Question

A uniform rod of mass m and length l is free to rotate in the vertical plane find the hinge reaction

Answer 1

Answer:

The answer will be 5/2 mg

Explanation:

According to the problem the length of the rod is l and mass is m.

As the rod rotates in the vertical plane we can say that the initial potential energy is now changed to the rotated kinetic energy.

hence,

mgl/2 = (1/2)iω^2 [ m is the mass, i is the moment of inertia,l is the length, g is  the gravitational force,ω is the angular velocity for every point in the rod ]

Now the moment of inertia of one end of the rod is i = 1/3ml^2

Therefore,

mgl=1/3ml^2ω^2

ω^2=3g/l.

Now we have to calculate the force to keep the rod circular with the same angular velocity.

Hence at any position of the rod of length dr and mass mldr at distance r the  force is  

df=(dm)rω^2=rω^2m/ldr.

Now to get the force for all position in the rod we need to integrate it.

f=  ∫ω^2m/l[1/2r^2]  (0 to l ) =3g/l m/l l^2/2 = 3/2mg.

Hence the normal reaction by the hinge at A is​

c =mg+3/2mg

 =5/2mg.