Question

A $40\text{kg}$ mass at the end of a rope of length $l$, oscillates in a vertical plane with angular amplitude ${\theta }_o$. What is the tension $T$ in the rope when it makes an angle $\theta$ with the vertical? If the breaking strength of the rope is $80\text{kgf}$, what is the maximum angular amplitude ${\theta }_{max}$ with which the mass can oscillate without the rope breaking?

• A. $T=mg(2\cos \theta -3\cos {\theta }_0),{\theta }_{max}={30}^{\circ }$
• B. $T=mg(3\cos \theta -2\cos {\theta }_0),{\theta }_{max}={60}^{\circ }$
• C. $T=mg(3\cos \theta -2\cos {\theta }_0),{\theta }_{max}={30}^{\circ }$
• D. $T=mg(2\cos \theta -3\cos {\theta }_0),{\theta }_{max}={60}^{\circ }$

Answer 1

The correct option is B $T=mg(3\cos \theta -2\cos {\theta }_0),{\theta }_{max}={60}^{\circ }$


The velocity at angular displacement $\theta$ is given by
$v=\sqrt{2gh}$ (from conservation of energy)
where $h=l(\cos \theta -\cos {\theta }_o)$

$\Rightarrow v=\sqrt{2gl(\cos \theta -\cos {\theta }_o})$

Tension at this instant in the rope is given by :
$T-mg\cos \theta =\frac{m{v}^2}{l}$

On substituting the value of $v$:
$T-mg\cos \theta =2mg(\cos \theta -\cos {\theta }_0)$
$\Rightarrow T=mg(3\cos \theta -2\cos {\theta }_0)$ ......(1)

Tension is maximum at mean position where $\theta =0^{\circ }$
As stated in the question, $T_{max}=80\text{kgf}=800\text{N}$.
$\therefore$ From equation (1)
$800=400(3-2\cos {\theta }_{max})$
$\Rightarrow \cos {\theta }_{max}=\frac{1}{2}$
$\Rightarrow {\theta }_{max}={60}^{\circ }$