Physics Quiz 10

[Ether]

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?

[Jared]

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.

[dellagd]

Originally Posted by Jared

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.

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.

[Nate Laverdure]

Originally Posted by Jared

I'm not sure how friction factors into this.

Look at Endnote 2 of Scenario 4.

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

[seg9585]

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.

[Bryce2471]

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.

[Jared]

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

[FrankJ]

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.

[Ether]

Originally Posted by Nate Laverdure

Look at Endnote 2 of Scenario 4.

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

Reps to you Nate.

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

Originally Posted by FrankJ

Just where do you define the tangential speed?

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.

[Ether]

Originally Posted by Ether

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

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).

[Ether]

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?

[dellagd]

.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.

Spoiler for Calculations:

Vcarpet = Vc = opposite Vrobot
F = friction (not the actual force but friction lies along this vector, this is technically the vector sum mentioned earlier)

2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -(28/12) / 2 #Known angle of vector sum/friction
Vcx/Vcy = (28/12) / 2 #Known angle of carpet velocity

Vcy = -.847
Vcx = -.988

Vc magnitude = Vrobot magnitude = 1.302 ft/s

Vrobot = r * omega

omega = 1.302 / (sqrt(2^2 + (28/12)^2) / 2) = .847 rad/s

*crosses fingers*

[Ether]

Originally Posted by dellagd

Vcy = -.847
Vcx = -.988

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

omega = 1.302 / (sqrt(2^2 + (28/12)^2) / 2) = .847 rad/s

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

[Jared]

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.

[dellagd]

Originally Posted by Ether

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

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

Spoiler for Solving:

2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -(28/12) / 2 #Known angle of vector sum/friction
Vcx/Vcy = (28/12) / 2 #Known angle of carpet velocity

Substitute in for Vcx and Fx:

2 + Vcy = Fy
((28/12)/2) * Vcy = (-2/(28/12)) * Fy

Substitute in for Fy:

2 + Vcy = ((-(28/12)^2)/(2^2)) * Vcy
Vcy = -.847

Into another equation:

Vcx/(-.847) = (28/12)/2
Vcx = -.988

Robot Velocity components are opposite of these.

Vry = .847
Vrx = .988

Vr = sqrt(.847^2 + .988^2) = 1.302 ft/s

Vtangential = r * omega

r = d/2 = sqrt(2^2 + (28/12)^2)/2 = 1.537 ft

omega = Vt / r = 1.302 / 1.537 = .847 rad/s

Originally Posted by Ether

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

Yes, via the equation Vt = r * omega