A rear engine automobile is travelling along a track of 100 metres mean radius. Each of the four road wheels has a moment of inertia of 2.5 kg - m ^{2} and an effective diameter of 0.6 m. The rotating parts of the engine have a moment of inertia of 1.2 kg - m ^{2}. The engine axis is parallel to the rear axle and the crankshaft rotates in the same sense as the road wheels. The ratio of engine speed to back axle speed is 3 : 1. The automobile has a mass of 1600 kg and has its centre of gravity 0.5 m above road level. The width of the track of the vehicle is 1.5 m.
Determine the limiting speed of the vehicle around the curve for all four wheels to maintain contact with the road surface. Assume that the road surface is not cambered and centre of gravity of the automobile lies centrally with respect to the four wheels.
Question : A rear engine automobile is travelling along a track of 100 ...
Given : R = 100 m ; I_{ W }=2.5 kg – m ^{2} ; d_{ W }=0.6 m \text { or } r_{ W }=0.3 m ; I_{ E }=1.2 kg – m ^{2} ; G=\omega_{ E } / \omega_{ W }=3 ; m=1600 kg ; h=0.5 m ; x=1.5 m
The weight of the vehicle (m.g) will be equally distributed over the four wheels which will act downwards. The reaction between the wheel and the road surface of the same magnitude will act upwards.
\therefore Road reaction over each wheel
= W/4 = m.g / 4 = 1600 × 9.81/4 = 3924 N
Let v = Limiting speed of the vehicle in m/s.
We know that angular velocity of the wheels,
\omega_{ W }=\frac{v}{r_{ W }}=\frac{v}{0.3}=3.33 v rad / sand angular velocity of precession,
\omega_{ P }=\frac{v}{R}=\frac{v}{100}=0.01 v rad / s\therefore Gyroscopic couple due to 4 wheels,
C_{ W }=4 I_{ W } \cdot \omega_{ W } \cdot \omega_{ P }=4 \times 2.5 \times \frac{v}{0.3} \times \frac{v}{100}=0.33 v ^{2} N – mand gyroscopic couple due to rotating parts of the engine,
C_{ F }=I_{ F } \cdot \omega_{ F } \cdot \omega_{ p }=I_{ F } \cdot G \cdot \omega_{ W } \cdot \omega_{ P } =1.2 \times 3 \times 3.33 v \times 0.01 v=0.12 v^{2} N – m\therefore Total gyroscopic couple,
C=C_{ W }+C_{ E }=0.33 v^{2}+0.12 v^{2}=0.45 v^{2} N – mDue to this gyroscopic couple, the vertical reaction on the rails will be produced. The reaction will be vertically upwards on the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at each of the outer or inner wheel be P/2 newtons.
\therefore \quad P / 2=C / 2 x=0.45 v^{2} / 2 \times 1.5=0.15 v^{2} NWe know that centrifugal force,
F_{ C }=m \cdot v^{2} / R=1600 \times v^{2} / 100=16 v^{2} N\therefore Overturning couple acting in the outward direction,
C_{ O }=F_{C} \times h=16 v^{2} \times 0.5=8 v^{2} N – mThis overturning couple is balanced by vertical reactions which are vertically upwards on the outer wheels and vertically downwards on the inner wheels. Let the magnitude of this reaction at each of the outer or inner wheels be Q/2 newtons.
\therefore Q / 2=C_{0} / 2 x=8 v^{2} / 2 \times 1.5=2.67 v^{2} N
We know that total vertical reaction at each of the outer wheels,
P_{ O }=\frac{W}{4}+\frac{P}{2}+\frac{Q}{2} …(i)
and total vertical reaction at each of the inner wheels,
P_{ I }=\frac{W}{4}-\frac{P}{2}-\frac{Q}{2}=\frac{W}{4}-\left(\frac{P}{2}+\frac{Q}{2}\right) …(ii)
From equation (i), we see that there will always be contact between the outer wheels and the road surface because W/4, P/2 and Q/2 are vertically upwards. In order to have contact between the inner wheels and road surface, the reactions should also be vertically upwards, which is only possible if
\frac{P}{2}+\frac{Q}{2} \leq \frac{W}{4}i.e. 0.15 v^{2}+2.67 v^{2} \leq 3924 or 2.82 v^{2} \leq 3924
\therefore v^{2} \leq 3924 / 2.82=1391.5
or v \leq 37.3 m / s =37.3 \times 3600 / 1000=134.28 km / h