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# Help needed!

The given question has multiple correct answers.

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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. · 3 months, 1 week ago

There is going to be torque by friction about the centre of the ring right? · 3 months ago

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. · 3 months ago

Yes. I read ur comment wrong ... I read radial acceleration as angular acceleration · 3 months ago

Ok... Give your best in jee adv.....All the best.... · 3 months ago

Thx! I hope things go right as well · 3 months ago

$$A,D$$ are correct.simple mechanics q. · 3 months, 1 week ago