Grade 12th passMechanics

Rod length l and mass m rotates about end A in vertical plane as shown.magnitude of normal reaction by the hinge at A?

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Tannu

6 Years agoGrade 12th pass

1 Answer

Vikas TU

6 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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Rod length l and mass m rotates about end A in vertical plane as shown.magnitude of normal reaction by the hinge at A?

Tannu

1 Answer

Vikas TU

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