Difficult problem from IPhO Selection Test

Физика
12
  1. To answer the first question you can firstly take a look at my post here:

The radius vector in rotating system of coordinate S’:

\vec r'=x'\vec{\hat x'}+y'\vec{\hat y'}+z'\vec {\hat z'}=r'\vec{\hat r'}

Then you need to take into account the fact that: (its obvious)

\vec a'=\ddot{\vec {\ r'}}

Firstly take the first derivative:

\dot{\vec r'}={\dot r'}\vec{\hat r'}+r'\dot {\vec{\hat r'}}

Then take the second derivative:

\ddot{\vec r'}={\ddot r'}\vec{\hat r'}+r'\ddot {\vec{\hat r'}}+2\dot{r'}\dot{\vec {\hat r'}}

You should use these equations:

\vec {\hat {x'}}=\vec {\hat {x}}\cos\omega_0t+\vec {\hat {y'}}\sin\omega_0t \quad \vec {\hat {y'}}=-\vec {\hat {x'}}\sin\omega_0t+\vec {\hat {y}}\cos \omega_0t

and take their derivatives
By the way, formulas of the unit vectors \vec{\hat x'}, \vec{\hat y'} are similar to the ones, which were in polar coordinates:

\vec {\hat r}=\vec {\hat x}\cos\theta+\vec{\hat y}\sin\theta \quad \vec{\hat \theta}=-\vec{\hat x}\sin\theta +\vec{\hat y}\cos\theta

(You’re able to finish it)
2. Answer to the 2nd question is simple. You need to use the equation:

\vec F'=m\vec a'=\vec F-m\vec \omega_0\times(\vec\omega_0\times \vec r')-2m\vec\omega_0\times \vec v'

Consider the projections of the force on axes x and y:

m\ddot y=F_y-(m\vec \omega_0\times(\vec\omega_0\times \vec r'))_y-(2m\vec\omega_0\times \vec v')_y
m\ddot x=F_x-(m\vec \omega_0\times(\vec\omega_0\times \vec r'))_x-(2m\vec\omega_0\times \vec v')_x

You also should use the equation of surface:

z=A_0(x')^2-B_0(y')^2

to find an expression for potential energy:

U(z)=mgz=mg(A_0(x')^2-B_0(y')^2)

The force vector is an anti-gradient of potential energy:

\vec F=-\nabla U=-(\frac{\partial U}{\partial x}{\hat {x}}+\frac{\partial U}{\partial y}{\hat y}+\frac{\partial U}{\partial z} {\hat z})\Rightarrow F_x=-(\frac{\partial U}{\partial x}), F_y=-(\frac{\partial U}{\partial y})

Using this and also completing some of the cross-products you will solve the problem
(You’re able to finish it)
3. You shouldn’t have any problems answering this question, because you just need to put the x’,y’ functions to the differential equations from question 2:

m\ddot x'-2m\omega_0\dot{y'}+m(2A_0g-\omega_0^2)x'=0
m\ddot y'+2m\omega_0\dot x'-m(2B_0g+\omega_0^2)y'=0
x'=x_0e^{-i\Omega t}, y'=y_0e^{-i\Omega t}

(You’re able to finish it)
I think that you’re capable of solving the 4th and 5th problems, but take into account that in a stable state, where oscillation occurs only around the origin you can use these kinds of approximations for displacements:

(1+\delta)^\gamma\approx1+\gamma \delta,\quad e^\delta\approx1+\delta

and also the force must be equal to zero in a stable equilibrium

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