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Grade: 11
        
Find tension at any distance x?  …..............................................................................................                                                          
9 months ago

Answers : (6)

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=Fcentripetal
T=mrw2 (r is the distance of that point from axis of rotation)
Since r=x
T=mxw2
Hope this helps
9 months ago
AJ
12 Points
							
@Agrata Singh
You cannot take r=x. x could be any distance from the end.
…..........................................................................................
9 months ago
Agrata Singh
208 Points
							
Well in that case this would be the procedure. I have attached the solution below....................
9 months ago
AJ
12 Points
							
i cant see the image. i think you need to re upload it.
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9 months 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 L
I got the answer as (Mw2x2)/2L
This is how tension depends on x
 
9 months 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=mldrω2r

So, T0dT=mlω2xlrdr (as,at r=l,T=0)

So, T=mω22l(x2l2)=mω22l(l2x2)

7 months ago
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