Question

An automobile has a total mass of 1100 kg. It accelerates from rest to 69 km/h in 12 s. Assume each wheel is a uniform 23 kg disk. Find, for the end of the 12 s interval, (a) the rotational kinetic energy of each wheel about its axle, (b) the total kinetic energy of each wheel, and (c) the total kinetic energy of the automobile.

(A) Rotational Kinetic Energy : Kr = (1/2)*I*(w^2)

I for the disc about its (1/2)*m*R^2

Kr = (1/2)*(1/2)*m*R^2*(w^2) = (1/4)*m*v^2 = (1/4)*23*(69*1000/60*60)^2 = 2112.3 J

(B) Kinetic energy = Kv = (1/2)*m*v^2 = (1/2)*23*(69*1000/60*60)^2 = 4224.6 J

So total KE = Kr + Kv = 6336.9 J

(C) Total Kinetic energy = (1/2)*m*V^2 = (1/2)*1100*(69*1000/60*60) ^2

= 202048.6 J

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