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A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity w about a vertical axis passing through one end.Find the tension in the rod at a distance x from the axis.
Dear student,
mass per unit length= m/l dr
Tension=integral mw2rdr/L=mw2r2/2L
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Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com
Deat student,
The solution for this has been posted by me. Please do not post the same doubt twice..
Askiitians Expert
consider a small element dx at a distance x from axis of rod
the centripetol force required is given by tension so,
dT=(dm)(x)(w2
dm=(m/l)dx
integrate it over the limits 0 to X
let AB is a rod of length l ,mass M & this rod is rotating about end A....
now , mark a point c anywhere on the rod , distance bw end A & C is x ...
now the rod is divided into two parts , AC & CB .......
tension of rod at point C is (T)= dmw2 r (dm is mass of part BC = m(l-x)/l )
=M(l-x)w2r/l
this r is the distance of center of mass of BC from end A so r = (x+l)/2
so
T = M(l-x)(l+x))w2 /2l
T = M(l2-x2)w2 /2l
this is the expression for tension at any point
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