Physics Quiz 10

[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