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5.3.2Torque and Rotational Dynamics

Parallel Axis Theorem — AP PHYSICS 1 Revision Notes & Topical Questions

5 min read 8 parts Lesson 13 of 18 1 quiz

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Lesson summary

5.3.2 Parallel Axis Theorem — AP PHYSICS 1 notes: Use the parallel axis theorem to find the rotational inertia about an axis parallel to one through the centre of mass.Standard formulas — such as ½MR2 for a disc and (1/12)ML2 for a rod, apply to an axis through the centre of mass…

  • Orientation
  • I = Icm + Md2
  • Always larger
  • Requirement
  • Rod about its end
  • Disc about its rim axis

Section 1 of 8The problem

Standard formulas
  • Standard formulas such as ½MR2 for a disc and (1/12)ML2 for a rod, apply to an axis through the centre of mass.
Real axes
  • Real axes a door hinged at its edge, a rod pivoted at its end, a wheel hung from its rim.
Orientation
  • Orientation rotational inertia changes with the axis;
    • the minimum is about the axis through the centre of mass.

Section 2 of 8The theorem

I
Icm + Md2
  • I about an axis a perpendicular distance d from a parallel axis through the centre of mass.
  • Icm is about the centre of mass.
  • M is the total mass.
  • d is the distance between the two axes, not the object length.
Always larger
  • Always larger moving the axis away from the centre of mass always adds Md2.
Requirement
  • Requirement the two axes must be parallel.

Section 3 of 8Checking known results

Rod about its end
  • Rod about its end d = L/2, so I = (1/12)ML2 + M(L/2)2 = (1/3)ML2.
Disc about its rim axis
  • I = ½MR2 + MR2 = (3/2)MR2.
Point mass
  • Icm = 0, so I = Md2.

Section 4 of 8Composite systems

Composite systems
Total I
  • Total I sum of the rotational inertias of each part about the same axis.
Parts with centre of mass off the axis
  • Parts with centre of mass off the axis use the theorem for each part.
Dumbbell
  • Dumbbell two spheres on a light rod: two sphere terms, each (2/5)MR2 + Md2, plus the rod's (1/12)mL2.
Rod with a disc at one end
  • Rod with a disc at one end rod's I about the pivot plus the disc's Icm plus Mdiscd2.
Example
  • Example spheres of 0.75 kg and radius 0.04 m at 0.22 m, with a 0.15 kg rod of length 0.36 m, give I about 0.075 kg·m2 with the rod contributing about 2%.
Whole-system caution
  • Whole-system caution the theorem applies to the whole system only if its own Icm and total mass are known.

Section 5 of 8Physical consequences

1Pendulum rod
  • Pendulum rod a rod swung from its end resists rotation more than the same rod pivoted at its centre.
2Choked-up bat
  • Choked-up bat moving the pivot toward the centre of mass reduces d and I, making the bat easier to rotate.
3Angular acceleration
  • Angular acceleration larger I gives smaller α = Στ/I for a given torque.
4Mass at large radius
  • 1.0 kg at 1.0 m has the same I as 4.0 kg at 0.50 m, so a tightrope walker's long pole and a flywheel's rim mass both raise I.
5Hinged beam
  • Hinged beam swings according to Στ = Iα with the weight torque at the centre of mass.

Section 6 of 8More from the specification

More from the specification
Rotational inertia of solid rigid bodies
  • Rotational inertia of solid rigid bodies calculating moment of inertia for solid rigid bodies around parallel axes uses the parallel axis theorem.
Linear
  • A graph that is a straight line through the origin, so resistance is constant.
Rigid system
  • System that holds its shape while rotating.
Rigid body
  • Body whose shape does not change when rotating.
Proportion
  • An equation stating two ratios are equal, solved by cross-multiplying.

Section 7 of 8Key terms

TermMeaning
Parallel axis theoremI = Icm + Md2.
IcmRotational inertia about an axis through the centre of mass.
Perpendicular distance dSeparation of two parallel axes.
Composite bodySystem of several parts, each contributing to I.

Section 8 of 8Exam technique

1
Using the theorem when the axes are not parallel.
2
Taking d as the object length instead of the distance between axes.
3
Adding d instead of d2, or omitting Md2.
4
Using an I that is not about the centre of mass as Icm.
5
Forgetting to add the rotational inertias of every part of the system.

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