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A rod of length l and mass m rotates about end a in vertical plane as shown. Magnitude of normal reaction by the hinge at a is.

A rod of length l and mass m rotates about end a in vertical plane as shown. Magnitude of normal reaction by the hinge at a is.

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Grade:12th pass

1 Answers

aswanth nayak
100 Points
4 years ago
The initial gravitational PE of the falling rod is converted into rotational KE :
MgL/2=(1/2)Iω2.
Here I=1/3ML2 is the moment of inertia about one end of the rod and ω is the angular velocity of every point on the rod about the pivot. So
MgL=1/3ML2ω2
ω2=3g/L.
Now we have to find the centripetal force required to keep the rod in circular motion about a pivot at one end, with angular velocity ω.
For an element of the rod of length dr and mass MLdr at distance r from the pivot, the centripetal force is 
dF=(dm)rω2=rω2M/Ldr.
To get the centripetal force for the whole rod we integrate : 
F= ω2M/L[1/2r2](0 to L ) =3g/L M/L L2/2 = 3/2Mg.
The normal reaction at the hinge is the sum of the weight of the rod and the centripetal force :
N=Mg+3/2Mg=5/2Mg.

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