Physics for Computing - Rotational Motion and Rotational Dynamics

PHYSICS FOR COMPUTING - ROTATIONAL MOTION AND ROTATIONAL DYNAMICS

Key Dates

  • 2nd June 2025: Lesson 7 & Lesson 8 (online)

  • 9th June 2025: Lesson 9, Group presentation 11 &12 (in-person)

  • 16th June 2025: Lesson 10, Brush 11 & 12 Solve Q&A (online)

  • 23rd June 2025: Group presentation lesson 13 & 14 (in-person)

  • 30th June 2025: Brush 13 & 14, Groups solve Q&A on treated topics (online)

  • 7th July 2025: Group presention lesson 15 & 16 (in-person)

  • 14th July 2025: Groups solve Q&A on treated topics (online)

  • 21th July 2025: Revision & Group studies (in-person)

  • 28th July 2025: Revision week (online)

  • 4th August 2025: Exams starts

Groups and Lessons

  • Group 1: LESSON 11: Wave Motion

  • Group 1: LESSON 12: Interference of Waves

  • Group 2: LESSON 13: Electrostatics

  • Group 3: LESSON 14: The Electric Field and the Electric Potential

  • Group 4: LESSON 15: Electric Current

  • Group 5: LESSON 16: Magnetic Fields and Electromagnetic Waves

Short Quiz 2

  • 10 MINS

  • HTTPS://FORMS.GLE/7WCP6QE8M9GYV3YE8

Learning Objectives

  • Describe rotation of a rigid body using angular displacement, angular velocity, and angular acceleration.

  • Analyze rigid-body rotation when angular acceleration is constant.

  • Relate rigid body rotation to linear velocity and linear acceleration of a point on the body.

  • Understand the meaning of moment of inertia and its relation to rotational kinetic energy.

  • Calculate the moment of inertia of various bodies.

  • Compare torques on an object caused by various forces.

  • Estimate the torque on an object caused by various forces compared to other situations.

  • Determine the angular acceleration of an object when an external torque or force is applied.

  • Analyze problems involving strings and massive pulleys

Rotational Motion

  • Rotational motion is the motion of an object around a circular path in a fixed orbit.

Key Terms:

  • Angular Position

  • Angular Displacement

  • Angular Velocity

  • Angular Acceleration

  • Frequency

  • Period

Angular Position (\theta)

  • Angular position refers to the location of a point on a rotating body, measured as an angle from a reference point.

  • It is quantified by measuring how far the body is rotated from the reference position.

  • It is measured in radians, degrees, or revolutions.

  • \theta = \frac{s}{r}, where:

    • \theta = angular displacement of the particle

    • s = distance moved on the circle by the particle

    • r = radius of the circle

Example Problem

  • A racer runs around a circular track of 49 m radius. He covers a distance of 312 m. Calculate the angular displacement of the runner.

  • Solution:

    • s = 312 \text{ m}

    • r = 49 \text{ m}

    • \theta = \frac{s}{r} = \frac{312 \text{ m}}{49 \text{ m}} = 6.37 \text{ radians}

Angular Displacement (\Delta\theta)

  • \Delta\theta = \theta - \theta_0 (Final Angle - Initial Angle)

  • The circumference of a circle is 2\pi r.

  • Conversion factors:

    • 1 \text{ revolution} = 360 \text{ degrees} = 2\pi \text{ radians}

    • 1 \text{ radian} = \frac{360^\circ}{2\pi} \approx 57.3^\circ

    • \pi \text{ rad} = 180^\circ

    • 1 \text{ rad} = 57.3^\circ = 0.159 \text{ rev}

  • Angle in radians is the ratio of arc s to radius r: \theta = \frac{s}{r}

Conversions

  • Converting 1 radian into degrees:

    • Using 2\pi \text{ rad} = 360^\circ

    • 1 \text{ rad} = x \implies x = 57.3^\circ

  • If 1 \text{ rad} = 57.3^\circ, then 1 \text{ rad} = 57.3^\circ = x \text{ revs}

Angular Velocity (\omega)

  • The rate at which a body rotates.

