Physics Quiz 10

[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.

[Ether]

Originally Posted by 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.

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.

[dellagd]

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.

[Ether]

Originally Posted by dellagd

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.

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.

It seems thats not what you used here.

That is what I used.

[dellagd]

Originally Posted by Ether

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.

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:

Spoiler for Solution:

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

Substitute in for Vcx and Fx:

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

Substitute in for Fy:

2 + Vcy = ((-2^2)/((28/12)^2)) * Vcy
Vcy = -1.153

Into another equation:

Vcx/(-1.153) = 2 / (28/12)
Vcx = -.988

Robot Velocity components are opposite of these.

Vry = 1.153
Vrx = .988

Vr = sqrt(1.153^2 + .988^2) = 1.519 ft/s

Vtangential = r * omega

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

omega = Vt / r = 1.519 / 1.537 = 84/85 = .988 rad/s


I confirmed this using the other method and my answer was the same down to 6 decimal places.

[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

[Ether]

Originally Posted by dellagd

Yes, via the equation Vt = r * omega

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

[Ether]

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.

[Jared]

Originally Posted by Ether

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.

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

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

[Ether]

Originally Posted by Jared

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

Work:

Reps to you !

[Ether]

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?

[Jared]

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: