A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate total kinetic energy of road roller.

A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate total kinetic energy of road roller. Correct Answer 55170 N-m

Given : m = 12 t = 12 000 kg ; m1 = 2 t = 2000 kg ; k1 = 0.4 m ; d1 = 1.2 m or r1 = 0.6 m ; m2 = 2.5 t = 2500 kg ; k2 = 0.6 m ; d2 = 1.5 m or r2 = 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m Kinetic energy of rotation of the wheels and axles We know that mass moment of inertia of the front roller, I1 = m1(k1)2 = 2000 (0.4)2 = 320 kg-m2 and mass moment of inertia of the rear axle together with its wheels, I2 = m2 (k2)2 = 2500 (0.6)2 = 900 kg -m2 Angular speed of the front roller, ω1 = v/r1 = 2.5/0.6 = 4.16 rad/s and angular speed of rear wheels, ω2 = v/r2 = 2.5/0.75 = 3.3 rad/s We know that kinetic energy of rotation of the front roller, E1 =1/2 I1 (ω1)2 = 1/2 × 320(4.16)2 2770 N-m and kinetic energy of rotation of the rear axle together with its wheels, E2 =1/2 I2 (ω2)2 = 1/2 × 900(3.3)2 4900 N-m ∴ Total kinetic energy of rotation of the wheels, E = E1 + E2 = 2770 + 4900 = 7670 N-m We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller, E3 = 1/2 mv2 = 1/2 x 1200 (2.5)2 = 37500 N-m This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered. ∴ Total kinetic energy of road roller, E4 = Kinetic energy of translation + Kinetic energy of rotation = E3 + E = 37 500 + 7670 = 45 170 N-m