Things that move in circles have angular velocity, measured in radians per second, so dimensionally [T-1]. This leads directly to the relationship v= rω or ω = v/r, using ω (lower case omega) for angular velocity. Using a dash, ‘, to indicate differentiating with respect to time, we can recognize that ø’=w is angular velocity and so ø”=w’ = angular acceleration.
Angular acceleration depends on a torque being applied. The relationship is direct proportion; if the torque is doubled, the angular acceleration is doubled. Angular acceleration depends on mass, on the axis of rotation and on where the rotation applies. Experimentally, it is soon found that bodies of the same mass experiencing the same torque behave differently; the changes depend on something more to do with the shape. This property the resistance to rotational motion, is called the rotational inertia of a body, represented by a capital letter, I.
If we have a light rod length r OP with a mass m at P, being acted upon by a force F so as to rotate about O, then The tangential force on the mass is, from F=ma, FT=mrω’=mrø”
So the resulting torque of applying F is the couple Fr, which is the moment of the force F about the centre O, Fr= mr²ø”
so somehow mr² represents the effective mass of the particle at the end of the rod. The tangential force is the couple (turning force) that causes the rotation. So where we have been writing F=mx” for Newton’s second law, we can now write C=Iø”, using I as the mass of inertia. For the single particle, I = mr².
The kinetic energy is ½ mv², or ½ mr²ω² or ½ I ω².
Hopefully you can see a complete swap of symbols in moving from linear motion to circular motion: for constant acceleration å, we have the faintly familiar equations
w₂=w₁ + åt; ø= w₁t + ½ åt²; w₂² = w₁² + 2åø
Furthermore, since the above is considering a single particle, the extension for the general case is straightforward: I = mr². This will lead directly to the use of integration in calculating formulae for the moment of inertia for two and the dimensional shapes.
See Bostock & Chandler Applied 2 Chapters 10-12
See Inertia 1 and other revision sheets on this site.
Moments of Inertia - a simple rod
Consider a uniform rod ABC of length 2r and mass M The centre of mass is at the midpoint, B. By treating the point as a point mass, then:
1 If the rod is rotated around its longitudinal axis the inertia is zero.
2 If the rod is parallel to the axis of rotation, displaced by d, the inertia must be I = Md².
3If the rod lies in the plane of rotation (the axis of rotation is perpendicular to the rod) then we must consider a small element of the rod, length dx at distance x from B and mass mdx, so M = 2rm. The moment of inertia of the element is mdx (x)² so adding these (integration) will give ⅓ Mr² when the centre of rotation is at B. Write this out for yourself.
4 If the axis is oblique to the rod at an angle ø, and still through B then the elemental part has a moment of inertia of mdx (x sinø)² so integration will give ⅓ Mr² sin²ø . Draw the diagram that convinces you of this. [or is that e ⅓ Mr² sin2ø ?]
5 If the axis is perpendicular to the rod and passes through A or C the integration is the same as 3 but the limits change from [-r,r] to [0,2r] moment of inertia must be I = 4/3 Mr². Write this for yourself.
6 The last answer suggests that the extra inertia gained from displacing the axis from B to A was Mr². So, if the axis is perpendicular to the rod and passes through any point P on AC at a distance x from B, the moment of inertia is likely to be I = ⅓ Mr² + Mx². Convince yourself of this by doing the integration across the interval [-r-x, r-x].
7 (hard) If the rod AMB is rotated about a perpendicular axis through the origin O when the rod lies between A (a, b-r) and C (a, b+r), show that the moment of inertia is I = ⅓ Mr² + Md² where d is length OB. This, when achieved, serves to prove the parallel axes theorem, which says:
if the moment of inertia of a body of mass M about an axis through its centre of mass is Mk² then the moment of inertia about a parallel axis a distance d from the first axis is Mk² + Md².
Note that the minimum such inertia is when the axis passes through the centre of mass; this is self-evident but frequently forgotten by students.
