Tension approaching infinity? 
I've been trying to solve this sum, where you have to find tension as a function of angle $\theta$. The three masses are on a frictionless surface, and the middle one is given an initial velocity $u$. It's a pretty long problem, but there is one thing in my process I'm unsure about.
Let the masses be $m_1$, $m_2$ and $m_3$ from left to right. The force on $m_2$ will be $2T \cos \theta$. Looking at $m_1$ in the frame of reference of $m_2$, it will have some velocity $v$ perpendicular to the string. Along with the tensile force, there will also be a pseudo force $2T \cos \theta$ in the downward direction. Thus, taking the component of the pseudo force in the radial direction, and writing the equation for centripetal acceleration, I get:
$$\frac {mv^2}{l} = T - 2T \cos ^2 \theta$$
$$T = -\frac {mv^2}{l \cos 2\theta}$$
But according to this equation, once the angle reduces to $45^\circ$, the tension will approach infinity. Where did I go wrong?
 A: 
Along with the tensile force, there will also be a pseudo force $2T\cos \theta$ in the downward direction.

From the reference frame of the observer, the net force acting on $m_2$ mass is $2T\cos \theta$ in the downwards direction. Thus the acceleration of the block will be $2T\cos \theta \over m$ in the downwards direction. The pseudo force acting on $m_1$ will be $2T\cos \theta$ in upwards direction from the reference frame of $m_2$, and the radial component of net force will be $T(1+2\cos ^2 \theta)$ instead of $T(1-2\cos ^2 \theta)$.
A: $T$ is an absolute value of a tensile force. This quantity cannot be negative. Thus, your last formula cannot be right.

Along with the tensile force, there will also be a pseudo force $2T\cos\theta$ in the downward direction.

Actually, a pseudo force $2T\cos\theta$ is directed upward. Hence, the correct form of  your second equation is
$$
\frac{mv^2}{l} = T + 2T\cos^2\theta
$$
A: Given, u, as an initial velocity, we can assume there are no external forces. Momentum and energy are conserved. Conservation of momentum gives the velocity of the center of mass as u/3. Combining this with the geometry of the system gives the x and y components of position for each of the masses in terms of  θ and t (taking the y axis in the direction of u).  Taking derivatives yields the components of velocity.  Applying conservation of energy to either the fixed or the center of mass frame yields:
$ω^2 =[(u^2)/(L^2)]/[3 – 2(sin^2θ)]$. (Where ω = dθ/dt).
Using L$ω^2$ as the centripetal acceleration in the $m_2$ system permits finding T as a function of θ and u.
