Physics Quiz 10

FIRST General Forum
21

***Physics Quiz 10 part C:

Given:

  • wheelbase=24 inches, trackwidth=30 inches

  • center of mass 150lb located 3 inches forward and 4 inches to the right of the center of geometry

  • assume flat level floor, and the bottom of all four wheels coplanar

Find the weight (normal force) at each wheel.**

22

Front left 34.375
Front right 59.375
Back left 20.625
Back right 35.625

Work:
http://i.imgur.com/MQ0xHY7.jpg

23

Reps to you !

24

***Physics Quiz 10 part D:

Assume the following:

  • wheelbase=24 inches, trackwidth=30 inches

  • center of mass 150lb located 3 inches forward and 4 inches to the right of the center of geometry

  • assume flat level floor, and the bottom of all four wheels coplanar

  • skid-steer vehicle with 4 standard non-treaded wheels, except the 2 front wheels are badly misaligned:
    — FrontRight is 0.22983514251446 radians CCW
    — FrontLeft is 0.65426122466113 radians CW

  • coefficient of kinetic friction 0.7, independent of speed and direction

  • left wheels being driven forward at tangential speed 2 ft/sec

  • right wheels being driven backward at tangential speed 2 ft/sec

  • at steady state, the vehicle’s center of rotation is located at the center of mass

At steady state, how fast is the robot rotating?**

25

This one is trickier…

Im not sure if this is right, especially because the robot must now be constrained to rotate around its center, but I got 0.1748 rotations per second = 1.098 rad/sec.

Work:

26

Yes it is. But this is all leading somewhere, and we’re almost there.

Im not sure if this is right, especially because the robot must now be constrained to rotate around its center, but I got 0.1748 rotations per second = 1.098 rad/sec.

Work:

There was no link to your work.

Here’s how you can check your answer:

  1. at each wheel, use your answer to calculate the carpet velocity vector at that wheel due to robot rotation around the center of mass

  2. at each wheel, combine (vector addition) the carpet velocity due to robot motion at that wheel (from step1) and the tangential velocity of that wheel to get the net carpet velocity at that wheel.

  3. compute the normal force at each wheel, and use that to calculate the magnitude of the kinetic friction at each wheel.

  4. at each wheel, use the net carpet velocity direction (from step2) at that wheel to split the kinetic friction magnitude at that wheel into X and Y components at that wheel

  5. Sum the X components from step4. The sum should be zero at steady state.

  6. Sum the Y components from step4. The sum should be zero at steady state.

  7. Sum the torques around the center of mass due to the X and Y components from step4. The sum should be zero at steady state.

  8. If any of the sums from steps 5, 6, or 7 are not zero, your answer is not correct.

27

I think now that my answer is wrong because I must have a net force in the x direction due to friction. For my answer I had used my previous method of breaking the velocity into two perpendicular components, with one of those components tangent to the circle centered at the robots center and intersecting the wheel.
I then multiplied each of these velocities by the ratio of weight over the wheel to total weight so that the wheel closer to the center of mass and with the most weight on top of it have a greater impact on total velocity.

To me, this seems like you might be leading Us toward forward kinematics for swerve…

28

Notice that you can use the above to find the answer.

Set up equations using steps 1 thru 7, then plot 5, 6, and 7 vs omega and locate the value of omega for which all 3 curves are zero.

Reps to the first person to post a graph with the answer (show your work).

29

This thread keeps getting hits so maybe there’s still some interest.

I’ll kick-start this, with Step#1 for the Front Right wheel:

*
*
> # rad/sec robot rotation around CoM:
> omega=1.098
      1.098
>
> # wheelbase:
> L=24/12
          2
>
> # trackwidth:
> W=30/12
        2.5
>
> # X coordinate of Center of Mass (relative to Center of Geometry):
> Xm=4/12
 0.33333333
>
> # Y coordinate of Center of Mass (relative to Center of Geometry):
> Ym=3/12
       0.25
>
> # X component of carpet motion at FrontRight wheel due to robot rotation:
> V1x=-omega*(L/2-Ym)
    -0.8235
>
> # Y component of carpet motion at FrontRight wheel due to robot rotation:
> V1y=omega*(W/2-Xm)
     1.0065
30

*Steps 1 & 2 for front right wheel.
*
*

steps1&2.pdf (12.3 KB)


steps1&2.pdf (12.3 KB)