Interactive Lecture
Torque, Moment of Inertia, and Angular Motion
by Dr Arnaldo Delli Carri
Rotational dynamics describes how objects rotate when subjected to torques. Just as forces cause linear acceleration in Newton's second law, torques cause angular acceleration. Rotational dynamics is what lets you analyse spinning wheels, rotating machinery, and pivoting structures.
Rotational motion is the angular analogue of linear motion. Every concept in linear dynamics — force, mass, acceleration, momentum — has a direct rotational counterpart: torque, moment of inertia, angular acceleration, angular momentum.
Before studying rotational dynamics, we must understand how to describe rotational motion:
Angular Position: \(\theta\) (radians)
Angular Velocity: \(\omega = \frac{d\theta}{dt}\) (rad/s)
Angular Acceleration: \(\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}\) (rad/s²)
For constant angular acceleration, the rotational kinematics equations mirror the linear SUVAT equations:
Angular Kinematics (Constant \(\alpha\)):
Where:
Torque (also called moment) is the rotational equivalent of force. It measures the turning effect of a force about a pivot point or axis of rotation.
Torque Definition:
\[\vec{\tau} = \vec{r} \times \vec{F}\]
Magnitude: \(\tau = rF\sin\theta\)
For perpendicular force: \(\tau = rF\) or \(\tau = Fd\)
Where:
Torque is a vector quantity with direction given by the right-hand rule. Conventionally, anti-clockwise rotation is positive (+), and clockwise rotation is negative (−).
Moment of inertia \(I\) is the rotational analogue of mass. It quantifies an object's resistance to changes in rotational motion and depends on both the mass distribution and the axis of rotation.
Moment of Inertia:
\[I = \sum m_i r_i^2 = \int r^2 \, dm\]
SI Unit: kg·m²
Where \(r\) is the perpendicular distance from the axis of rotation to the mass element \(dm\).
Point Mass: \(I = mr^2\)
Thin Rod (about centre): \(I = \frac{1}{12}mL^2\)
Thin Rod (about end): \(I = \frac{1}{3}mL^2\)
Solid Disk/Cylinder (about centre axis): \(I = \frac{1}{2}mr^2\)
Hollow Cylinder (about centre axis): \(I = mr^2\)
Solid Sphere (about diameter): \(I = \frac{2}{5}mr^2\)
Hollow Sphere (about diameter): \(I = \frac{2}{3}mr^2\)
Rectangular Plate (about centre, perpendicular to plate): \(I = \frac{1}{12}m(a^2 + b^2)\)
Here \(L\) is the rod's length. Later in this lecture the same letter \(L\) stands for angular momentum (\(L = I\omega\)); the meaning is set by context.
The parallel axis theorem allows us to calculate the moment of inertia about any axis parallel to an axis through the centre of mass.
Parallel Axis Theorem:
\[I = I_{\text{cm}} + md^2\]
Where:
The rotational analogue of Newton's second law \(F = ma\) relates torque to angular acceleration:
Rotational Dynamics Equation:
\[\sum \tau = I\alpha\]
Where:
Just as a larger mass requires more force to achieve the same acceleration, a larger moment of inertia requires more torque to achieve the same angular acceleration.
A rotating object possesses kinetic energy due to its rotation:
Rotational Kinetic Energy:
\[E_{\text{rot}} = \frac{1}{2}I\omega^2\]
This is analogous to linear kinetic energy: \(E_{\text{lin}} = \frac{1}{2}mv^2\)
Rotational Work: \(W = \tau \theta\)
Rotational Power: \(P = \tau \omega\)
Work-Energy Theorem: \(W = \Delta E_{\text{rot}} = \frac{1}{2}I\omega^2 - \frac{1}{2}I\omega_0^2\)
Angular momentum is the rotational analogue of linear momentum:
Angular Momentum:
\[L = I\omega\]
SI Unit: kg·m²/s
Conservation of Angular Momentum:
If the net external torque is zero: \(\sum \tau_{\text{ext}} = 0\)
Then: \(L = I\omega = \text{constant}\)
Conservation of angular momentum explains why ice skaters spin faster when they pull their arms in — decreasing \(I\), so \(\omega\) must increase to keep \(L\) constant — and why divers tuck to increase rotation speed.
For a point at distance \(r\) from the rotation axis:
Linear velocity: \(v = r\omega\)
Tangential acceleration: \(a_t = r\alpha\)
Centripetal acceleration: \(a_c = r\omega^2 = \frac{v^2}{r}\)
Watch all four shapes rotate under the same applied torque. Notice how different mass distributions lead to different angular accelerations and velocities, even with identical mass and radius!
