Premed · Premed · Physics 1
Lecture 10: Rotational Kinematics
Physics I — Mechanics & Thermodynamics
Learning Objectives
By the end of this lecture, students will be able to:
- Define angular position, angular displacement, angular velocity, and angular acceleration
- Apply the rotational kinematic equations for constant angular acceleration
- Relate angular and linear (tangential) quantities for a point on a rotating body
- Distinguish between tangential and centripetal acceleration
- Calculate the moment of inertia for point masses, systems, and common shapes
Lecture Content
I. Angular Position and Displacement
Angular position (theta) measures the angle of a reference line on a rotating body relative to a fixed axis. It is measured in radians (rad), defined by the relation theta = s / r, where s is arc length and r is the radius. A full revolution corresponds to 2 pi rad or 360 degrees, and 1 rad equals approximately 57.3 degrees.
Angular displacement (Delta theta) is the change in angular position: Delta theta = theta_f - theta_i. By convention, positive angular displacement corresponds to counterclockwise rotation and negative to clockwise. Although radians are dimensionless (being a ratio of two lengths), the label "rad" is retained for clarity.
II. Angular Velocity and Angular Acceleration
Average angular velocity is omega_avg = Delta theta / Delta t, and instantaneous angular velocity is omega = d theta / dt, measured in rad/s. It is positive for counterclockwise rotation and negative for clockwise rotation. Average angular acceleration is alpha_avg = Delta omega / Delta t, and instantaneous angular acceleration is alpha = d omega / dt = d^2 theta / dt^2, measured in rad/s^2.
Angular velocity can also be represented as a vector directed along the axis of rotation, with its sense determined by the right-hand rule: curl the fingers in the direction of rotation, and the thumb points along the angular velocity vector.
III. Rotational Kinematic Equations
For constant angular acceleration, the rotational kinematic equations are direct analogs of their linear counterparts: (1) omega = omega_0 + alpha t, (2) theta = theta_0 + omega_0 t + (1/2) alpha t^2, (3) omega^2 = omega_0^2 + 2 alpha (theta - theta_0), and (4) theta = theta_0 + (1/2)(omega_0 + omega) t. These have identical form to the linear kinematic equations with the substitutions x -> theta, v -> omega, and a -> alpha. All the same problem-solving strategies apply: identify the knowns, select the appropriate equation, and solve algebraically.
<image>A side-by-side comparison table of linear and rotational kinematic equations. Left column: linear quantities (x, v, a) and the four kinematic equations. Right column: rotational analogs (theta, omega, alpha) with the corresponding rotational equations. Arrows connect corresponding quantities across the columns. Below the table, a spinning disk is shown with angular position theta, angular velocity omega, and angular acceleration alpha labeled.</image>
IV. Relating Angular and Linear Quantities
For a point located at distance r from the axis of rotation, the angular and linear quantities are related by simple proportionalities. The arc length traversed is s = r theta (with theta in radians). The tangential velocity is v_t = r omega, directed tangent to the circular path. The tangential acceleration is a_t = r alpha, arising from changes in angular speed. The centripetal (radial) acceleration is a_c = v_t^2 / r = r omega^2, directed toward the axis and arising from the continuous change in direction.
The total linear acceleration is the vector sum of the tangential and centripetal components: |a| = sqrt(a_t^2 + a_c^2). These two components are perpendicular to each other. An important consequence of these relationships is that points farther from the axis move faster (larger v_t), even though all points on a rigid body share the same angular velocity omega. This is why the outer edge of a spinning disk moves faster than points near the center.
V. Rolling Without Slipping
A wheel of radius R rolling without slipping on a surface satisfies several important constraints. The contact point has zero velocity relative to the ground, leading to the fundamental relationship v_cm = R omega, where v_cm is the translational speed of the center. Similarly, a_cm = R alpha and the distance traveled is d = R theta.
Rolling combines translation of the center of mass with rotation about the center. The velocity of any point on the wheel is the vector sum of the translational velocity and the rotational velocity about the center. At the contact point, these two contributions exactly cancel, giving zero net velocity. At the top of the wheel, they add together, giving a speed of 2v_cm.
<image>A wheel of radius R rolling to the right without slipping. The center of mass moves with velocity v_cm. At the contact point with the ground, the rotational velocity (R omega, directed to the left) exactly cancels the translational velocity (v_cm, directed to the right), giving zero net velocity. At the top of the wheel, both add to give 2v_cm. Velocity vectors are drawn at the top, center, bottom, and front of the wheel, each decomposed into translational and rotational contributions.</image>
VI. Moment of Inertia
The moment of inertia (I) is the rotational analog of mass, quantifying an object's resistance to angular acceleration. For a system of point masses, I = Sum of m_i r_i^2, where r_i is the perpendicular distance from each mass to the axis of rotation. For continuous bodies, I = integral of r^2 dm.
The moment of inertia depends on the total mass, how that mass is distributed relative to the axis, and the choice of axis itself. Common moments of inertia include: solid cylinder or disk (axis through center), I = (1/2)MR^2; hollow cylinder (thin-walled), I = MR^2; solid sphere (axis through center), I = (2/5)MR^2; hollow sphere (thin-walled), I = (2/3)MR^2; thin rod (axis through center), I = (1/12)ML^2; and thin rod (axis through one end), I = (1/3)ML^2.
VII. Parallel-Axis Theorem
If the moment of inertia about an axis through the center of mass is I_cm, then the moment of inertia about a parallel axis displaced by distance d is I = I_cm + Md^2. This is always larger than I_cm, confirming that the center-of-mass axis gives the minimum moment of inertia among all parallel axes. The theorem is useful for finding I about off-center or edge axes without having to re-integrate.
As an example, consider a thin rod about one end. Starting with I_cm = (1/12)ML^2 for the axis through the center, the end is displaced by d = L/2, giving I_end = (1/12)ML^2 + M(L/2)^2 = (1/3)ML^2, which matches the result obtained by direct integration.
<image>A reference sheet showing six common geometric shapes (solid disk, hollow cylinder, solid sphere, hollow sphere, thin rod through center, thin rod through end) each with the axis of rotation drawn as a dashed line and the moment of inertia formula written below. Beside the shapes, a diagram illustrates the parallel-axis theorem: a shape with the CM axis and a parallel axis displaced by distance d, with the formula I = I_cm + Md^2.</image>
