Unit 1: Rotational Dynamics

1. Equation of Angular Motion

Just as linear motion describes the movement of objects in a straight line, angular motion describes the rotation of objects around an axis. We define analogous quantities for rotational motion:

• Angular Displacement (theta): The angle through which a point or line has been rotated about an axis. Measured in radians.
• Angular Velocity (omega): The rate of change of angular displacement. It is the rotational equivalent of linear velocity. omega = d theta/dt

  where d theta is the infinitesimal angular displacement and dt is the infinitesimal time interval. Units: radians per second (rad/s).

• Angular Acceleration (alpha): The rate of change of angular velocity. It is the rotational equivalent of linear acceleration. alpha = d omega/dt

  where d omega is the infinitesimal change in angular velocity. Units: radians per second squared (rad/s²).

For situations with constant angular acceleration, we have a set of kinematic equations analogous to those for linear motion:

• First Equation: Relates final angular velocity to initial angular velocity, angular acceleration, and time. omega = omega_0 + alpha t

  where omega is the final angular velocity, omega_0 is the initial angular velocity, alpha is the constant angular acceleration, and t is the time.

• Second Equation: Relates angular displacement to initial angular velocity, angular acceleration, and time. theta = omega_0 t + 1/2 alpha t^2

  where theta is the angular displacement.

• Third Equation: Relates final angular velocity, initial angular velocity, angular acceleration, and angular displacement, without involving time. omega^2 = omega_0^2 + 2alpha theta

Example: A bicycle wheel initially at rest begins to rotate with a constant angular acceleration of 2 rad/s². After 5 seconds, what is its angular velocity and angular displacement? Using omega = omega_0 + alpha t, omega = 0 + (2 rad/s²)(5 s) = 10 rad/s. Using theta = omega_0 t + 1/2 alpha t^2, theta = 0 + 1/2 (2 rad/s²)(5 s)² = 25 rad.

2. Relation Between Linear and Angular Kinematics

When a rigid body rotates, every point on the body moves in a circle. The linear motion of any point on the body is related to the angular motion of the body. For a point at a distance r from the axis of rotation:

• Arc Length (s): The linear distance traveled by a point on the circumference of a circle. s = r theta

  where r is the radius of the circular path and theta is the angular displacement in radians.

• Tangential Speed (v): The linear speed of a point tangent to its circular path. v = omega r

  where omega is the angular velocity in rad/s.

• Tangential Acceleration (a_t): The component of linear acceleration tangent to the circular path, responsible for changing the magnitude of the tangential velocity. a_t = alpha r

  where alpha is the angular acceleration in rad/s².

• Centripetal (Radial) Acceleration (a_c): The component of linear acceleration directed towards the center of the circular path, responsible for changing the direction of the tangential velocity. a_c = omega^2 r = v^2/r

  This acceleration is always present when an object moves in a circular path, even at constant angular speed.

Example: A point on the edge of a spinning wheel has a radius of 0.5 m and an angular velocity of 4 rad/s. Its tangential speed is v = (4 rad/s)(0.5 m) = 2 m/s. Its centripetal acceleration is a_c = (4 rad/s)²(0.5 m) = 16 * 0.5 = 8 m/s².

3. Kinetic Energy of Rotation

A rotating body possesses kinetic energy due to its rotation. This is known as rotational kinetic energy, analogous to translational kinetic energy (1/2 mv²).

• Rotational Kinetic Energy (KE_rot): KE_rot = 1/2 I omega^2

  where I is the moment of inertia (rotational inertia) of the body and omega is its angular velocity. Units: Joules (J).

• Total Kinetic Energy: For a body that is both translating (moving linearly) and rotating (e.g., a rolling object), its total kinetic energy is the sum of its translational and rotational kinetic energies. Total KE = KE_trans + KE_rot = 1/2 mv^2 + 1/2 I omega^2

  where m is the total mass of the body and v is the speed of its center of mass.

Example: A solid sphere of mass M and radius R rolls without slipping with a linear speed v. Its moment of inertia is I = (2/5)MR². Since it rolls without slipping, v = omega R, so omega = v/R. KE_rot = 1/2 I omega^2 = 1/2 ((2/5)MR²)(v/R)² = 1/2 (2/5)MR² (v²/R²) = 1/5 Mv². Total KE = 1/2 Mv² + 1/5 Mv² = (7/10)Mv².

4. Moment of Inertia and Radius of Gyration

Moment of Inertia (I)

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass of the object and how that mass is distributed relative to the axis of rotation. It is the rotational analogue of mass in linear motion.

• Definition for a System of Point Masses: I = Sigma m_i r_i^2

  where m_i is the mass of the i-th particle and r_i is its perpendicular distance from the axis of rotation. For continuous bodies, this summation becomes an integral.

• Parallel Axis Theorem: This theorem is used to find the moment of inertia about any axis parallel to an axis passing through the center of mass. I = I_cm + Md^2

  where I is the moment of inertia about the new axis, I_cm is the moment of inertia about a parallel axis through the center of mass, M is the total mass of the body, and d is the perpendicular distance between the two parallel axes.

• Perpendicular Axis Theorem: This theorem applies to laminar (planar) bodies and helps find the moment of inertia about an axis perpendicular to the plane of the body. I_z = I_x + I_y

  where I_z is the moment of inertia about an axis perpendicular to the plane of the body, and I_x and I_y are the moments of inertia about two mutually perpendicular axes lying in the plane of the body and intersecting the z-axis.