  • The rate of change of angular displacement with respect to time.

  • Denoted by \omega (Omega).

  • \omega = \frac{\theta2 - \theta1}{t2 - t1}, where:

    • \theta_2 = final angular position

    • \theta_1 = initial angular position

    • t2 - t1 = total time interval

  • Expressed in rad/s, rev/s, or degree/s.

Angular Acceleration (\alpha)

  • The rate of change of angular velocity with respect to time.

  • Denoted by \alpha (alpha).

  • Unit of angular acceleration: radian per second-squared (\text{rad/s}^2) or revolution per second-squared (\text{rev/s}^2).

Example Problem

  • A wheel starts rotating at 10 rad/s and attains an angular velocity of 100 rad/s in 15 s. What is the angular acceleration in rad/s²?

  • Solution:

    • \alpha = \frac{\omegaf - \omega0}{t} = \frac{100 - 10}{15} = \frac{90}{15} = 6 \text{ rad/s}^2

Linear vs. Angular Equations

Linear

Angular

v = v_0 + at

\omega = \omega_0 + \alpha t

x = v_0t + \frac{1}{2}at^2

\theta = \omega_0t + \frac{1}{2}\alpha t^2

v^2 = v_0^2 + 2ax

\omega^2 = \omega_0^2 + 2\alpha\theta

v = \frac{v + v_0}{2}

\omega = \frac{\omega + \omega_0}{2}

Linear and Angular Measures Relationship

Quantity

Linear

Angular

Relationship

Displacement

d (m)

\theta (rad)

d = r\theta

Velocity

v (m/s)

\omega (rad/s)

v = r\omega

Acceleration

a (m/s²)

\alpha (rad/s²)

a = r\alpha

Frequency vs. Period

  • Frequency (f): The number of cycles per second. Unit is in Hertz (Hz).

    • f = \frac{1}{T}

  • Period (T): The reciprocal of frequency. The time it takes to make 1 revolution.

    • T = \frac{1}{f}

Orbital Speed from \omega

  • v = \frac{\text{arc length}}{t} = \frac{r\theta}{t} = r\omega

  • Periodic time, T = time for one revolution

  • T = \frac{1}{f}

  • \omega = \frac{\theta}{t}

  • For one revolution, \theta = 2\pi, so \omega = \frac{2\pi}{T}

Rigid Body

  • An object or system of particles in which the distances between particles are fixed and remain constant.

  • Solids can be considered to be rigid bodies for the purpose of analyzing rotational motion. Fluids are not.

  • Motion consists of translational and rotational components.

    • Example: A car driving (translation) versus a car tire spinning (rotation).

Translational Motion

  • Moving an object from one location to another without changing its orientation or shape.

  • Involves moving all points of an object by the same distance and direction, keeping the object's orientation and shape intact.

  • Example: Sliding a box across a room.

  • Examples: A car moving forward on a road, a block sliding on a table, or moving a piece of furniture from one room to another.

Rotational Motion

  • Involves turning an object around a fixed point (the center of rotation).

  • Example: Spinning a record on a turntable.

  • Examples: A spinning top, a wheel rotating, or the Earth's rotation around its axis.

Rolling Motion

  • Combination of translational and rotational motion without slipping.

Torque (\tau)

  • An action that causes objects to rotate.

  • Required to rotate an object, just as a force is required to move an object in a line.

  • Sign convention:

    • Torque that rotates clockwise (CW) is negative.

    • Torque that rotates counterclockwise (CCW) is positive.

  • SI unit: Newton-meter (Nm)

Force vs. Torque

  • Forces cause accelerations; torques cause angular accelerations.

  • Factors determining the effectiveness of a force in opening a door:

    • Magnitude of the force

    • Position of the application of the force

    • Angle at which the force is applied

  • Torque depends on where the force is applied and the point about which the object rotates.

  • A door pushed at its handle will easily turn and open, but a door pushed near its hinges will not move as easily.

Center of Rotation

  • The point or line about which an object turns.

  • Example: A door’s center of rotation is at its hinges.

