Question

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) 25170 N-m

(b) 35170 N-m

(c) 45170 N-m

(d) 55170 N-m

The question was posed to me in a national level competition.

The above asked question is from Numericals On Kinetics Of Motion and Loss of Kinetic Energy in division Kinematics & Kinetics of Motion of Machine Kinematics

Answer 1

The correct answer is (d) 55170 N-m

Best explanation: 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-m^2

and mass moment of inertia of the rear axle together with its wheels,

I2 = m2 (k2)^2 = 2500 (0.6)^2 = 900 kg -m^2

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 mv^2 = 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