# 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?

• How about showing where you got the cos(2θ). Oct 14 '20 at 14:26
• @R.W.Bird I have edited the question to innclude the derivation Oct 14 '20 at 14:35
• You need to show what direction you are writing the forces Oct 14 '20 at 15:49
• " tangential to the string" pretty sure the velocity is not tangential Oct 14 '20 at 15:50
• The velocity has to be tangential for the length of the string to remain constant, which is assumed. Note that the equations are written in the frame of reference of the central mass, which means that it is assumed to be stationary. That is why, the velocity of can only be tangential. Oct 14 '20 at 16:00

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 $$T(1-2\cos ^2 \theta)$$.

$$T$$ 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 $$\frac{mv^2}{l} = T + 2T\cos^2\theta$$

• Ohh right, i made a very idiotic mistake apparently. Thanks for pointing it out! Oct 15 '20 at 6:20

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.