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#Mechanics

Note by Aneesh Kundu
4 years, 1 month ago

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Comments

Let mm be the mass of the ring so 2m2m is the mass of each loop.Now by symmetry its clear that the ring moves downwards so let its acceleration be aa then each of the loops moves with acceleration w w .But consider a loop+ring.Both must have a common acceleration along the common normal (here radius).So w/2=a/2w/√2=a/√2 so w=aw=a.The normal reaction between them beN N and friction be ff then we have (N/2f)/2m=Rα=f/2m(N/√2-f)/2m=R\alpha =f/2m.so f=N/22f=N/2√2,a=w=N/42ma=w=N/4√2m.For the ring mg2N=mamg-√2N=ma.Yields N=42mg/9N=4√2mg/9.So normal given by ground if F=2mg+Ncosπ/4=22mg/9F=2mg+Ncosπ/4=22mg/9.And accn of the ringa=g/9a=g/9.Friction is f=2mg/9=2/9f=2mg/9=2/9.Note that we don't take into account the radial acceleration since t=0t=0 there is nt any given to the system.

Spandan Senapati - 4 years, 1 month ago

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There is going to be torque by friction about the centre of the ring right?

Sumanth R Hegde - 4 years ago

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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 - 4 years ago

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@Spandan Senapati Yes. I read ur comment wrong ... I read radial acceleration as angular acceleration

Sumanth R Hegde - 4 years ago

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@Sumanth R Hegde Ok... Give your best in jee adv.....All the best....

Spandan Senapati - 4 years ago

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@Spandan Senapati Thx! I hope things go right as well

Sumanth R Hegde - 4 years ago

A,DA,D are correct.simple mechanics q.

Spandan Senapati - 4 years, 1 month ago
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