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There are two blocks of masses `m_(1)` and `m_(2) , m_(1)` is placed on `m_(2)` on a table which is rotating with an angular velocity `omega` about the vertical axis . The coefficent of friction between the block is `mu_(1)` and between `m_(2)` and table is `mu_(2) (mu_(1) lt mu_(2))` if block are placed at distance R from the axis of ratation , for relative sliding between the surface in contact , find the
a. friction force at the contacting surface
b. maximum angular speed `omega`

Text Solution

Verified by Experts

(a) `f_(1) - f_(1) = m_(2) omega^(2) R`
`f_(2) - f_(1) = m_(2) omega^(2) R`
`f_(2) = f_(1) + m_(2) omega^(2) R = (m_(1) +m_(2)) omega^(2) R`
If there is no sliding between `m_(1) andm)=_(2)` therefore

`f_(1) le f_(1 max)`
`m_(1) omega^(2) R le mu_(1) m_(1) g`
`rArr omega le sqrt(mu_(1) g)/(R )`
If there is no sliding between `m_(2)` and ground
`f_(2) le mu_(1) N_(2)`
`(m_(1) + m_(2)) omega ^(2)R le mu_(2) (m_(1) + m_(2))g`
`omega le sqrt((mu_(2)g)/(R))`
As `mu_(1) lt mu_(2)` hence maximum angular velocity will be `sqrt((mu_(1)g)/(R))`
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