All shapes have the same mass and radius, but different moments of inertia due to mass distribution. With the rod modelled at length \(L = 2r\), the hollow ring accelerates slowest (largest \(I\)), while the thin rod accelerates fastest (smallest \(I\)).
| Concept | Equation | Description |
|---|---|---|
| Angular Velocity | \(\omega = \frac{d\theta}{dt}\) | Rate of change of angular position |
| Angular Acceleration | \(\alpha = \frac{d\omega}{dt}\) | Rate of change of angular velocity |
| Torque | \(\tau = rF\sin\theta\) | Rotational effect of a force |
| Moment of Inertia | \(I = \sum m_i r_i^2\) | Rotational inertia |
| Solid Disk/Cylinder | \(I = \frac{1}{2}mr^2\) | About central axis |
| Hollow Cylinder | \(I = mr^2\) | Thin-walled, about central axis |
| Solid Sphere | \(I = \frac{2}{5}mr^2\) | About diameter |
| Thin Rod (centre) | \(I = \frac{1}{12}mL^2\) | About centre, perpendicular to rod |
| Parallel Axis Theorem | \(I = I_{\text{cm}} + md^2\) | Shift axis from centre of mass |
| Rotational Dynamics | \(\sum \tau = I\alpha\) | Newton's 2nd Law for rotation |
| Rotational KE | \(E_{\text{rot}} = \frac{1}{2}I\omega^2\) | Kinetic energy of rotation |
| Angular Momentum | \(L = I\omega\) | Rotational momentum |
| Rotational Power | \(P = \tau\omega\) | Rate of doing rotational work |
| Linear-Angular Link | \(v = r\omega\) | Tangential velocity |
A solid disk of mass 5 kg and radius 0.3 m is free to rotate about its central axis. A force of 20 N is applied tangentially to the rim of the disk.
Show Solution
Given: \(m = 5\) kg, \(r = 0.3\) m, \(F = 20\) N (tangential)
a) Moment of inertia:
For a solid disk: \(I = \frac{1}{2}mr^2\)
\[I = \frac{1}{2}(5)(0.3)^2 = \frac{1}{2}(5)(0.09) = 0.225 \text{ kg·m}^2\]
b) Torque:
Since force is tangential: \(\tau = rF\)
\[\tau = (0.3)(20) = 6 \text{ N·m}\]
c) Angular acceleration:
Using: \(\tau = I\alpha\)
\[\alpha = \frac{\tau}{I} = \frac{6}{0.225} = 26.67 \text{ rad/s}^2\]
d) Angular velocity after 3 seconds:
Using: \(\omega = \omega_0 + \alpha t\) with \(\omega_0 = 0\)
\[\omega = 0 + (26.67)(3) = 80 \text{ rad/s}\]
Answer: \(I = 0.225\,\text{kg·m}^2\), \(\tau = 6\,\text{N·m}\), \(\alpha = 26.67\,\text{rad/s}^2\), \(\omega = 80\,\text{rad/s}\).
A thin uniform rod of mass 2 kg and length 1.2 m is pivoted about a point 0.3 m from one end.
Show Solution
Given: \(m = 2\) kg, \(L = 1.2\) m, pivot at 0.3 m from one end
a) Moment of inertia about centre:
For a thin rod about centre: \(I_{\text{cm}} = \frac{1}{12}mL^2\)
\[I_{\text{cm}} = \frac{1}{12}(2)(1.2)^2 = \frac{1}{12}(2)(1.44) = 0.24 \text{ kg·m}^2\]
b) Moment of inertia about pivot:
Centre of mass is at \(L/2 = 0.6\) m from either end
Distance from pivot to centre of mass: \(d = 0.6 - 0.3 = 0.3\) m
Using parallel axis theorem: \(I = I_{\text{cm}} + md^2\)
\[I = 0.24 + (2)(0.3)^2 = 0.24 + 0.18 = 0.42 \text{ kg·m}^2\]
c) Angular acceleration:
Using: \(\tau = I\alpha\)
\[\alpha = \frac{\tau}{I} = \frac{8}{0.42} = 19.05 \text{ rad/s}^2\]
Answer: \(I_{\text{cm}} = 0.24\,\text{kg·m}^2\), \(I_{\text{pivot}} = 0.42\,\text{kg·m}^2\), \(\alpha = 19.05\,\text{rad/s}^2\).
A student sits on a rotating stool holding a 3 kg dumbbell in each hand. With arms extended, the student rotates at 1.5 rad/s with a total moment of inertia (student + dumbbells) of 6 kg·m². The student then pulls the dumbbells inward, reducing the total moment of inertia to 3 kg·m².
Show Solution
Given: \(I_1 = 6\) kg·m², \(\omega_1 = 1.5\) rad/s, \(I_2 = 3\) kg·m²
a) Initial angular momentum:
\[L = I_1\omega_1 = (6)(1.5) = 9 \text{ kg·m}^2\text{/s}\]
b) Final angular velocity:
By conservation of angular momentum: \(L_1 = L_2\)
\[I_1\omega_1 = I_2\omega_2\]
\[\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{(6)(1.5)}{3} = 3 \text{ rad/s}\]
c) Change in kinetic energy:
\[E_1 = \frac{1}{2}I_1\omega_1^2 = \frac{1}{2}(6)(1.5)^2 = 6.75 \text{ J}\]
\[E_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(3)(3)^2 = 13.5 \text{ J}\]
\[\Delta E = E_2 - E_1 = 13.5 - 6.75 = 6.75 \text{ J}\]
d) Energy source:
The extra kinetic energy came from the student's muscular work done in pulling the dumbbells inward against the centrifugal effect. Angular momentum is conserved, but mechanical energy is not conserved when internal forces do work.
Answer: \(L = 9\,\text{kg·m}^2\text{/s}\), \(\omega_2 = 3\,\text{rad/s}\), \(\Delta E = +6.75\,\text{J}\) (supplied by the student's muscular work).