- #1
johnhuntsman
- 76
- 0
The disc brakes of a high performance car are often made of carbon fiber instead of iron, thereby reducing the mass. If both types of discs are of the same size and shape, and each iron disc has a mass of 4 kg and each carbon disc has a mass of 1 kg, what is the reduction in rotational kinetic energy at 72 km/h if all the four iron discs in the car are replaced with carbon discs?
M1 = 4 kg
M2 = 1 kg
v = 72 km / h = 20 m / s
I = (1/2)MR2
K = (1/2)Iω2 + (1/2)Mv2
ω = v / R
ΔK = KFe - Kcf
My equation:
4[.5(.5M1R2)(v2 / R2) + .5M1v2] - 4[.5(.5M2R2)(v2 / R2) + .5M2v2]
The R's should cancel and and from that point on it's just plug and chug. Hoever I get 3600 J as the difference in KE. What's wrong with it?
[Edit] The answer is 1200 J by the way. That's why I don't like the 3600 J. [Edit]
M1 = 4 kg
M2 = 1 kg
v = 72 km / h = 20 m / s
I = (1/2)MR2
K = (1/2)Iω2 + (1/2)Mv2
ω = v / R
ΔK = KFe - Kcf
My equation:
4[.5(.5M1R2)(v2 / R2) + .5M1v2] - 4[.5(.5M2R2)(v2 / R2) + .5M2v2]
The R's should cancel and and from that point on it's just plug and chug. Hoever I get 3600 J as the difference in KE. What's wrong with it?
[Edit] The answer is 1200 J by the way. That's why I don't like the 3600 J. [Edit]