Question 12.2: A four-wheeled automobile car has a total mass of 1000 kg. T...

\text { Given For car, } \quad m=1000 kg \quad v_{1}=100 km / h .

Deceleration = (0.5g)
For wheels, I = 0.5 kg-m² R = 0.35 m
For engine and the transmission system,
I = 2.5 kg-m²
speed = 5 (wheel speed)
Step I Energy absorbed by each brake
(i) KE of the car

v_{1}=100 km / h =\frac{100 \times 10^{3}}{60 \times 60} m / s .

=27.78 m / s \quad \text { and } \quad v_{2}=0 .

KE =\frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2}\right)=\frac{1}{2}(1000)(27.78)^{2} .

=385802.44 J .

(ii) KE of the wheels

\omega_{1}=\frac{v_{1}}{R}=\frac{27.78}{0.35}=79.37 rad / s \text { and } \omega_{2}=0 .

text { KE of four wheels }=4\left[\frac{1}{2} I\left(\omega_{1}^{2}-\omega_{2}^{2}\right)\right] .

=4\left[\frac{1}{2}(0.5)(79.37)^{2}\right]=6298.81 J .

(iii) KE of the engine and transmission system

\omega_{1}=5(79.37)=396.83 rad / s \text { and } \omega_{2}=0 .

KE =\frac{1}{2} I\left(\omega_{1}^{2}-\omega_{2}^{2}\right)=\frac{1}{2}(2.5)(396.83)^{2} .

=196837.97 J .

The energy absorbed by the four brakes consists of the kinetic energy of the car, the kinetic energy of the wheel and the kinetic energy of the engine and the transmission system.

E=\frac{1}{4}(385802.44+6298.81+196837.97) .

=147234.8 J               (i).

Step II Torque capacity of brake
The braking time t is given by

\frac{v_{1}-v_{2}}{t}=0.5 g \quad \therefore \frac{27.78-0}{t}=0.5(9.81) .

∴          t = 5.66 s
The average velocity during the braking time is

\theta=\left(\frac{\omega_{1}}{2}\right) t=\left(\frac{79.37}{2}\right)(5.66)=224.6 rad .

M_{t}=\frac{E}{\theta}=\frac{147234.8}{224.6}=665.54 \text { N-m }             (ii).