Physics Quiz 10
 

Assume the following:

- skid-steer vehicle with 4 standard non-treaded wheels

- square drivebase: trackwidth = 2 ft, wheelbase = 2ft

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

- center of mass 150 pounds is located at center of geometry

- 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, how fast is the robot rotating?

Re: Physics Quiz 10
 

I'm not sure how friction factors into this. A really wide wheel will have much more turning scrub than a thin or donut shaped wheel.

Here's what I've got so far:
I broke the wheel velocity into two components, one in the direction from the wheel to the center of the bot, and the other perpendicular to this one. The inward pointing one is all lost to friction. The other one has a magnitude of 0.707 * 2 = 1.414 feet/second.

The distance from the center to the wheel is 2.828 feet. This means one rotation is 2*pi*r=17.768 feet. This means it's rotating at 1.414/17.768 = 0.079 rotations per second.

I think I'm missing something eight the friction.

Re: Physics Quiz 10
 

I'm not getting the friction part yet either, though I think you may have miscalculated the radius you used, 2.828 is the distance from wheel to wheel diagonally, making the rotation .159 rev/s. With the friction I'm guessing it will be less than this.

Re: Physics Quiz 10
 

Look at Endnote 2 of Scenario 4.

I also got dellagd's answer. 1 rad/s

Re: Physics Quiz 10
 

I would think the coefficient of kinetic friction (and robot mass, for that matter) comes into play when calculating torque required to continue to rotate at the determined steady-state speed. But this was not asked.

Re: Physics Quiz 10
 

My first thought was 2rad/sec, but I'm pretty sure that's with 4 omni wheels. In order to solve this you need to know how much linear distance is traveled, when 2ft of tire tread is pulled by.

Re: Physics Quiz 10
 

Oops. I change my answer to .158 rotations per second.

Re: Physics Quiz 10
 

Just where do you define the tangential speed? Tangent to the circle subscribed through the wheels center line? if so rotation is 2ft/s* C/8.884ft=.2251rev/sec.

If "Tangential speed" is the slip between the wheel & the ground normal to the wheel axis, then life becomes more complicated & you need weight & friction. For instance a coefficient of 0 rotational speed is uncoupled from wheel speed.

Re: Physics Quiz 10
 

Reps to you Nate.

Given the assumptions in the problem statement, the answer is 1 rad/sec.



The wheel radius times the wheel radians/sec, tangent to the circumference of the wheel at the contact patch, in the plane of the wheel.

Re: Physics Quiz 10
 

At steady-state, the net torque acting on the bot around its center of mass must be ZERO.

The only forces acting on the bot in the horizontal plane are the four friction forces (one at each wheel). The direction of these forces will start out aligned with the plane of each wheel, but will change as the robot begins to spin. Given the assumption about CoF being independent of speed and direction, the friction at each wheel will always point in the direction of the relative motion between the carpet and the wheel at the contact patch. This relative motion is the vector sum of the wheel tangential speed plus the carpet speed at the wheel due to the robot's motion.

When the robot spin reaches the point where the net torque around the center of rotation is zero, the spin will stop changing.

The robot weight (150 lbs) and CoF (0.8) were red herrings. Given the assumptions, their values do not affect the answer (as long as neither is zero).

Re: Physics Quiz 10
 

Physics Quiz 10 part B

Now, assume the drivebase is rectangular instead of square:
wheelbase = 2 feet, trackwidth = (28/12) feet. Everything else the same.

At steady state, how fast is the robot rotating?

Re: Physics Quiz 10
 

.85 rad/s

The rotation speed will be achieved when the net torque on the robot becomes zero (after it was non-zero while the robot is initially accelerating). Also, there is only one force coming from each wheel, the frictional force provided by the carpet on the wheels. Therefore, the only way bring this force to zero (since it has a non-zero magnitude) is to have its line of action be in line with the center of rotation. (such that cos(theta) = cos(180) = 0 in the torque calculation).

The frictional force will also be in line with the vector sum of the wheel's tangential speed and the carpets speed.

Knowing the tangential speed of the wheel, the tangential direction of the robot's spin and the direction of the frictional force at the wanted equilibrium, the speed of the robot's spin can be solved for.

Re: Physics Quiz 10
 

What formula did you use to calculate the above? Show your work please.

Does that value of omega agree with your Vcy and Vcx calculations above??

Re: Physics Quiz 10
 

cos(tan^-1(12/14) = 0.7593

0.7593*2 = 1.5186 ft/second

Length of diagonal is 3.0731 feet.

Circumference is 3.0731*pi = 9.6544

1.5186/9.6544 =0.1572 rotation/sec = around 1 rad/sec.

Re: Physics Quiz 10
 

Its a system of 4 equations, I just solved for Vcy and and then got Vcx from the ratio of Vcy/Vcx.



Yes, via the equation Vt = r * omega

Re: Physics Quiz 10
 

Take your value omega=0.847 rad/sec and use it to calculate the x and y components of carpet speed:

Vcx = omega*(L/2) = 0.847*(2/2) = 0.847 ≠ 0.988

Vcy = 2-omega*(W/2) = 2-0.847*((28/12)/2) = 1.012 ≠ 0.847


Now try omega = 84/85:

Vcx = omega*(L/2) = (84/85)*(2/2) = 0.988

Vcy = 2-omega*(W/2) = 2-(84/85)*((28/12)/2) = 0.847

Re: Physics Quiz 10
 

In this particular problem, you can get the correct answer -- as you have done -- by simply projecting the wheel speed unto the direction the wheel is moving.

But as the problem gets more general (and more complicated), that will no longer work.

Re: Physics Quiz 10
 

Oh, it would seem that I made a stupid error. I swapped what I used for Wheelbase and Trackwidth. I wasnt sure, so I looked it up online and people were saying that trackwidth is the distance between wheels on a vehicle. It seems thats not what you used here.

I swapped my values in the calculations and my method gave me the correct answer. When I get home I'll repost a correct explanation.

Re: Physics Quiz 10
 

Yes, trackwidth is the distance between the 2 front wheels. (Or the 2 rear wheels). Here's an easy way to remember: It's the width of the "track" left by the vehicle when traveling forward in a straight line.

That is what I used.

Re: Physics Quiz 10
 

Ah, well even if I did find out the dimensions correctly I must have set up the proportion between the carpets velocity and the robot's orientation inversely. The solution below works out, so unless a mathematical equality that makes this give the right answer for the wrong reason is working here and I haven't realized what it is, I must have accidentally flipped that value somehow. It was late, I probably used the wrong angle for being tangential to the robots rotation which lead me to the wrong proportion.

Updated solution:

Re: Physics Quiz 10
 

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.

Re: Physics Quiz 10
 

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

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

Re: Physics Quiz 10
 

Reps to you !

Re: Physics Quiz 10
 

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?

Re: Physics Quiz 10
 

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:

Re: Physics Quiz 10
 

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

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.

Re: Physics Quiz 10
 

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...

Re: Physics Quiz 10
 

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).

Re: Physics Quiz 10
 

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:

Re: Physics Quiz 10
 

Steps 1 & 2 for front right wheel.