  • A force applied far from the center of rotation produces a greater torque than a force applied close to the center of rotation.

Line of Action of Force

  • Torque is created when the line of action of a force does not pass through the center of rotation (pivot point).

  • Line of Action: The imaginary line that extends through the force vector, indicating where the force is acting.

  • If the line of action does pass through the center of rotation, the force will cause only a translation (linear motion) of the object, and no rotation (torque).

  • When a force acts through the axis of rotation, the torque is zero.

Lever Arm

  • The perpendicular distance between the line of action of the force and the center of rotation.

  • Torque: A rotational force that causes an object to rotate.

General Definition of Torque

  • Let F be a force acting on an object, and r be a position vector from a rotational center to the point of application of the force.

  • \tau = rF\sin\theta

    • \theta = 0^\circ \text{ or } \theta = 180^\circ: torque is zero

    • \theta = 90^\circ \text{ or } \theta = 270^\circ: torque is maximum

  • Torque direction:

    • Counterclockwise turning tendency: positive torque

    • Clockwise turning tendency: negative torque

Net Torque

  • If more than one torque acts on an object, the torques are combined to determine the net torque.

  • Torques tending to make an object spin in the same direction are added together.

  • Torques tending to make the object spin in opposite directions are subtracted

  • \Sigma \tau = \tau1 + \tau2 = F1d1 - F2d2

  • If \Sigma \tau \neq 0, the object starts rotating.

  • If \Sigma \tau = 0, the rotation rate does not change.

  • The sum of clockwise moments is equal to the sum of anticlockwise moments.

    • +ve CCW, -ve CW

Example Problem

  • A force of 50 newtons is applied to a wrench that is 30 centimeters long. Calculate the torque if the force is applied perpendicular to the wrench so the lever arm is 30 cm.

  • Given:

    • Force F = -50 \text{ N}

    • lever arm r = 0.3 \text{ m}

  • Equation:

    • \tau = rF

  • Solution:

    • \tau = (-50 \text{ N})(0.3 \text{ m}) = -15 \text{ Nm}

Example Problem 2

  • A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m from the center of rotation. If the boy has a mass of 50 kg, where should he sit so that the seesaw will balance?

Example problem 3

  • A 20-centimeter wrench is used to loosen a bolt. The force is applied 0.20 m from the bolt. It takes 50 newtons to loosen the bolt when the force is applied perpendicular to the wrench.

  • How much force would it take if the force was applied at a 30-degree angle from perpendicular?

Rotational Dynamics

  • \tau = I\alpha

    • \tau = torque

    • I = moment of inertia

    • \alpha = angular acceleration

Moment of Inertia

  • I = \Sigma m_i r^2

    • I = moment of inertia

    • m_i = point mass

    • r = distance of point mass from axis

Moment of Inertia of Point Mass

  • For a single particle, the definition of moment of inertia is

    • I = mr^2

      • m is the mass of the single particle

      • r is the rotational radius

  • SI units of moment of inertia are kg·m²

  • Moment of inertia and mass of an object are different quantities

  • It depends on both the quantity of matter and its distribution (through the r^2 term)

Moment of Inertia of a hoop

  • I = Mr^2

    • I = moment of inertia

    • M = mass of hoop

    • r = radius of hoop

Moment of Inertia of Composite Particle

  • For a composite particle, the definition of moment of inertia is:

    • I = \Sigma mi ri^2 = m1r1^2 + m2r2^2 + m3r3^2 + m4r4^2 + …

      • m_i is the mass of the ith single particle

      • r_i is the rotational radius of ith particle

  • SI units of moment of inertia are kg·m²

Moment of Inertia for some other common shapes

  • Long, thin rod with rotation axis through center:

    • I_{CM} = \frac{1}{12}ML^2

  • Solid sphere:

    • I_{CM} = \frac{2}{5}MR^2

  • Long, thin rod with rotation axis through end:

    • I = \frac{1}{3}ML^2

  • Thin spherical shell:

    • I_{CM} = \frac{2}{3}MR^2

Moments of Inertia of Homogeneous Rigid Objects with Different Geometries

  • Hoop or thin cylindrical shell:

