# Releasing a rod tied to a string which was held at an angle

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1. Feb 25, 2016

### FreezingFire

1. The problem statement, all variables and given/known data
"Find the tension in the right string, when the left string is cut, if the mass of the rod is $m$ and its length is $l$, and $\sin \theta = 3/5$. The rod is initially horizontal." Please refer the diagram below (sorry for its clumsiness!).

2. Relevant equations
$$I_{rod} = \frac{ml^2}{12}$$
$$\tau _{CM} = I_{rod,CM} \cdot \alpha _{CM}$$ $\tau$ represents torque, CM to denote "in the centre of mass frame", $\alpha$ is the angular acceleration.

The Newton's Laws of Motion.

3. The attempt at a solution
First, only looking at the translational motion of the centre of mass, we get two equations:
$$mg - T \sin \theta = ma_x \qquad\text{...(i)}$$
where $T$ represents the tension in the string and $a$ represents acceleration.
$$T \cos \theta = ma_y \qquad \text{...(ii)}$$
Now, using the rotational dynamics formula,
$$T \sin \theta \cdot l = \frac{ml^2}{12} \cdot \alpha \qquad \text{...(iii)}$$
I have these three equations, which i think are correct. However i doubt my fourth equation:
$$a_y - \frac{l\alpha}{2} = a_x \cot \theta \qquad \text{...(iv)}$$
as acceleration of the point attached to the string, along the string, must be zero. However, on solving them, i get,
$$T=\frac{mg}{2 \sin \theta - 3 \sin \theta \tan \theta}=\frac{-20mg}{3}$$
which seems to be wrong (i don't have the correct answer to it). Where am i going wrong? Is the fourth equation really correct? If not, how do i correct it? Please guide me.

2. Feb 25, 2016

### haruspex

Your eqn (iv) looks ok, but the use of x and y there (standard) conflicts with your use in earlier equations (non-standard).
Also, the left hand side of (iii) is out by a factor.

3. Feb 25, 2016

### FreezingFire

Ah, i get it now! The first equation should have had $a_y$ and the second, $a_x$ (i did it by mistake). And the 3rd equation should have been:
$$T \sin \theta \cdot \frac l2 = \frac{ml^2}{12} \cdot \alpha \qquad \text{...(iii)}$$
$$T=\frac{mg}{4 \sin \theta + \cos \theta \cot \theta}=\frac{15mg}{52}$$
Is it correct now?