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`        Find tension at any distance x?  …..............................................................................................                                                          `
one year ago

Agrata Singh
208 Points
```							Whenever a body moves in a circular path it can have two types of acceleration. First is radia accn which is provided by the centripetal force and is responsible for changing the direction of motion of particle thus moving it in a circle. Second is the tangential accn which is responsible for the change in magnitude of velocity and angular velocity.If a body is rotating with uniform angular velocity, it has no tangential acceleration. The only acceleration is radial due to centripetal force. Now this centripetal force is not a new force. It is our old well known forces like tension, gravitational forces, etc. Here in your question the centripetal force is provided by the tension. Therefore tension at any point is equal to the centripetal force on that particle T=FcentripetalT=mrw2 (r is the distance of that point from axis of rotation)Since r=xT=mxw2Hope this helps
```
one year ago
AJ
12 Points
```							@Agrata SinghYou cannot take r=x. x could be any distance from the end.…..........................................................................................
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one year ago
Agrata Singh
208 Points
```							Well in that case this would be the procedure. I have attached the solution below....................
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one year ago
AJ
12 Points
```							i cant see the image. i think you need to re upload it.…............................................................................
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one year ago
Agrata Singh
208 Points
```							See if this link works. Elsewise I am really sorry bcus it's quite impossible to type all the integrals involved. I ll tell you the method of approach though.I took a small differential element dm of length dx at a distance of x from axis of rotation dm=(M/L)dx where M is mass of rod and L is the length dT=dm.x.w2(centrifugal force)I put in the value of dm and integrated it from limit 0 to LI got the answer as (Mw2x2)/2LThis is how tension depends on x
```
one year ago
Ojas
20 Points
```							Considering a small portion of dr in the rod at a distance r from the axis of the rod.So,mass of this portion will be dm=mldr (as uniform rod is mentioned)Now,tension on that part will be the Centrifugal force acting on it, i.e dT=−dmω2r (because,tension is directed away from the centre whereas,r is being counted towards the centre,if you solve it considering Centripetal force,then the force will be positive but limit will be counted from r to l)Or, dT=−mldrω2rSo, ∫T0dT=−mlω2∫xlrdr (as,at r=l,T=0)So, T=−mω22l(x2−l2)=mω22l(l2−x2)
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions