a weightless rod oflength l with a small load of mass m at the end is hinged at a point as shown in the figure and occupies a strictly vertical position , touching a body of mass M. A light jerk sets the system in motion . For what mass ratio will the rod form a angle 30 with the horizontal at the moment of seperation from the body.Ans. 4

When the rod makes an angle θ with the horizontal, let the velocity of the block M be V while its acceleration be aM. At this instant, let the angular velocity of the rod about the hinge A be ω.Since the bob is in contact with the block, the horizontal velocity of the bob must be V. Therefore\omega \ell\sin\theta =VNext, energy conservation givesmg\ell(1-\sin\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}m\ell^2\omega^2Eliminating V givesmg\ell(1-\sin\theta)=\dfrac{1}{2}M\omega^2\ell^2\sin^2\theta+\dfrac{1}{2}m\ell^2\omega^2=\dfrac{1}{2}\omega^2\ell^2(m+M\sin^2\theta)which gives\omega^2=\dfrac{2mg(1-\sin\theta)}{\ell(m+M\sin^2\theta)}Differentiate w.r.t. t and use the fact that \omega=-\dfrac{\mathrm d\theta}{\mathrm dt}, we get the angular acceleration \dfrac{\mathrm d\omega}{\mathrm dt}=\dfrac{g}{\ell}\cdot\dfrac{m\cos\theta(m+M\sin\theta(2-\sin\theta))}{(m+M\sin^2\theta)^2}Hence the tangential acceleration of the bob isa​t​​ =g⋅​(m+Msin​2​​ θ)​2​​ ​​mcosθ(m+Msinθ(2−sinθ))​​ Also, the acceleration aM of the block isa_M=\dfrac{\mathrm dV}{\mathrm dt}=\ell\left(\sin\theta \dfrac{\mathrm d\omega}{\mathrm dt}+\omega \cos\theta\dfrac{\mathrm d\theta}{\mathrm dt}\right)=\ell\left(\sin\theta \dfrac{\mathrm d\omega}{\mathrm dt}-\omega^2 \cos\theta\right)Solving for the tangential acceleration at in terms of aM and V we get\boxed{a_t = \dfrac{a_M\sin\theta + (V^2/\ell)\cos\theta}{\sin^2\theta}}as the answer to the first part.