- #1
FreezingFire
- 12
- 1
Homework Statement
"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!).
Homework 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.
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.
Thanks in advance!