A pendulum consisting of a small sphere of mass $m$ suspended by an inextensible and massless string of length $l$ is made to swing in a vertical plane. If the breaking strength of the string is $2mg$ , then the maximum angular amplitude of the displacement from the vertical can be ....... $^o$
Diffcult
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Let the maximum angular amplitude be $\theta .$ When the pendulum bob moves from $\mathrm{B}$ to $\mathrm{A}$, the decrease in potential energy $=$ the increase in kinetic energy at $A$
${\mathop{\rm mg}\nolimits} ({\rm{PA}}) = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}{\rm{;mgh}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}$
$v^{2}=2 g(P A)=2 g(O A-O P)$
${v^2} = 2g(l - l\cos \theta )$
${T-m g=\frac{m v^{2}}{l}} $
${T-m g=\frac{m}{l} \cdot 2 g((1-\cos \theta)}$
${T-m g=2 m g(1-\cos \theta)}$ ......$(i)$
At $A, \quad T=T_{\max }=2 m g$
On putting this value of $\mathrm{T}$ in eq. $(i).$ we get
${2 m g-m g=2 m g(1-\cos \theta)} $
or ${1-\cos \theta=\frac{1}{2}}$
$ \cos \theta =\frac{1}{2} $
$ \theta =60^{\circ} $
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