Answer:
Let θ be the angle the other end and the vertical from center subtends at the center.
Let λ be the mass per unit length of the chain=lm.
Gravitational potential energy of the chain=∫0θλRdθg(Rcosθ)
=lmR2gsinθ
Since R>l, θ=Rl
⟹GPE=lmR2gsinRl
Tq fr ur answer mate...
I hav done my best hope this is correct, n plzz mention if it's wrong....
c that step...hope u got...
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Answers & Comments
Answer:
Let θ be the angle the other end and the vertical from center subtends at the center.
Let λ be the mass per unit length of the chain=lm.
Gravitational potential energy of the chain=∫0θλRdθg(Rcosθ)
=lmR2gsinθ
Since R>l, θ=Rl
⟹GPE=lmR2gsinRl
Answer:
Tq fr ur answer mate...
I hav done my best hope this is correct, n plzz mention if it's wrong....
☺✌
c that step...hope u got...