Additional: Work, Energy, and Power in Rotation

Work Done by a Torque ($ W = \int \tau \, d\theta $)

Just as a force doing work in translational motion transfers energy, a torque doing work in rotational motion also transfers energy to the rotating body. The work done by a torque is the rotational equivalent of the work done by a force.

Definition of Work Done by Torque

Consider a rigid body rotating about a fixed axis. A torque $\vec{\tau}$ acts on the body, causing an infinitesimal angular displacement $d\vec{\theta}$ about the same axis. The infinitesimal work done ($dW$) by the torque is defined as the dot product of the torque vector and the angular displacement vector:

$ dW = \vec{\tau} \cdot d\vec{\theta} $

For rotation about a fixed axis (say, the z-axis), both $\vec{\tau}$ and $d\vec{\theta}$ are directed along the axis. Their dot product simplifies to the product of their magnitudes (or components along the axis, with appropriate sign):

$ dW = \tau_z \, d\theta_z $

Let's use scalar notation $\tau$ and $d\theta$ for components along the axis, with direction handled by the sign (e.g., positive for counterclockwise rotation and torque).

$ dW = \tau \, d\theta $

The total work done by a torque during a finite angular displacement from $\theta_1$ to $\theta_2$ is the integral of $dW$ over the angular displacement:

$ W = \int_{\theta_1}^{\theta_2} dW = \int_{\theta_1}^{\theta_2} \tau \, d\theta $

where $\tau$ is the torque acting about the fixed axis, and the integration is performed over the angular displacement $\Delta\theta = \theta_2 - \theta_1$.

Work Done by a Constant Torque

If the torque $\tau$ is constant during the angular displacement $\Delta\theta = \theta_2 - \theta_1$, the integral simplifies:

$ W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = \tau \int_{\theta_1}^{\theta_2} d\theta = \tau (\theta_2 - \theta_1) $

$ W = \tau \Delta\theta $

This is the rotational analogue of the work done by a constant force ($W = F \Delta x$).

The units of work done by torque are the same as those for work done by force: Joules (J). Note that torque has units of N·m, and angular displacement is in radians (a dimensionless unit), so N·m $\times$ radians = N·m = J.

Rotational Kinetic Energy ($ KE_{rot} = \frac{1}{2}I\omega^2 $)

As discussed in the previous section, a rigid body rotating about a fixed axis with angular velocity $\omega$ possesses rotational kinetic energy. This energy arises from the motion of all the particles making up the rigid body.

Derivation from Linear Kinetic Energy

Consider a rigid body rotating about a fixed axis with angular speed $\omega$. A particle of mass $m_i$ at a perpendicular distance $r_i$ from the axis has a linear speed $v_i = r_i \omega$. The kinetic energy of this particle is $KE_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2$.

The total kinetic energy of the rigid body rotating about this axis is the sum of the kinetic energies of all its particles:

$ KE_{total} = \sum_i KE_i = \sum_i \frac{1}{2} m_i r_i^2 \omega^2 $

Since $\omega$ is the same for all particles in a rigid body, we can factor it out:

$ KE_{total} = \frac{1}{2} \left(\sum_i m_i r_i^2\right) \omega^2 $

The term in the parenthesis, $\sum_i m_i r_i^2$, is the moment of inertia $I$ of the rigid body about the fixed axis of rotation.

Thus, the rotational kinetic energy is:

$ KE_{rot} = \frac{1}{2} I \omega^2 $

This formula applies specifically to the rotational kinetic energy about the axis of rotation. For rolling motion, where there is both translation of the centre of mass and rotation about an axis through the CM, the total kinetic energy is the sum of translational and rotational kinetic energies, $KE_{total} = KE_{trans} + KE_{rot, CM} = \frac{1}{2}Mv_{CM}^2 + \frac{1}{2}I_{CM}\omega^2$.

The units of rotational kinetic energy are Joules (J).

Work-Energy Theorem for Rotation

Similar to the Work-Energy Theorem for translational motion, there is a Work-Energy Theorem for rotational motion. It states that the net work done by external torques on a rigid body rotating about a fixed axis is equal to the change in its rotational kinetic energy.

Statement of the Work-Energy Theorem for Rotation

The net work done by all external torques acting on a rigid body as it rotates about a fixed axis is equal to the change in its rotational kinetic energy.

