A four-wheeled automobile car has a total mass of 1000 kg. The moment of inertia of each wheel about a transverse axis through its centre of gravity is 0.5 kg-m^2. The rolling radius of the wheel is 0.35 m. The rotating and reciprocating parts of the engine and the transmission system are

Question

A four-wheeled automobile car has a total mass of 1000 kg. The moment of inertia of each wheel about a transverse axis through its centre of gravity is 0.5 kg-m². The rolling radius of the wheel is 0.35 m. The rotating and reciprocating parts of the engine and the transmission system are equivalent to a moment of inertia of 2.5 kg-m², which rotates at five times the road-wheel speed. The car is traveling at a speed of 100 km/h on a plane road. When the brakes are applied, the car decelerates at 0.5 g. There are brakes on all four wheels. Calculate:
(i) the energy absorbed by each brake; and
(ii) the torque capacity of each brake.

Accepted Answer

[latex] ext { Given For car, } \quad m=1000 kg \quad v_{1}=100 km / h [/latex].

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

[latex] v_{1}=100 km / h =\frac{100 imes 10^{3}}{60 imes 60} m / s [/latex].

[latex] =27.78 m / s \quad ext { and } \quad v_{2}=0 [/latex].

[latex] KE =\frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2} ight)=\frac{1}{2}(1000)(27.78)^{2} [/latex].

[latex] =385802.44 J [/latex].

(ii) KE of the wheels

[latex] \omega_{1}=\frac{v_{1}}{R}=\frac{27.78}{0.35}=79.37 rad / s ext { and } \omega_{2}=0 [/latex].

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

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

(iii) KE of the engine and transmission system

[latex] \omega_{1}=5(79.37)=396.83 rad / s ext { and } \omega_{2}=0 [/latex].

[latex] KE =\frac{1}{2} I\left(\omega_{1}^{2}-\omega_{2}^{2} ight)=\frac{1}{2}(2.5)(396.83)^{2} [/latex].

[latex] =196837.97 J [/latex].

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.

[latex] E=\frac{1}{4}(385802.44+6298.81+196837.97) [/latex].

[latex] =147234.8 J [/latex] (i).

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

[latex] \frac{v_{1}-v_{2}}{t}=0.5 g \quad herefore \frac{27.78-0}{t}=0.5(9.81) [/latex].

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

[latex] \left(\omega_{1}+\omega_{2} ight) / 2 ext { or }\left(\omega_{1} / 2 ight) ext {. Therefore, } [/latex]

[latex] heta=\left(\frac{\omega_{1}}{2} ight) t=\left(\frac{79.37}{2} ight)(5.66)=224.6 rad [/latex].

[latex] M_{t}=\frac{E}{ heta}=\frac{147234.8}{224.6}=665.54 ext { N-m } [/latex] (ii).

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions