Automotive Disk Brakes The driver of a 1200-kg automobile traveling at 100 km/h applies the brakes and brings the vehicle to a complete stop. The vehicle has front and rear disk brakes, and the braking system is balanced so that the front set provides 75% of the total braking capacity. Through

Question

Automotive Disk Brakes

The driver of a 1200-kg automobile traveling at 100 km/h applies the brakes and brings the vehicle to a complete stop. The vehicle has front and rear disk brakes, and the braking system is balanced so that the front set provides 75% of the total braking capacity. Through friction between the brake pads and the rotors, the brakes convert the automobile’s kinetic energy into heat (Figure 7.9). If the two front 7-kg brake rotors are initially at a temperature of [latex] 25^°C [/latex], how hot are they after the vehicle stops? The specific heat of cast iron is [latex] c = 0.43 kJ/(kg .^°C).[/latex]

Approach
As the vehicle comes to a stop, a portion of its initial kinetic energy is lost through air drag, the rolling resistance of the tires, and the wear of the brake pads, but we will neglect those extraneous factors at the first level of approximation. The automobile’s kinetic energy [Equation (7.3)]

[latex] U_{k}=\frac{1}{2} m v^2 [/latex] (7.3)

decreases as work is performed on it, and the heat produced will cause the temperature of the brake rotors to
increase. That temperature rise can be found by applying Equation (7.7).

[latex] Q = mc(T - T_ 0) [/latex] (7.7)

Accepted Answer

In dimensionally consistent units, the initial velocity is

[latex]v=\left(100 \frac{km}{h} ight)\left(1000 \frac{m}{km} ight)\left(\frac{1}{3600}\frac{h}{s} ight) [/latex]

[latex] = 27.78 \left(\frac{\cancel {km}}{\cancel {h}} ight)\left(\frac{m}{\cancel{km}} ight)\left(\frac{\cancel{h}}{s} ight) [/latex]

[latex]=27.78 \frac{m}{s}[/latex]

The automobile’s kinetic energy decreases by the amount

[latex]\Delta U_{k}=\frac{1}{2} (1200 kg)(27.78\frac{m}{s} )^2 \longleftarrow \left[U_{k}=\frac{1}{2}m v^2 ight] [/latex]

[latex]= 4.630 imes 10^5\left(\frac{kg.m}{s^2} ight)(m) [/latex]

[latex]= 4.630 imes 10^5 N.m[/latex]

[latex]= 4.630 imes 10^5 J[/latex]

[latex]= 463.0 kJ [/latex]

Because the front brakes provide three-quarters of the braking capacity, the quantity

Q = (0.75)(463.0 kJ)

= 347.3 kJ

of heat flows into the front rotors. Their temperature rises according to

[latex] 347.3 kJ = 2(7 kg)\left(0.43\frac{kJ}{kg.^\omicron C} ight)(T-25^\omicron C) \longleftarrow \left[Q=mc(T-T_{0}) ight] [/latex]

where the factor of 2 accounts for both front rotors. The equation is dimensionally consistent, and the final temperature becomes [latex]T = 82.69^°C.[/latex]

Discussion
The brakes convert the automobile’s kinetic energy to heat, which in turn is stored as thermal energy by virtue of the rotors’ temperature rise. This temperature rise is an upper bound because we assumed that all the kinetic energy is converted to heat and not lost in other forms.

[latex]T = 82.69^°C[/latex]

Question 7.10: Automotive Disk Brakes The driver of a 1200-kg automobile tr...

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