12u/R=ω
ye bhi galat hi hoga .............
A diwali cracker known as sudarshan chakra works on the principle of thrust. Consider such a toy the centre of which is hinged.the initial mass of the toy is M0 and the radius is R. The toy is in the shape of a spiral the turns of which are very close[it can be assumed as a disc]. The gases are ejectyed tangentially from the end of the toy with a constant velocity `u` relative to the toy. Find the angular velocity of the toy when mass remains half.
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8 Answers
1pls find the mistake in my solution,
I1ω1=I2ω2
(M0R2/2)(u/2R)=(M0R2/8)ω
ω=2u/R
39i donno yaar but the correct answer is not of what u said.
ω=4u[√2 -1]/R
1here is the method(prtty easy ppoblem)
let at any moment of time radius =x
then mass= ÏÏ€x2
then extra angu;lar momentum delivered by gas going of mass dm=dmux
this adds to the angular momentum of the disc
hence
Idw=dmux
write dm=Ï2Ï€xdx and I=mx2/2 with m=as early mentioned
put limits of x from R to R/√2
(as mass gets halved)
you will get answer as
4(√2-1)u/R
and dimensions u cant conserve angular momentum as gas is ejected tangentially