    • I_{CM} = MR^2

  • Hollow cylinder:

    • I{CM} = \frac{1}{2}M(R1^2 + R_2^2)

  • Solid cylinder or disk:

    • I_{CM} = \frac{1}{2}MR^2

  • Rectangular plate:

    • I_{CM} = \frac{1}{12}M(a^2 + b^2)

Torque on a Rotating Object

  • Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force F_t

  • The tangential force provides a tangential acceleration: Ft = mat

  • Multiply both sides by r: rFt = mrat

  • Since at = r\alpha, we have rFt = mr^2\alpha

  • We can rewrite it as \tau = mr^2\alpha, but I = mr^2

  • \tau = I\alpha

Torque on a Solid Disk

  • Consider a solid disk rotating about its axis.

  • The disk consists of many particles at various distances from the axis of rotation.

  • The torque on each one is given by \tau = mr^2\alpha

  • The net torque on the disk is given by

    • \Sigma \tau = (\Sigma mr^2)\alpha

  • A constant of proportionality is the moment of inertia, I = \Sigma mr^2 = m1r1^2 + m2r2^2 + m3r3^2 + …

  • So, we can rewrite it as \Sigma \tau = I\alpha

Newton’s Second Law for a Rotating Object

  • When a rigid object is subject to a net torque (\neq 0), it undergoes an angular acceleration.

  • The angular acceleration is directly proportional to the net torque.

  • The angular acceleration is inversely proportional to the moment of inertia of the object.

  • The relationship is analogous to \Sigma F = ma

    • \Sigma \tau = I\alpha

Useful Equations in Rotational and Translational Motion

Rotational Motion About a Fixed Axis

Translational Motion

Angular speed \omega = \frac{d\theta}{dt}

Translational speed v = \frac{dx}{dt}

Angular acceleration \alpha = \frac{d\omega}{dt}

Translational acceleration a = \frac{dv}{dt}

Net torque \Sigma \tau = I\alpha

Net force \Sigma F = ma

If \alpha = constant

If a = constant

\omegaf = \omegai + \alpha t

vf = vi + at

\thetaf = \thetai + \omega_i t + \frac{1}{2} \alpha t^2

x = xi + vi t + \frac{1}{2} at^2

\omegaf^2 = \omegai^2 + 2\alpha(\thetaf - \thetai)

vf^2 = vi^2 + 2a(x - x_i)

Work W = \int \tau d\theta

Work W = \int F dx

Rotational kinetic energy K_R = \frac{1}{2}I\omega^2

Kinetic energy K = \frac{1}{2}mv^2

Power P = \tau \omega

Power P = Fv

Angular momentum L = I\omega

Linear momentum p = mv

Net torque \Sigma \tau = \frac{dL}{dt}

Net force F = \frac{dp}{dt}

Conceptual Questions

  • A door is pushed on by two forces, a smaller force at the door knob and a larger force nearer the hinge as shown. The door does not move. The force exerted on the door by the hinge… (correct answer is not provided).

  • Three forces labeled A, B, C are applied to a rod which pivots on an axis through its center. Which force causes the largest magnitude torque?

    • \tau_A = F(L/4)

    • \tau_B = F(L/2) \cos(45^\circ) = F(L/2)/\sqrt{2}

    • \tau_C = 2F(L)

  • A mass M is placed on a very light board supported at the ends. The free-body diagram shows directions of the forces, but not their correct relative sizes. What is the ratio FL/FR? (Hint: consider the torque about the mass M).

  • A planet in elliptical orbit about the Sun is in the position shown. With the origin located at the Sun, the vector torque on the planet…

  • Two light (massless) rods, labeled A and B, each are connected to the ceiling by a frictionless pivot. Rod A has length L and has a mass m at the end of the rod. Rod B has length L/2 and has a mass 2m at its end. Both rods are released from rest in a horizontal position.

    • Which one experiences the larger torque?

    • Which one falls to the vertical position fastest? Hint \alpha = \frac{\tau}{I}