 # 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.

12 years ago

Dear student,

Take an element of length dr.

mass per unit length= m/l dr

Tension=integral mw2rdr/L=mw2r2/2L

All the best.

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Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com 12 years ago

Deat student,

The solution for this has been posted by me. Please do not post the same doubt twice..

All the best.

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Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

12 years ago

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

12 years ago

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

5 years ago
For length l mass is is m therefor For dx length mass will be m/ldx Now force observed by this element will be m/l xdx w2 Now integrate with limit 0,to l to get the answer Kushagra Madhukar
2 years ago
Dear student,

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       (Integrating)
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

Hope this helps.
Thanks and regards,
Kushagra