Explanation
Here, h=Lsinθ
For rotating body,
given that:
Moment of inertia = $$I$$
We know that,
Angular momentum $$L= I\omega $$
Now,
Rotational kinetic energy = $$\dfrac{\left( I{{\omega }^{2}} \right)}{2}$$
So, rotational kinetic energy = $$\left( \dfrac{{{L}^{2}}}{2I} \right)$$
Hence, $$\left( \dfrac{{{L}^{2}}}{2I} \right)$$ is the rotational kinetic energy.
Given,
Mass of rod $$=M$$
Length of Rod $$=L$$
Distance of rods edge from center $$=\dfrac{L}{2}$$
Moment of inertia of rod at center $${{I}_{c}}=\dfrac{1}{12}M{{L}^{2}}$$
For moment of inertia at edge apply parallel axis theorem. $${{I}_{e}}={{I}_{c}}+M{{\left( \dfrac{L}{2} \right)}^{2}}$$
$$I=\dfrac{1}{12}M{{L}^{2}}+M{{\left( \dfrac{L}{2} \right)}^{2}}=\dfrac{1}{3}M{{L}^{2}}$$
moment of inertia at edge is $$\dfrac{1}{3}M{{L}^{2}}$$
Vertical displacement, $$s=r(1-\cos \theta )$$
Apply kinematic equation,
$${{V}^{2}}={{u}^{2}}+2as=\left[ {{\left( 0.5\sqrt{gr} \right)}^{2}}+2gr(1-\cos \theta ) \right]$$
Kinetic energy = Potential Energy
$$ \dfrac{1}{2}m{{v}^{2}}=mgr\left( 1-\cos \theta \right) $$
$$ {{v}^{2}}=2gr(1-\cos \theta )\,\,.........\,\,(1) $$
Normal reaction on block
$$ N=mg\cos \theta -\dfrac{m{{v}^{2}}}{r} $$
$$ 0=mg\cos \theta -\dfrac{m{{v}^{2}}}{r} $$
$$ 0=mg\cos \theta -\dfrac{m{{\left( 0.5\sqrt{gr} \right)}^{2}}+2gr(1-\cos \theta )}{r} $$
$$ 0=mg\cos \theta -2.25mg+2mg\cos \theta $$
$$ \cos \theta =\dfrac{3}{4} $$
$$ \theta ={{\cos }^{-1}}\left( \dfrac{3}{4} \right) $$
Hence, at angle $${{\cos }^{-1}}\left( \dfrac{3}{4} \right)$$ , block will leave the surface.
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