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This article is a topic within the subject Higher Physics 1A.


Moment of Inertia

[1]Moment of inertia measures and object's resistance to changes in rotational motion (just as mass measures an object's resistance to changes in translational motion). Moment of inertia depends on the distribution of mass in an object, and varies depending on the axis of rotation (the line around which the object will rotate). The minimum moment of inertia occurs when the axis of rotation passes through the centre of the object. Moment of inertia is measured in kilogram metres squared (kgm2).

I = Σmiri2

Moment of inertia (I) is found by multiplying each mass in the body (mi) with the square of its displacement from the axis of rotation (ri2), and taking the sum of all these.

For continuous bodies, moment of inertia must be calculated using integration:

I = ʃr2dm

The minimum moment of inertia occurs when the axis of rotation passes through the centre of the object. Moment of inertia may be increased by moving the object's mass further from the axis of rotation, and vice versa.

Radius of gyration

Radius of gyration is the radius of a hoop with the same moment of inertia as an object.[2] For an object with mass m and moment of inertia I, the radius of gyration (k) is found using the following formula:

I = mk2

Rotational Kinetic Energy

[3]Rotating objects have rotational kinetic energy. As an object rotates, the mass making up the object is moving around the axis of rotation, so it has kinetic energy.

Krotation = ½Iω2

Rotational kinetic energy is calculated in a similar way to regular kinetic energy, except mass and velocity are replaced with moment of inertia and angular velocity respectively.


 A massless, equilateral, triangular plate rotates around one of its corners. The other two corners
 are each attached to a lead ball weighing 50g, and the side length of the triangle is 8cm. What is
 the rotational kinetic energy of the triangle if is has an angular velocity of 120 radians per
 Rotational kinetic energy.jpg
 First, we must calculate the moment of inertia. The triangular plate is massless, so we can ignore
 it. We will assume the masses have negligible radius. Each mass weighs 0.05kg and lies 0.08m from 
 the centre of rotation.
 I = Σmiri2
 I = (0.05*0.08) + (0.05*0.08)
 I = 0.008kgm2
 Now, the rotational kinetic energy may be calculated.
 Krot = ½Iω2
 Krot = 0.5*0.008*120
 Krot = 0.48J
 The rotational kinetic energy is 0.48 Joules.

It is important to note that friction is often negligible between a rolling object and the surface on which it is rolling. This means that non-conservative forces do no work on rolling objects (unless they act in come other way, such as friction in a rusty bearing).

Rolling objects

[4]If an object is undergoing both translational and rotational motion (for example a rolling ball), it will have both rotational and translational kinetic energy. The total kinetic energy is the sum of these.

Ktotal = Ktranslation + Krotation Ktotal = ½mvcm2 + ½Icrω2

The translational kinetic energy is calculated using the velocity of the centre of mass. The rotational kinetic energy is calculated using the moment of inertia of the centre of rotation. For a rolling object, the centres of mass and rotation are the same.

One consequence of this is that rolling objects use energy for translation and rotation. An object with a high moment of inertia will need more energy to rotate at a given speed (Krot = ½Iω2) and so will have less energy available for translation. This means it will roll slower than an object with a lower moment of inertia.

Another important observation is that the velocities of individual points on an object are affected by both the rotation and translation of the object.

 The net velocity at any point on an object can be found by taking the vector sum of component
 For an object undergoing rotation and translation, there will be a rotational and translational
 component to the velocity of any point on the object.
 vtotal = vtrans + vrot (1)
 For a rigid object, vtrans = vcm (velocity of the centre of mass) (2)
 vrot = rω (the product of angular velocity and displacement from axis of rotation) (3)
 Substituting (2) and (3) into (1) gives:
 vtotal = vcm + rω
 (note that the directions of the components will need to be accounted for; this is a vector sum).
 Rolling velocity.jpg 
 For a rolling object, the magnitudes of vrot and vtrans are equal at object's
 surface. This means they cancel out at the bottom (opposite directions) and combine at the top
 (same direction) to give vtot = 2vcm.

Rotational Kinematics

Rotational kinematics is very similar to regular kinematics, except it focuses on angular motion, rather than translational motion. Angular motion is measured anticlockwise using radians (not degrees).[5] Angular motion has already been introduced, in uniform circular motion.

  • Angular displacement, θ, is measured in radians.
    • θ = s/r. Angular displacement is displacement divided by the radius of the circle.
  • Angular velocity, ω, is measured in radians per second.
    • ω = v/r = dθ/dt. Angular displacement is velocity divided by radius. It is also the derivative of angular displacement with respect to time.
  • Angular acceleration, α, is measured in radians per second squared.
    • α = a/r = dω/dt. Angular displacement is acceleration divided by radius. It is also the derivative of angular velocity with respect to time.

The kinematics equations for angular motion are identical to the equivalent translational kinematics equations[6]:

ωt = ω0 + αt

Angular velocity at time t is equal to initial velocity (ω0) plus angular acceleration times time (αt).

θt = θ0 + ωx0t + ½αt2

Angular displacement at time t is the sum of initial displacement (θ0), angular displacement due to initial angular velocity (ωx0t), and angular displacement due to constant angular acceleration (½αt2)

θt = θ0 + ½(ω0 + ωt)t

Angular displacement at time t, expressed in terms of initial angular displacement (θ0), initial angular velocity (ω0), final angular velocity (ωt) and time (t).

ωt2 = ω02 +2α(θt0) Angular velocity at time t, expressed in terms of initial angular velocity, acceleration, and displacement.


[7]Rotational acceleration is caused by torque (τ). [8] Torque is the turning effect caused by a force acting on an object, measured in Newton metres (Nm). Torque is a vector, calculated as the cross product of a force and its distance from the axis of rotation.

τ = F x r

For most problems in this course, only the magnitude of the torque is needed. From the definition of a cross product:

τ = Frsinθ , where θ is the angle between the force (F) and its displacement from the axis of rotation (r).

Only the force component perpendicular to r will have any turning effect - this is the reason for the sinθ term in the torque equation.


Alternatively, torque can be expressed as the product of force and moment arm. The moment arm is the perpendicular distance from the force's line of action tot the axis of rotation.

Moment arm = rsinθ

τ = F*(moment arm)

Torque 2.jpg

Torque and angular acceleration

[9]The external torque on a rigid body is proportional to the angular acceleration of that body.

τext = Iα

Torque is equal to the product of moment of inertia and angular acceleration. This equation is essentially Newton's second law, for rotation.


This is the end of this topic. Click here to go back to the main subject page for Higher Physics 1A.


Textbook refers to Serway & Jewett, Physics for Scientists and Engineers (Brooks/Cole , 8th ed, 2010)
(Slides) refers to those distributed by Wolfe, J (2012) on his First Year Physics site

  1. Textbook, pp284-285
  2. (Slides), Rotation, p2
  3. Textbook, pp284-285
  4. Textbook, pp299-301
  5. Textbook, pp. 278-279
  6. Textbook, pp280-281
  7. Textbook, pp290-291
  8. Textbook, pp291-293
  9. Textbook, pp291-293
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