The magic cart (part 2)

This problem is a continuation of this problem . Information from that question has been provided here as well.

A magic cart is placed at the origin of the complex plane. It has two small motors, built into each side of it, which propel it throughout the plane using thrust force. These motors are programmed to function in a way such that the sum of the vectors of the cart's experienced force, momentum and position on the plane, z z , is always equal to a certain function, f ( t ) f(t) . The rate at which this sum changes is proportional, by a constant k k , to the sum's value at any given time.

A magic rod, connecting the origin to the center of mass of the cart, is added to the system. This rod magically changes length depending on the distance of the cart from the origin. The cart returns to the origin and carries out the same journey again, this time with the magic rod.

What is the moment (torque), in N m \ce {N m} , generated about the origin after 2 π 2\pi seconds?

Details and Assumptions :

  • When the cart begins its journey, it has a velocity of ( 0.25 0 ) \begin{pmatrix}-0.25\\0\end{pmatrix} .
  • The initial sum of the force, momentum, and position is equal to ( 1 0 ) \begin{pmatrix}-1\\0\end{pmatrix} .
  • The cart has a mass of 3 k g 3kg and each motor has a mass of 0.5 k g 0.5kg .
  • The rod has negligible mass and thickness.
  • k = λ i k=\lambda i for some λ > 0 \lambda >0 .
  • i i denotes the imaginary unit, i = 1 i=\sqrt{-1} .
  • Ignore effects of friction and gravity; the only force acting on the cart is the thrust force.

Part 3


The answer is 0.

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1 solution

Charley Shi
Mar 8, 2020

After calculating both y ( 2 π ) y(2\pi) and y ( 2 π ) y''(2\pi) , we can see that both are purely imaginary. This means that force vector and the position vector both point down the imaginary axis, so no moment is generated.

When I computed the force, I found that it was directed upward in the complex plane: 4y''(2pi) = +i/2 [ e^-pi (1 - pi) + 1 ]. There's still no torque in part 2, but the sign of the force vector makes a difference when solving part 3!

Matt Janko - 11 months ago

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