A really awesome situation

Solving differential equations to solve physical problems is always fun, but it is never as fun as the situation i am going to present in this note,

Think about the situation shown in the figure, a tiny block is moving upon and along a rod that is hinged at the middle and can rotate in the vertical plane only,

Firstly you will observe that the rod rotates because the blocks weight has a net torque about it (when it is not at centre)

Let the distance along the rod be measured as 'x' and the angle the rod makes with the horizontal is 'a'

Qualitatively here are the Three situations you will expect would happen

1) The block moves too fast (at t=0) and so it slides off the rod very soon

2) The block moves too slow and comes to rest at the middle (ofcourse this can never practically happen, imagine the limiting case)

3) The block exactly reaches the centre when the rod becomes horizontal and both also reach the extremum at the same time, which means the system will be in oscillatory motion forever

Ok now let us see what The Math of the situation says,,

First let me make some assumptions so the differential equation we have to solve comes out to be simple

namely

1) The angle the rod rotates is very small, and thus this also tells that the systems initial energy is very small and hence we can neglect coriolis and centripetal forces in rotating frame (or their accelerations in ground frame)

2) All motions very slow

While the assumptions greatly limit us, i am forced to do so but the results are still awesome and Diverse

So, Let us analyse -

The torque on the rod is

$Nx\quad =\quad I\ddot { a }$

(worry not the directions and signs, they all depend on initial conditions and we can use negative numbers)

(also 'a' is the angle)

Now, for small angles, the normal reaction is roughly the weight of the block itself,

hence

$Nx\quad =\quad I\ddot { a } =mgx$

Now the force on the block along the rod assuming no friction is

$mgsin(a)-m{ w }^{ 2 }x=m\ddot { x } \simeq \quad mga$

Now to solve the two equations, i differentiate second one twice and substitute to get

$\\ m{ g }^{ 2 }x\quad =\quad I\ddddot { x } \\ \\ m{ g }^{ 2 }a\quad =\quad I\ddddot { a }$

Solving them yields me

$x\quad =\quad A{ e }^{ wt }+B{ e }^{ wt }+Csin(wt+D)\\ \\ a\quad =\quad A{ e }^{ wt }+B{ e }^{ wt }+Csin(wt+D)$

where $w=\left( \frac { m{ g }^{ 2 } }{ I } \right) ^{ \frac { 1 }{ 4 } }$

Which tell me all the information i need,

Now observe, if A and B are 0 as per initial conditions,

, Then the rod executes simple harmonic motion along with the block ,

If C and D and B are 0, then Quickly the block acquires enormous speed and distance and the block flies off or slides off (technically this is wrong as under my approximation this is impossible)

and motion dies off if A=C=D=0 as it exponentially decays

and there are several other cases, by setting some initial conditions, you can easily check it.