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.

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet \(m\) 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.

Log in to reply

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

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

\(A,D\) are correct.simple mechanics q.

Log in to reply