CENTRE OF PERCUSSION & CENTRE OF OSCILLATION - <- Google these terms and you find essentially the same problem as in Sketch-2.
For the School Bus the Centre of Percussion is the point M2, namely the point on the LoA of the force Ft acting on the (non-rotating) body, with the body's CG being perpendicular to the LoA at this point. We take it as given here that the distance from CG to Ft's LoA (or M2) is "A".
We seek the Centre of Oscillation (this term coming from pendulum theory, and = AP here), namely the point about which the whole Body (= bus) starts to rotate when first acted on by the force. We take it as given that the Body has total mass = Mt, Radius of Gyration = K, and lives in Flatland.
Step 1. We first convert the single linear force Ft into an "equivalent force system" of another force Ft', of same magnitude and direction as Ft but acting on a LoA through the CG, together with a couple T = Ft' x A acting anywhere on the Body.
Step 2. By Euler's Rigid Body Equations (student-special "Flatland-lite" version), force Ft' causes rate-of-change of Linear Momentum (Mt x V) of the Body, and couple T causes rate-of-change of Rotational Momentum (Mt x K^2 x W) of Body. Or, assuming Mt and K are unchanging, and taking "=>" to mean "causes",
Ft' => Mt x V-dot, and
T (= Ft' x A) => Mt x K^2 x W-dot.
In words:
2a. Force Ft' Linearly Accelerates every point of the whole bus, in the direction of Ft', at rate A = V-dot = Ft'/Mt (like top-left of Sketch-5).
2b. Couple T Rotationally Accelerates every point of the whole bus, in Clockwise direction about CG, at rate W-dot = (Ft' x A)/(Mt x K^2) (like top-second-from-left of Sketch-5).
Step 3. The acceleration of any point of the bus (wrt absolute space...) is simply the vector addition of the above two patterns. Since, in rotating motion V-dot = W-dot x R (where R is the radius vector, from-centre to-given-point, and this much dumbed down here...), the pattern of 2b above gives at any point a linear acceleration of V-dot = R x (Ft' x A)/(Mt x K^2).
We seek the point where the acceleration is zero (ie. the AP), which is the point where the two vectors in 2a and 2b are equal and opposite. So, we seek Ro such that,
Ft'/Mt = -Ro x (Ft' x A)/(Mt x K^2).
The force Ft' and the total mass Mt cancel out (so easy!), leaving K^2 = -Ro x A. And writing -Ro as the more usual "B" gives,
K^2 = A x B... ALWAYS!
Or..., the position "B" of the Acceleration Pole is dependant ONLY on the distance "A" from CG to the acting force (= ~distance from CG to front-axle of the bus), and on the Radius of Gyration "K" of the bus.
Or..., the position of the AP is NOT influenced by how hard the bus driver pulls on the steering-wheel, because the bus ALWAYS starts rotating about the same point near the rear-axle, regardless of the size of the front-wheel forces. (Err..., not considering changing live loads, or boisterous school kids!).
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