Calculating the Energy Absorbed by a Spring in a Car Suspension System

[Ebby]

Homework Statement

What will be the maximum instantaneous deformation of the springs if the car is lifted by a crane and dropped on the street from a height of 0.8 m?

Relevant Equations

W = 0.5*kx^2
K.E. = 0.5*mv^2
G.P.E. = mgh

I approach this by considering the four springs in parallel each with spring constant $k$ as one spring with four times the spring constant $k' = 4k$. The car is dropped and at the moment its tyres touch the ground I assume that the spring is in its resting position. As the car continues to travel in a downward direction the spring is loaded up, absorbing the car's kinetic and gravitational potential energy.

The governing equations seems to be:$$P.E._{spring} = K.E._{car}+\,G.P.E._{car}$$
$$k'y^2 = mv^2 + 2mgy$$
So we have a quadratic equation involving $y$:$$k'y^2 - 2mgy - mv^2 = 0$$
Where: $v^2 = 2gs = 2(9.81)( 0.8)$. So:$$280000y^2 - 2(1200)(9.81)y - 1200(2)(9.81)( 0.8) = 0$$
$$280000y^2 - 23544y - 18835.2 = 0$$

This has a positive solution at $y = 0.31$. However, I know that the correct answer is $0.26$ (metres). I cannot see what I've done wrong. I am pretty sure I have got the governing equation right, but... Any ideas?

 

[erobz]

Homework Statement: What will be the maximum instantaneous deformation of the springs if the car is lifted by a crane and dropped on the street from a height of 0.8 m?
Relevant Equations: W = 0.5*kx^2
K.E. = 0.5*mv^2
G.P.E. = mgh

View attachment 330445
I approach this by considering the four springs in parallel each with spring constant $k$ as one spring with four times the spring constant $k' = 4k$. The car is dropped and at the moment its tyres touch the ground I assume that the spring is in its resting position. As the car continues to travel in a downward direction the spring is loaded up, absorbing the car's kinetic and gravitational potential energy.

The governing equations seems to be:$$P.E._{spring} = K.E._{car}+\,G.P.E._{car}$$
$$k'y^2 = mv^2 + 2mgy$$
So we have a quadratic equation involving $y$:$$k'y^2 - 2mgy - mv^2 = 0$$
Where: $v^2 = 2gs = 2(9.81)( 0.8)$. So:$$280000y^2 - 2(1200)(9.81)y - 1200(2)(9.81)( 0.8) = 0$$
$$280000y^2 - 23544y - 18835.2 = 0$$

This has a positive solution at $y = 0.31$. However, I know that the correct answer is $0.26$ (metres). I cannot see what I've done wrong. I am pretty sure I have got the governing equation right, but... Any ideas?

For Starters: How much kinetic energy does the car have the instant maximum deflection of the springs has occurred?

It looks like there are other issues with your application of C.o.E. as well. Why is the deflection of the spring $y$ and the initial height it is dropped from the same variable?

 

[mjc123]

I also get 0.26 m if I use
k'y² = 2mg(0.8)
i.e. if I ignore the change in GPE after the tyres touch the ground and the spring starts to compress. This involves the simplifying assumption that y << 0.8, and the first thing to do after getting an answer is to check whether the assumption is valid - which it obviously isn't in this case.

 

[jack action]

This could be very simple ... or very complicated. But there is one piece of information missing for a complete analysis.

Let us imagine that, once in the air, the spring is fully extended. This means that on the ground, under the car's weight, the springs are compressed by $y_i = \frac{mg}{k}$. ($0.042\ m$ in our case.)

So, if you raise the car by $h$ the springs decompress by $y_i$. (Assuming $h > y_i$.) After letting the car go, once it dropped a distance $h$, the energy left to dissipate is:
$$mgh - \frac{1}{2}ky_i^2$$
This must be absorbed by the springs which will be compressed an additional amount $y$. That height must also be added to the potential energy, thus:
$$mg(h+y) = \frac{1}{2}k(y+y_i)^2$$
Or
$$\frac{k}{2mg}(y+y_i)^2- y - h = 0$$
This gives us $y = 0.256\ m$ in our case. Note that - IMHO - "the maximum instantaneous deformation of the springs" would be $y+y_i$ or $0.298\ m$ in our case.

But then the springs may be preloaded. Let us assume the springs are already pre-compressed by $y_i$. This means that on the ground, under the car's weight, the springs do NOT compress at all.

So, if you raise the car by $h$, the springs do NOT decompress and the wheel is effectively a distance $h$ from the ground. After letting the car go, once it dropped a distance $h$, the energy left to dissipate is:
$$mgh$$
Following the same logic as in the previous case:
$$mg(h+y) = \frac{1}{2}ky^2$$
Or
$$\frac{k}{2mg}y^2- y - h = 0$$
This is the same equation as in the OP and it gives us $y = 0.305\ m$ in our case. I'm not sure if "the maximum instantaneous deformation of the springs" would include the preload ($y+y_i$). But the wheel effective distance traveled with respect to the car would be only $y$.

And then, the preload on the springs could be even higher than $y_i$. The effect would be the same, i.e. the springs do NOT compress under the car's weight. This is more complicated to evaluate as the springs will not begin to compress as soon as the wheels touch the ground. If the preload is high enough it might not compress at all because the wheels and the rest of the car are effectively a single block. In this extreme case, we may assume the true stiffness is that of a solid metal bar ($k = \frac{EA}{L_0}$; source).

And then there are all the cases in between.

 

[haruspex]

The statement of height of lift is unclear. My best guess is that the mass centre of the car is raised $\Delta y=0.8m$ above its rest position.
Suppose that if the car were raised just enough that the springs were relaxed its mass centre would be at Y above the ground. If its rest position is $y_0$ above the ground then
$mg=k'(Y-y_0)$
If, at maximum spring compression, the mass centre is $y_1$ above the ground then the drop distance is $y_0+\Delta y-y_1$. At that point, the spring compression, the value we are asked to find, is $Y-y_1$:
$mg(y_0+\Delta y-y_1)=\frac 12k'(Y-y_1)^2$
For ease of typing, $A=mg/k'=\frac{1200\cdot 9.81}{4\cdot 7\cdot 10^4}$ m=0.042m.
$A=Y-y_0$
$2A(y_0+\Delta y-y_1)=(Y-y_1)^2$.
Writing $y=Y-y_1$,
$A=y+y_1-y_0$
$2A(y_0+\Delta y-y_1)=y^2 =2A(y-A+\Delta y)$
$(y-A)^2=A^2+2A\Delta y=0.069$m$^2$
$y=0.305$m.

 

[Ebby]

Thank you all. I think the question, as set, could have been a little clearer.