1i think you made some mistakes above
1) m= r dα λ and not m= 4 r dα λ
2) N cosθ +2T dα/2 = mw2r when you make assumption and not
N cosθ +2T dα = mw2r
please correct me if i am wrong.......................
An circular Iron rod of radius r is put into an inverted cone which rotates with an angular velocity 'w'. If the mass of the ring me M find the tension in the the Iron rod if it rotates with the the cone without sliding. Neglect any friction.
Given Semi angle =( theta )
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6 Answers
62draw the fbd of a small element which subtends and angle dα at the center...
The length of the element will be r.dα
Now the acceleration of the rod is w2r towards the center..
balance the forces to get mg=Nsinθ
N cos θ + 2Tsin(dα/2) = m w2r
From the above two equations...
mg cotθ +2T dα = m w2r
Also, 4r dα λ = m (where λ is the mass per unit length)
So, 4r dα λ g cotθ +2T dα = 4r dα λ w2r
T = 2r λ w2r - 2rλ g cotθ
Where λ*2πr = M
So T= (w2r - g cotθ)/π
71So the tension is the same as the force on any arbitrarily small part of the chain (Of course due to Cone). That's the idea I got it.
11for the concept just remem. how we derived speed of a trnasverse wave the same concept is applied here...
71derived speed of a trnasverse wave the same concept is applied here!
Haven't read that yet!
71@ abhishek. I don't know the answer. Basically I just ask for questions and the answers to anyone. So i can't explain whether it is matching or not.
But I think you're correct since why should be put 4 there.?