension. The moment arm for the tension is $d = R$ , and the torque is positive because the string turns the cylinder ccw. Thus $τ_{string} = T R$ and Newton’s second law for the rotational motion is

α = \frac{τ_{net}}{I} = \frac{T R}{\frac{1}{2} M R^{2}} = \frac{2 T}{M R}

The moment of inertia of a cylinder rotating about a center axis was taken from Table 12.2.

The last piece of information we need is the constraint due to the fact that the string doesn’t slip. Equation 12.33 relates only the absolute values, but in this problem $α$ is positive (ccw acceleration) while $a_{y}$ is negative (downward acceleration). Hence

a_{y} = - α R

Using $α$ from the cylinder’s equation in the constraint, we find

a_{y} = - α R = - \frac{2 T}{M R} R = - \frac{2 T}{M}

Thus the tension is $T = - \frac{1}{2} M a_{y}$ . If we use this value of the tension in the bucket’s equation, we can solve for the acceleration:

\begin{array}{l} m a_{y} = - \frac{1}{2} M a_{y} - m g \\ a_{y} = - \frac{g}{(1 + M/2m)} = - 7.84 {m/s}^{2} \end{array}

The time to fall through $Δ y = - 1.0 m$ is found from kinematics:

\begin{array}{l} Δ y = \frac{1}{2} a_{y} {(Δ t)}^{2} \\ Δ t = \sqrt{\frac{2 Δ y}{a_{y}}} = \sqrt{\frac{2 (- 1.0 m)}{- 7.84 {m/s}^{2}}} = 0.50 s \end{array}

REVIEW The expression for the acceleration gives $a_{y} = - g$ if $M = 0$ . This makes sense because the bucket would be in free fall if there were no cylinder. When the cylinder has mass, the downward force of gravity on the bucket has to accelerate the bucket and spin the cylinder. Consequently, the acceleration is reduced and the bucket takes longer to fall.