$ W_{net, ext} = \Delta KE_{rot} $

where:

• $W_{net, ext}$ is the total work done by all external torques about the fixed axis.
• $\Delta KE_{rot} = KE_{rot, final} - KE_{rot, initial} = \frac{1}{2} I \omega_{final}^2 - \frac{1}{2} I \omega_{initial}^2$.

So, $ W_{net, ext} = \frac{1}{2} I \omega_{final}^2 - \frac{1}{2} I \omega_{initial}^2 $

Derivation (using Constant Net Torque)

Consider a rigid body rotating about a fixed axis with constant net external torque $\tau_{net}$. This torque causes a constant angular acceleration $\alpha = \tau_{net}/I$. Let the initial angular velocity be $\omega_0$ and the final angular velocity be $\omega$ after an angular displacement $\Delta\theta$.

From rotational kinematics with constant angular acceleration, we have the equation $\omega^2 = \omega_0^2 + 2\alpha \Delta\theta$.

Rearranging this equation to solve for $\alpha \Delta\theta$:

$ \omega^2 - \omega_0^2 = 2\alpha \Delta\theta $

$ \alpha \Delta\theta = \frac{\omega^2 - \omega_0^2}{2} $

The net work done by the constant net torque is $W_{net, ext} = \tau_{net} \Delta\theta$. Substitute $\tau_{net} = I\alpha$:

$ W_{net, ext} = (I\alpha) \Delta\theta = I (\alpha \Delta\theta) $

Now, substitute the expression for $\alpha \Delta\theta$:

$ W_{net, ext} = I \left(\frac{\omega^2 - \omega_0^2}{2}\right) $

$ W_{net, ext} = \frac{1}{2} I \omega^2 - \frac{1}{2} I \omega_0^2 $

Recognizing the terms as the final and initial rotational kinetic energies, we get:

$ W_{net, ext} = KE_{rot, final} - KE_{rot, initial} = \Delta KE_{rot} $

This confirms the Work-Energy Theorem for rotation about a fixed axis under constant net torque. It is generally valid even for variable torques (where the work is calculated by integration) and holds true as long as the axis of rotation is fixed.

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Power in Rotational Motion ($ P = \tau \omega $)

Power is the rate at which work is done. For rotational motion, the instantaneous power delivered by a torque is the product of the torque and the instantaneous angular velocity.

Definition of Power in Rotation

From the definition of instantaneous power $P = dW/dt$ and the infinitesimal work done by torque $dW = \vec{\tau} \cdot d\vec{\theta}$, we have:

$ P = \frac{d}{dt} (\vec{\tau} \cdot \vec{\theta}) $ (Caution: this is not always the case for time-varying torque and angular displacement vectors. Let's use the infinitesimal work form)

$ P = \frac{dW}{dt} = \frac{\vec{\tau} \cdot d\vec{\theta}}{dt} $

Since the instantaneous angular velocity is $\vec{\omega} = \frac{d\vec{\theta}}{dt}$, assuming the angular displacement is along the axis of rotation, we get:

$ P = \vec{\tau} \cdot \vec{\omega} $

For rotation about a fixed axis (say, the z-axis), where $\vec{\tau}$ and $\vec{\omega}$ are parallel to the axis, the dot product becomes the product of the components along the axis:

$ P = \tau_z \omega_z $

Using scalar notation for the components along the axis:

$ P = \tau \omega $

where $\tau$ is the magnitude of the torque about the axis and $\omega$ is the magnitude of the angular velocity about the axis. The power is positive if the torque and angular velocity are in the same rotational direction (e.g., both counterclockwise), meaning energy is being transferred to the body, increasing its rotational kinetic energy. The power is negative if they are in opposite directions, meaning energy is being removed from the body.

The units of power are Joules per second (J/s), which is Watts (W). $\tau \omega$ has units of (N·m) $\times$ (rad/s) = N·m/s = J/s = W. This is consistent.

Relation to Linear Power ($ P = \vec{F} \cdot \vec{v} $)

The formula $P = \tau \omega$ is the rotational analogue of the linear power formula $P = \vec{F} \cdot \vec{v}$ or $P = Fv$ when force and velocity are parallel. It tells us the rate at which a torque is doing work on a rotating body.

This formula is useful for calculating the power output of rotating machinery, such as the power delivered by a motor to a rotating shaft, or the power generated by a turbine, given the torque and angular speed.

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