Two block tied with a massless string of length `3m` are placed on a rotating table as shown. The axis of rotation is `1m` from `1kg` mass and `2m` from `2kg` mass. The angular speed `omega=4rad//s` . Ground below `2kg` block is smooth and below `1kg` block is rough. `(g=10m//s^(2))`
(a) Find tension in the string, force of friction on `1kg` block and its direction.
(b) If coefficient of friction between `1kg` block and groung is `mu=0.8`. Find maximum angular speed so that neither of the blocksd slips.
(c) If maximum tension in the string can be `100N` , then maximum angular speed so that neither of the blocks slips.

Answer

A, B, C, D

Explanation

`2KgrarrT`
`T=m_(2)r_(2)omega^(2)=(2)(2)(4)^(2)=64N`
Centripetal force required to `1Kg` block is

`F_(c)=m_(1)r_(1)omega^(2)=(1)(1)(4)^(2)=16N`
But available tensions is `64N` . So, the extra force of `48N` is balance by friction acting radially outward from the centre.
(b) `f_(max)=mu_(1)g=0.8xx1xx10=8N`

`T=m_(2)r_(2)omega^(2)=(2)(2)omega^(2)=4omega^(4)`
`T-8=m_(1)r_(1)omega^(2)=(1)(1)(omega^(2))=omega^(2)`
Solving Eqs. `(i)` and `(ii)` we get,
`omega=1.63rad//s`
(c) `T=m_(2)r_(2)omega^(2)`
`2KgrarrT`
`rarrm_(2)r_(2)omega^(2)`
:. `100=(2)(2)omega^(2)`
or `omega=5rad//s`