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.

Answer 1

(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