Question

A 500 kg electromagnet is initially at rest while holding 400 kg load of steel scrap. The combined mass has been observed to have a vertical natural frequency of 28.1 rad/sec. At time t =0, the magnet is turned off and the scrap is released. Assuming the crane and cable to be a single vertical spring, determine 1. a. The spring constant of the crane and cable. b. The natural frequency of vibration of the electromagnet without the scrap. c. The resulting motion of the electromagnet. d. The maximum tension developed in the cable during the motion. At what point does it occur?

I NEED HELP WITH C AND D!

Answer 1

Given,

Mass of electromagnet, m₁ = 500 kg

Mass of steel scrap, m₂ = 400 kg

Natural frequency of vibration, $\omega _{n}$ = 28.1 rad/sec

Total mass, M = 500+400 900 kg

a. We know,

$\omega _{n} = \sqrt{\frac{k}{M}}$                                                                                     ......(i)

$k = M*\omega ^{2}$

${\color{Red} k = 710649\: N/m}$

b. mass without scrap = mass of electromagnet = 500 kg

From equation (i)

$\omega _{n} = \sqrt{\frac{710649}{500}}$

${\color{Red} \omega _{n}=37.7 \: rad/sec}$

c. As the mass is reduced, the natural frequency of vibration increases. Thus the spring (cable) will start vibrating with more frequency. This same as what happens with automobiles. The suspension seems more comfortable when the car is loaded completely, while the suspension seems very poor while the car is empty.

d.The static deflection while the cable is supporting both the electromagnet and the scrap will be more.

It is given by,

$\delta = \frac{M*g}{k }$

$\delta = \frac{900*9.81}{710649}$

$\delta = 0.01242\: m$

There will also be deflection due to induced vibrations

We will use energy equillibrium to find out the amplitude of the vibration

At initial state:

x = 0 (distance from static deflection point)

$E_{spring} =\frac{1}{2}*k*x^{2}$

E_spring = 0

Potential Energy

Kinetic Energy, K.E.₁ = 0 (as initial velocity is 0)

Total energy at initial state

E₁ = E_spring + P.E.₁ + K.E.₁

E₁ = 0 + 8829 x + 0

E₁ = 8829 x                                                                                                   ......(ii)

At final state:

$E_{spring} =\frac{1}{2}*k*x^{2}$

$E_{spring} =\frac{1}{2}*710649*x^{2}$

Potential Energy

Kinetic Energy, K.E.₂ = 0 (as final velocity is 0)

Total energy at initial state

E₂ = E_spring + P.E.₂ + K.E.₂

E₂ = 355324.5 x² + 0 + 0

E₂ = 355324.5 x²                                                                                                   ......(iii)

From (ii) and (iii)

8829 x = 355324.5 x²

Therefore x = 0.02484 m

This is total amplitude of vibration about $\delta$

The maximum toatl deflection of spring will be

$X=\delta + \frac{x}{2}$

$X=0.01242+\frac{0.02484}{2}$

Tension = Force in spring

${\color{Red} T=17652.52\: N=17.652\: kN}$