Explanation
Here, h=Lsinθ
For rotating body,
given that:
Moment of inertia = I
We know that,
Angular momentum L=Iω
Now,
Rotational kinetic energy = (Iω2)2
So, rotational kinetic energy = (L22I)
Hence, (L22I) is the rotational kinetic energy.
Given,
Mass of rod =M
Length of Rod =L
Distance of rods edge from center =L2
Moment of inertia of rod at center Ic=112ML2
For moment of inertia at edge apply parallel axis theorem. Ie=Ic+M(L2)2
I=112ML2+M(L2)2=13ML2
moment of inertia at edge is 13ML2
Vertical displacement, s=r(1−cosθ)
Apply kinematic equation,
V2=u2+2as=[(0.5√gr)2+2gr(1−cosθ)]
Kinetic energy = Potential Energy
12mv2=mgr(1−cosθ)
v2=2gr(1−cosθ).........(1)
Normal reaction on block
N=mgcosθ−mv2r
0=mgcosθ−mv2r
0=mgcosθ−m(0.5√gr)2+2gr(1−cosθ)r
0=mgcosθ−2.25mg+2mgcosθ
cosθ=34
θ=cos−1(34)
Hence, at angle cos−1(34) , block will leave the surface.
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