39ya a perfect solution.
A weightless rod with a length `l` is rigidly attached to a weightless pulley with radius `r`. The end of the rod carries a mass `m` .A string to whose free end a mass `M` is fastened is passed over the pulley . in what condition will the motion of the system be oscillatory if the angle `α` between the rod and the vertical is zero at the initial moment.
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66If the mass M went down by x, then we have by conservation of energy:
Mgx = mg\ell(1-\cos\alpha)+ \dfrac{1}{2}MV^2+\dfrac{1}{2}mv^2
where V and v are respectively the speed of M and m. If \omega be the angular speed of the pulley, then V=\omega r, and v=\omega \ell. So the previous equation takes the form
\dfrac{\omega^2}{2}(m\ell^2+Mr^2)=Mgr\alpha -mg\ell(1-\cos\alpha)
Solving for \omega, we get
\omega =\sqrt{\dfrac{2\left(Mgr\alpha-2mg\ell\sin^2\dfrac{\alpha}{2}\right)}{m\ell^2+Mr^2}}
The system will undergo oscillatory motion, provided the angular speed becomes zero at some value of \alpha. That will happen when Mgr\alpha = 2mg\ell\sin^2\dfrac{\alpha}{2}. Or setting \dfrac{Mr}{m\ell}=k, whenk\,\dfrac{\alpha}{2}=\sin^2\dfrac{\alpha}{2}.
For each k, there exists a maximum value of \alpha, for which oscillations are possible.
Now this is a transcendental equation and is not solved easily. However, by graphical or iterative means, we can obtain the solution. We can look at the graphs of two curves:
y=k\dfrac{\alpha}{2} and y=\sin^2\dfrac{\alpha}{2}
The maximum value of k will be that one where the first graph is tangent to the second one. It is found that k_\mathrm{max}\approx .73, and the corresponding value of \alpha \approx 133^\circ
Hence, oscillatory motion is possible only if k=\dfrac{Mr}{m\ell}\leq k_\mathrm{max}.
