Mechanics> rotational...

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.

SOHAN SARANGI , 15 Years ago

Grade 12

6 Answers

SAGAR SINGH - IIT DELHI

Dear student,

Take an element of length dr.

mass per unit length= m/l dr

Tension=integral mw2rdr/L=mw2r2/2L

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Sagar Singh

B.Tech, IIT Delhi

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Last Activity: 15 Years ago

SAGAR SINGH - IIT DELHI

Deat student,

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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

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Last Activity: 15 Years ago

Sai Theja K

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

Last Activity: 15 Years ago

vikas askiitian expert

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

T = M(l-x)(l+x))w² /2l

T = M(l²-x²)w² /2l

this is the expression for tension at any point

Last Activity: 15 Years ago

Aryan

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

Last Activity: 8 Years ago

Kushagra Madhukar

Dear student,

Please find the attached solution to your problem below.

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

T = M(l-x)(l+x))w² /2l

T = M(l²-x²)w² /2l

Hope this helps.

Thanks and regards,

Kushagra

Last Activity: 5 Years ago