The magic cart (part 1)

Classical Mechanics Level 3

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.

If z ( 2 π ) = − i a [ e − π ( b + π ) + b ] z(2\pi)=\frac{-i}{a}[e^{-\pi}(b+\pi)+b] , where a a and b b are integers, calculate a + b a+b .

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 .
  • Effects of gravity and friction can be ignored; the only force on the cart is the thrust force.
  • k = λ i k=\lambda i for some λ > 0 \lambda >0 .
  • i i denotes the imaginary unit, i = − 1 i=\sqrt{-1} .


The answer is 3.

Charley Shi by Charley Shi , 19, New Zealand

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Charley Shi
Mar 7, 2020

From the information given, we can see that 4 z ′ ′ + 4 z ′ + z = f ( t ) 4z''+4z'+z=f(t) .

Let S = 4 z ′ ′ + 4 z ′ + z S=4z''+4z'+z . Then d S d t = k S \frac{dS}{dt}=kS , so S = A e k t S=Ae^{kt} . A A and k k are constants.

From the information given S ( 0 ) = − 1 S(0)=-1 , so A = − 1 A=-1 . Therefore, S = f ( t ) = − e k t S=f(t)=-e^{kt} .

Solving the resulting differential equation, 4 z ′ ′ + 4 z ′ + z = − e k t 4z''+4z'+z=-e^{kt} yields z ( t ) = e − t 2 ( C 1 + C 2 t ) − e k t 4 k z(t)=e^{\frac{-t}{2}}(C_1+C_2t)-\frac{e^{kt}}{4k} .

Substituting in initial conditions of z ( 0 ) = 0 z(0)=0 and z ′ ( 0 ) = − 1 4 z'(0)=-\frac{1}{4} , we get that C 1 = 1 4 k C_1=\frac{1}{4k} and C 2 = 1 8 k C_2=\frac{1}{8k} .

Now, we can use the fact that z ′ ′ ( 0 ) = 0 z''(0)=0 to see that k k must satisfy the quadratic equation of k 2 + 1 4 = 0 k^2+\frac{1}{4}=0 . This results in k = i 2 k=\frac{i}{2} .

Substituting this into z ( t ) z(t) , we get z ( t ) = − i 2 ( e − t 2 ( 1 + t 2 ) − e i t 2 ) z(t)=\frac{-i}{2}(e^{\frac{-t}{2}}(1+\frac{t}{2})-e^{\frac{it}{2}}) .

Therefore, z ( 2 π ) = − i 2 ( e − π ( 1 + π ) + 1 ) z(2\pi)=\frac{-i}{2}(e^{-\pi}(1+\pi)+1) .

This is a brief solution and I have skipped intermediate steps. Also, I just realised that k k should have two solutions...

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Thank you for the solution. It is indeed a nice problem. My approach lead me to the third order ODE which was turning into a tedious exercise.

Karan Chatrath - 1 year, 3 months ago

I started part 3, got stuck, had to go back to part 1.

I believe the denominator in the last summand in line 4 in your solution is wrong.

As a solution to the d.e. I got z ( t ) = ( C 1 + C 2 t ) e − t / 2 − e k t ( 2 k + 1 ) 2 z(t)=(C_1+C_2t)e^{-t/2}-\frac{e^{kt}}{(2k+1)^2} z ′ ( t ) = ( C 2 − 1 2 C 1 − 1 2 C 2 t ) e − t / 2 − k e k t ( 2 k + 1 ) 2 z'(t)=(C_2-\frac12 C_1- \frac12 C_2t)e^{-t/2}-k\frac{e^{kt}}{(2k+1)^2} z ′ ′ ( t ) = ( 1 4 C 1 − C 2 + 1 4 C 2 t ) e − t / 2 − k 2 e k t ( 2 k + 1 ) 2 z''(t)=(\frac14 C_1 - C_2 + \frac14 C_2t)e^{-t/2}-k^2\frac{e^{kt}}{(2k+1)^2}

So that 4 z ′ ′ + 4 z ′ + z = ( 0 + 0 t ) e − t / 2 − ( 1 + 4 k + 4 k 2 ) e k t ( 2 k + 1 ) 2 = − e k t = f ( t ) 4z''+4z'+z=(0+0t)e^{-t/2}-(1+4k+4k^2) \frac{e^{kt}}{(2k+1)^2}= -e^{kt} = f(t) .

From here, finding C 1 = 1 ( 2 k + 1 ) 2 C_1=\frac{1}{(2k+1)^2} and C 2 = 1 − 2 k 8 k + 4 C_2=\frac{1-2k}{8k+4} is doable, but the rest is not as straightforward... you get z ′ ′ ( 0 ) = k 2 ( 2 k + 1 ) 2 − k 2 ( 2 k + 1 ) 2 = 0 z''(0)=\frac{k^2}{(2k+1)^2} -\frac{k^2}{(2k+1)^2}=0 , so this yields no new information. And filling in z ( 2 π ) = ( 1 ( 2 k + 1 ) 2 + 1 − 2 k 4 k + 2 π ) e − π − e 2 k π ( 2 k + 1 ) 2 z(2π)=(\frac{1}{(2k+1)^2}+\frac{1-2k}{4k+2}π)e^{-π}-\frac{e^2kπ}{(2k+1)^2}

If k = λ i k=λi this writes as z ( 2 π ) = ( 1 + π 2 ( 1 + 4 λ 2 ) ) e − π − e 2 π λ i 1 − 4 λ 2 + 4 λ i z(2π)=\frac{(1+\frac{π}{2}(1+4λ^2))e^{-π}-e^{2πλi}}{1-4λ^2+4λi} which has to be imaginary to satisfy the given form of the answer. Very little perspective to get an answer of the given form.

K T - 12 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...