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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.

```
9 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.
Win exciting gifts by                                                answering the questions  on            Discussion        Forum.    So      help         discuss        any                query   on        askiitians  forum   and        become  an       Elite            Expert    League            askiitian.
Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com ```
9 years ago
```							Deat student,
The solution for this has been posted by me. Please do not post the same doubt twice..

All the best.
Win exciting gifts by                                                answering the questions  on            Discussion        Forum.    So      help         discuss        any                query   on        askiitians  forum   and        become  an       Elite            Expert    League            askiitian.
Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com
```
9 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
```
9 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
```
9 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
```
2 years ago 605 Points
```							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)/2soT = M(l-x)(l+x))w2 /2l   T = M(l2-x2)w2 /2l Hope this helps.Thanks and regards,Kushagra
```
4 months ago
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