Just as forces do work in linear motion, torques do work in rotational motion. When a torque acts through an angular displacement, it transfers energy to the rotating object. The rate at which this work is done is the rotational power.
When torque is not constant, we use calculus to find the total work done and instantaneous power in rotational motion.
This sums up the small amounts of work done by torque over each tiny angular displacement.
Instantaneous power is the product of torque and angular velocity at each moment.
Work: When you apply a torque (like turning a wrench), you do work equal to the torque times the angle through which you turn. This work can increase the rotational kinetic energy of the object.
Power: The rate at which work is done. A motor that applies a constant torque while rotating at a constant angular velocity delivers constant power.
Linear | Rotational |
---|---|
Work: $W = F \Delta x$ | Work: $W = \tau \Delta \theta$ |
Power: $P = Fv$ | Power: $P = \tau \omega$ |
Force ($F$) | Torque ($\tau$) |
Displacement ($\Delta x$) | Angular displacement ($\Delta \theta$) |
Velocity ($v$) | Angular velocity ($\omega$) |
Explore how torque impulse affects angular momentum and angular velocity.
A mechanic applies a torque of $25\,\text{N} \cdot \text{m}$ to tighten a bolt, rotating it through $0.5\,\text{rad}$. How much work is done?
Solution:
$W = \tau \Delta \theta = 25 \times 0.5 = 12.5\,\text{J}$
The mechanic does 12.5 Joules of work to tighten the bolt.
An electric motor applies a constant torque of $15\,\text{N} \cdot \text{m}$ while rotating at $120\,\text{rad/s}$. What is the motor's power output?
Solution:
$P = \tau \omega = 15 \times 120 = 1800\,\text{W} = 1.8\,\text{kW}$
The motor delivers 1.8 kilowatts of power.
A wind turbine blade experiences a torque that varies with time: $\tau(t) = 50t\,\text{N} \cdot \text{m}$ for $0 \leq t \leq 10\,\text{s}$. If it rotates at constant $\omega = 2\,\text{rad/s}$, what is the average power over 10 seconds?
Solution:
Average torque: $\tau_{\text{avg}} = \frac{1}{10} \int_0^{10} 50t \, dt = \frac{1}{10} \times 2500 = 250\,\text{N} \cdot \text{m}$
Average power: $P_{\text{avg}} = \tau_{\text{avg}} \omega = 250 \times 2 = 500\,\text{W}$
The work-energy theorem applies to rotation: the net work done by torques equals the change in rotational kinetic energy.
We will go into more detail about energy conservation in the next concept.