# Help needed!

The given question has multiple correct answers.

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Note by Aneesh Kundu
1 year, 2 months ago

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## Comments

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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.

- 1 year, 2 months ago

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

- 1 year, 1 month 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.

- 1 year, 1 month ago

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Yes. I read ur comment wrong ... I read radial acceleration as angular acceleration

- 1 year, 1 month ago

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Ok... Give your best in jee adv.....All the best....

- 1 year, 1 month ago

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Thx! I hope things go right as well

- 1 year, 1 month ago

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

- 1 year, 2 months ago

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