+1 vote
in Machine Kinematics by (38.0k points)
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 kinetic energy of rotation of the wheels and axles at a speed of 9 km/h.

(a) 7670 N-m

(b) 8670 N-m

(c) 9670 N-m

(d) 6670 N-m

The question was posed to me in an online interview.

My question is taken from Numericals On Kinetics Of Motion and Loss of Kinetic Energy in section Kinematics & Kinetics of Motion of Machine Kinematics

1 Answer

0 votes
by (106k points)
The correct option is (a) 7670 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

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