Problem — Temperature of a car's brakes

Problem

A $1000\;\mathrm{kg}$ car travelling at $50\;\mathrm{km/h}$ brakes sharply to a stop. Where does the initial kinetic energy end up? Estimate the temperature reached by the brake discs if their total mass is $20\;\mathrm{kg}$ ($c_\text{acciaio} \approx 450\;\mathrm{J/(kg\cdot K)}$).

Solution

Where the energy goes. During braking, friction between the pads and discs converts the car’s kinetic energy into thermal energy: the discs heat up. (A smaller part goes into the air, the tyres and the road; here we estimate the case where almost all of it ends up in the discs.)

Initial kinetic energy. $v = 50\;\mathrm{km/h} = \dfrac{50}{3{,}6}\;\mathrm{m/s} \approx 13{,}9\;\mathrm{m/s}$. $E_k = \frac{1}{2}\,m\,v^2 = \frac{1}{2}\cdot 1000\;\mathrm{kg}\cdot (13{,}9\;\mathrm{m/s})^2 \approx 9{,}6\cdot 10^4\;\mathrm{J}$

Temperature rise of the discs. The heat absorbed is $Q = m_\text{d}\,c\,\Delta T$. Setting $Q = E_k$: $\Delta T = \frac{E_k}{m_\text{d}\,c} = \frac{9{,}6\cdot 10^4\;\mathrm{J}}{20\;\mathrm{kg}\cdot 450\;\mathrm{J/(kg\cdot K)}} = \frac{9{,}6\cdot 10^4}{9000}\;\mathrm{K}$ $\ev{\Delta T \approx 11\;\mathrm{K}}$

Discussion. A single braking event heats the discs by about ten degrees: not much. But on a long descent, with repeated braking, the heat builds up and the brakes can reach hundreds of degrees, risking fading (loss of effectiveness). This is why on a descent it is better to use engine braking.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Energia termica da attrito · Calore specifico e capacità termica · Energia cinetica Competenze: Stime di Fermi