2

My question relates to finding the vertical component of force due to a rigid wheel hitting a rigid step. I know the details of the step and I know the horizontal velocity of the wheel. I'm using the result to try and determine a damper of some sort to react this vertical force.

With reference to the image below, the shallow angle of the step is, a = 9.1° with the sides of the triangle being, x = 62.4 mm, y = 10 mm. Horizontal velocity of the wheel is, $V_x = 8,33 \frac{m}{s} [8330 \frac{mm}{s}]$ The unsprung mass on this wheel is, M = 150 kg.

Time taken to cover the distance of the x dimension: $t = \frac{62,43mm}{8330\frac{mm}{s}} = 0,0075 s $

Acceleration, $A_x = \frac{V_x}{t} = \frac{8,33\frac{m}{s}}{0,0075s} = 1110,67\frac{m}{s^2}$

So I reckon the horizontal force component is, $F_x = MA_x = 166 kN$

And so the vertical force component should be, $F_y = F_x\sin(9,1°) = 26,25 kN$

Please review my method and determine if this is a plausible method.

If I've made any errors I would appreciate any advice on how to correct this.

enter image description here

Mark
  • 5,343
  • 14
  • 47
richyo1000
  • 65
  • 7
  • 1
    Your calculation of acceleration is incorrect - Acceleration is change in velocity over time, but you've just put the initial velocity... I'd take an energy approach to this... – Jonathan R Swift Aug 30 '18 at 16:39
  • 1
    Agree with @JonathanRSwift - I was going to recommend conservation of angular momentum (since no moments are being applied because the impulse force points towards the center of the rolling wheel), you can look at the angular momentum of the center when the instant center of rotation (https://en.wikipedia.org/wiki/Instant_centre_of_rotation) was at the first contact point, and look at it again with the instant center of rotation at the second contact point, and find the new velocities. I'm commenting because I want to run it myself first to make sure it works before answering... – Mark Aug 30 '18 at 16:46
  • 1
    I'm voting to close this question as off-topic because this is a basic Physics question, not engineering – Carl Witthoft Aug 30 '18 at 17:49
  • 3
    @CarlWitthoft I don't understand - if I'm an engineering professional needing to use physics to design something for engineering, when does that become off-topic? All of engineering is simply applied physics. By asking for a question about an application in physics, it becomes an engineering question. More to the point, the relevant meta post disagrees (https://engineering.meta.stackexchange.com/a/25/2015) - a world class engineering site should be answering the tricky physics issues. – Mark Aug 30 '18 at 19:43
  • Hi Carl, this is an actual engineering problem I’m trying to solve-I’ve tried to simplify things and present the question in as textbook format as possible, for my own benefit. I’m still struggling to see the simplicity but now have been pointed in a more relevant direction I’ll give that a go. – richyo1000 Aug 30 '18 at 20:10
  • @richyo1000 - It looks like it can be done using this method, but I have to go. At the point of impact, for an instant of time, angular momentum is conserved, so the angular velocity decreases. The weight causes a reverse torque, so the angular acceleration of the impact point can be determined, allowing computations of the instantaneous acceleration of the center point. From there it seems like a diffeq to solve the response with time as the wheel goes over the hump. I'll see if I can summarize at some point, but if you do it yourself, feel free to answer your own question! – Mark Aug 30 '18 at 20:26

1 Answers1

3

Mark has indicated that he's going to write an answer relating to conservation of angular momentum - this is very likely the best way to approach the problem.

I'm going to answer very specifically the "Please review my method and determine if this is a plausible method." part of your question, and leave the final answer to Mark, since I'm in a rush!

So:

The kinetic energy of the wheel with mass $150\text{kg}$, and velocity $8.33\text{m/s}$, before it reaches the step is:

$$E_0=\frac{1}{2}mv_0^2=\frac{1}{2}*150 *8.33^2=5208.33\text{J}$$

The work done against gravity to lift the wheel up the step of height $10\text{mm}=0.01\text{m}$ is:

$$E_g=mg\Delta h=150*9.81*0.01=14.72\text{J}$$

The final horizontal velocity of the wheel, after the step, therefore, can be calculated using its new kinetic energy, $E_1=E_0-E_g=5208.33-14.72=5193.61\text{J}$:

$$v_1=\sqrt{\frac{2E_1}{m}}=\sqrt{\frac{2*5193.61}{150}}=8.32\text{m/s}$$

Clearly then, the horizontal speed is not constant, so much of your original working, which relies upon a time calculation for the climb up the step, is not valid.

It's worth stressing, that the Forces in the $x$ and $y$ directions will not be constant during the 'climbing' phase, with each decaying sinusoidally until there is no force (other than reaction to the mass of the wheel), i.e. the step has been fully climbed.

Jonathan R Swift
  • 7,771
  • 1
  • 15
  • 28
  • Hi Jonathan, thanks for the advice (and correction!) as you can see, my physics is very rusty! I’ll see what I can learn from my dynamics book this evening. Many thanks for taking the time to answer-the problem seems a little more complicated than I’d initially thought : ) – richyo1000 Aug 30 '18 at 19:25