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Two separate cylinders of masses m(=1 kg...

Two separate cylinders of masses `m(=1 kg ) & 4m` radii `R(=10cm)` and `2R` rotating in clockwise direction with `omega_(1)=100rad //sec` and `omega_(2)=200 rad//sec` respectively. Now they are held in contact with each other as in figure. Determine their angular velocities after the slipping after the slipping between the cylinders. stops.

Text Solution

Verified by Experts

The correct Answer is:
`300 rad//sec., 150 rad//sec`.
Final direction of motions are shown by `omega_(1f) & omega _(21)`

Now,
`alpha_(1)=(omega_(f1)+epsilon_(i1))/(t),alpha_(2)=(omega_(i2)+epsilon_(f2))/(t)`
and `FR_(1)+I_(1)alpha_(2)(` torque equation of friction `)`
`FR_(2)=I_(2)alpha_(2)`
Dividing `(R_(1))/(R_(2))=(l_(1)alpha_(1))/(l_(2)alpha_(2))`
`rArr(I_(1))/(I_(2)).(omega_(f_(i))+omega_(i_(1)))/(omega_(i_(2))-omega_(f_(2)))=(R_(1))/(R_(2))`
For `pt. ` of contach when slipping stops `R_(1)omega_(f_(1))=R_(2)omega_(f_(2))`
`(mu_(1)R_(2)^(2)l_(2))/(mu_(2)R_(2)^(2)l_(2)).(omega_(f_(1))+omega_(i_(1)))/(omega_(f_(2))-(R)/(R_(2)))=(R_(1))/(R_(2))`
`rArr omega_(f_(1))=(mu_(2)R_(2)omega_(i_(2))-mu_(1)R_(1)omega_(i_(1)))/(mu_(2)R_(1)+mu_(1)R_(1))`
`=(4xx.2xx200-1xx.1xx100)/(.4xx+.1)=300r//s`
`omega_(f_(2))=(R_(1)omega_(f_(1)))/(R_(2))=(R)/(2R)xx300=150rad//sec.`
`[Ans. : 300 rad//sec., 150 rad//sec. ]`
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