Example: Consider a thin square plate of mass M and side L. If I_x and I_y are moments of inertia about axes passing through the center and parallel to the sides, then I_x = I_y = ML²/12. By the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the plate and passing through its center is I_z = I_x + I_y = ML²/12 + ML²/12 = ML²/6.

Radius of Gyration (k)

The radius of gyration is a conceptual distance from the axis of rotation at which the entire mass of an object could be concentrated to have the same moment of inertia. It provides an effective distance of mass from the axis.

• Definition: k = sqrt(I/M)

  where I is the moment of inertia and M is the total mass of the body. Units: meters (m).

Example: For a uniform rod of mass M and length L rotating about its center, I = ML²/12. Its radius of gyration is k = sqrt((ML²/12)/M) = sqrt(L²/12) = L/(2*sqrt(3)).

5. Moment of Inertia of Uniform Rod

For specific common geometries, moments of inertia are often pre-calculated. For a uniform rod of mass M and length L:

• About its center (perpendicular to the rod): I = ML^2/12
• About one end (perpendicular to the rod): I = ML^2/3

  This can also be derived using the parallel axis theorem: I = I_cm + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = 4ML²/12 = ML²/3.

6. Torque and Angular Acceleration

Torque is the rotational equivalent of force. It is what causes an object to undergo angular acceleration.

• Definition of Torque (tau): Torque is the rotational effect of a force. It depends on the magnitude of the force, the distance from the pivot point (axis of rotation), and the angle at which the force is applied. tau = r x F (vector cross product)

  The magnitude of torque is tau = r F sin(theta), where r is the lever arm (distance from the axis to the point where force is applied), F is the magnitude of the force, and theta is the angle between the position vector r and the force vector F. Units: Newton-meters (Nm).

• Relation to Angular Acceleration (Rotational Newton's Second Law): tau = I alpha

  This equation states that the net torque acting on a body is equal to the product of its moment of inertia and its angular acceleration. This is analogous to Newton's second law F = ma.

• Relation to Angular Momentum: Torque is also defined as the rate of change of angular momentum. tau = dL/dt

  where L is angular momentum.

Example: A force of 10 N is applied tangentially to the edge of a wheel with a radius of 0.2 m. The torque produced is tau = r F sin(90°) = (0.2 m)(10 N)(1) = 2 Nm. If the wheel has a moment of inertia of 0.5 kg m², the angular acceleration will be alpha = tau/I = 2 Nm / 0.5 kg m² = 4 rad/s².

7. Work and Power in Rotational Motion

Work and power concepts also have their rotational analogues.

• Work Done (W) by a Torque: When a constant torque acts on a body and causes an angular displacement, work is done. W = tau theta

  where tau is the constant torque and theta is the angular displacement in radians. This is analogous to W = Fd. Units: Joules (J).

• Power (P) in Rotational Motion: Power is the rate at which work is done. P = tau omega

  where tau is the torque and omega is the angular velocity. This is analogous to P = Fv. Units: Watts (W).

Example: A motor applies a constant torque of 50 Nm to a shaft rotating at a constant angular velocity of 10 rad/s. The power delivered by the motor is P = (50 Nm)(10 rad/s) = 500 W. If the motor operates for 60 seconds, the work done is W = P * t = 500 W * 60 s = 30,000 J.

8. Angular Momentum and Conservation

Angular Momentum (L)

Angular momentum is the rotational analogue of linear momentum. It is a measure of the amount of rotation an object has, taking into account its mass, shape, and speed.

• Definition: L = I omega

  where I is the moment of inertia and omega is the angular velocity. Units: kg m²/s or J s.

  For a single particle, angular momentum about a point is L = r x p = r x (mv), where r is the position vector from the point to the particle, and p is its linear momentum.

Conservation of Angular Momentum

One of the most fundamental principles in physics, the conservation of angular momentum states that if there is no external torque acting on a system, the total angular momentum of that system remains constant.

• Condition: If no external torque, L = constant.

  This implies that I_1 omega_1 = I_2 omega_2, where I_1 and omega_1 are the initial moment of inertia and angular velocity, and I_2 and omega_2 are the final values.

• Applications:
  • Ice Skater Spinning: When an ice skater pulls her arms and legs inward while spinning, she decreases her moment of inertia (I). Since angular momentum L = I omega must be conserved, her angular velocity (omega) increases dramatically, causing her to spin faster.
  • Neutron Star Formation: When a massive star collapses to form a neutron star, its radius (and thus its moment of inertia) decreases drastically. To conserve angular momentum, the neutron star spins at an incredibly high angular velocity, often making hundreds of rotations per second.
  • Planetary Orbits: For a planet orbiting the sun, the angular momentum of the planet-sun system is conserved. As a planet gets closer to the sun (smaller r, thus smaller I), its orbital speed increases to maintain constant angular momentum.

Example: An ice skater has an initial moment of inertia of 4 kg m² and spins at 2 rad/s. When she pulls in her arms, her moment of inertia decreases to 1 kg m². What is her new angular velocity? By conservation of angular momentum: I_1 omega_1 = I_2 omega_2. (4 kg m²)(2 rad/s) = (1 kg m²) omega_2. 8 kg m²/s = (1 kg m²) omega_2. omega_2 = 8 rad/s. Her angular velocity increases fourfold.