Dumbbell In a Cylindrical Cavity

Consider the following problem:

At any general time $t$, the configuration of the system is as shown in the figure below:

• Let the reaction force on the mass $A$ be $N_1$. It is directed along $AO$ and towards $O$.

• Let the reaction force on the mass $B$ be $N_2$. It is directed along $BO$ and towards $O$.

• Let the tension developed on the massless rod be $T$. The tension acts on mass $A$ along $AB$ directed towards $B$ and acts on mass $B$ along $BA$ directed towards $A$.

From the diagram, it is apparent that:

$\phi + \theta = \frac{\pi}{4}$

Let the coordinates of point mass $A$ be:

$x_A = r \cos{\theta}$ $y_A = -r\sin{\theta}$

Let the coordinates of point mass $B$ be:

$x_B = -r\sin{\theta}$ $y_B = -r\cos{\theta}$

The following equations can be obtained by drawing appropriate free body diagrams for each mass. I have left out the free body diagrams from this analysis.

$m\ddot{x}_A = -N_1 \cos{\theta} - T\cos{\phi}$ $m\ddot{y}_A = N_1 \sin{\theta}-mg - T\sin{\phi}$ $m\ddot{x}_B = N_2 \sin{\theta} + T\cos{\phi}$ $m\ddot{y}_B = N_2 \cos{\theta}-mg + T\sin{\phi}$

The acceleration expressions can be found by differentiating $x_A$, $y_A$, $x_B$ and $y_B$ twice with respect to time. Note that the system is released from rest, so:

$\theta(0) = 0$ $\dot{\theta}(0) = 0$

And let:

$\ddot{\theta}(0) = \alpha$

Also, when the rod is just released, one can compute the following results:

$\phi = \frac{\pi}{4}$ $\ddot{x}_A = 0$ $\ddot{y}_A = - r \alpha$ $\ddot{x}_B = -r\alpha$ $\ddot{y}_B = 0$

Plugging these all these expressions into the equations above, and simplifying gives:

$0 = -N_1 -\frac{T}{\sqrt{2}}$ $- r \alpha = -mg -\frac{T}{\sqrt{2}}$ $- r \alpha = \frac{T}{\sqrt{2}}$ $0 = \frac{T}{\sqrt{2} }+ N_2 - mg$

Solving for $N_2$ gives:

$\boxed{N_2 = \frac{3mg}{2}}$

The above is the reaction force computed when the system is just released from rest.