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Work and Power in Rotation

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.

Work Done by Torque

$$W = \tau \Delta \theta$$

Power from Torque

$$P = \tau \omega$$
engine torque

Calculus-Based Definitions

When torque is not constant, we use calculus to find the total work done and instantaneous power in rotational motion.

$$ W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta $$

This sums up the small amounts of work done by torque over each tiny angular displacement.

$$ P(t) = \frac{dW}{dt} = \tau(t) \omega(t) $$

Instantaneous power is the product of torque and angular velocity at each moment.

Physical Meaning

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.

Compare: Linear vs. Rotational Work and 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 ($\omega$)

Interactive Simulation: Rotational Impulse and Angular Momentum

Explore how torque impulse affects angular momentum and angular velocity.

Angular Impulse: 0 N·m·s
Angular Momentum: 10.0 kg·m²/s
Angular Impulse Bar:

Example 1: Work Done by a Wrench

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.

Example 2: Motor Power

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.

Example 3: Variable Torque

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: $P_{\text{avg}} = \tau_{\text{avg}} \omega = 250 \times 2 = 500\,\text{W}$

wind turbine

Real-World Applications

common applications

Energy Conservation

The work-energy theorem applies to rotation: the net work done by torques equals the change in rotational kinetic energy.

$$W_{\text{net}} = \Delta K_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$$

We will go into more detail about energy conservation in the next concept.

Summary / Takeaways