Question

A haulage rope winds on a drum of radius 500 mm, the free end being attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the rope tightens. By considering the change in linear momentum of the truck and in the angular momentum of the drum, find the speed of the truck when the motion becomes steady.

(a) 0.502 m/s

(b) 0.602 m/s

(c) 0.702 m/s

(d) 0.802 m/s

I got this question in an interview.

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

Answer 1

Correct answer is (a) 0.502 m/s

To explain: Given : r = 500 mm = 0.5 m ; m1 = 500 kg ; m2 = 1250 kg ; k = 450 mm = 0.45 m ; u = 0.75 m/s

We know that mass moment of inertia of drum,

I2 = m2.k^2 = 1250 (0.45)^2 = 253 kg-m^2

Let v = Speed of the truck in m/s, and

F = Impulse in rope in N-s.

We know that the impulse is equal to the change of linear momentum of the truck. Therefore

F = m1.v = 500 v N-s

and moment of impulse = Change in angular momentum of drum

i.e. F x r = I1 (ω2 − ω1) = I2(u – v/r)

500v x 0.5 = 253(0.75 – v/0.5)

or, 250v = 380 − 506v

∴ 250 v + 506 v = 380

or v = 380/756 = 0.502 m/s