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Note by Aneesh Kundu
3 months, 1 week ago

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Let \(m\) be the mass of the ring so \(2m\) is the mass of each loop.Now by symmetry its clear that the ring moves downwards so let its acceleration be \(a\) then each of the loops moves with acceleration \( w\) .But consider a loop+ring.Both must have a common acceleration along the common normal (here radius).So \(w/√2=a/√2\) so \(w=a\).The normal reaction between them be\( N\) and friction be \(f\) then we have \((N/√2-f)/2m=R\alpha =f/2m\).so \(f=N/2√2\),\(a=w=N/4√2m\).For the ring \(mg-√2N=ma\).Yields \(N=4√2mg/9\).So normal given by ground if \(F=2mg+Ncosπ/4=22mg/9\).And accn of the ring\(a=g/9\).Friction is \(f=2mg/9=2/9\).Note that we don't take into account the radial acceleration since \(t=0\) there is nt any given to the system. Spandan Senapati · 3 months, 1 week ago

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@Spandan Senapati There is going to be torque by friction about the centre of the ring right? Sumanth R Hegde · 3 months ago

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@Sumanth R Hegde I have included that.If the friction you are taking about is due to ground then I have included it.And if it's friction between both ring and hoop then the problem says to neglect that. Spandan Senapati · 3 months ago

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@Spandan Senapati Yes. I read ur comment wrong ... I read radial acceleration as angular acceleration Sumanth R Hegde · 3 months ago

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@Sumanth R Hegde Ok... Give your best in jee adv.....All the best.... Spandan Senapati · 3 months ago

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@Spandan Senapati Thx! I hope things go right as well Sumanth R Hegde · 3 months ago

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\(A,D\) are correct.simple mechanics q. Spandan Senapati · 3 months, 1 week ago

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