Two blocks of mass `m_(1)=15` kg and `m_(2)=10kg` connected to each other by a massless inextensible string of length 0.4 m are placed along a diameter of turn table rotating with an angular velocity of `15rads^(-1)` about a vertical axis passing through its centre O. Mass `m_(1)` is at a distance of 0.225 from O. The coefficient of friction between the table and `m_(1)` is 0.6 while there is no friction between `m_(2)` and the table. The masses are observed to be at rest with respect to an observer on the turn table. Calculate the frictional force in newton on `m_(1)`.

Answer

`365.62`

Explanation

`m_(1)=15kg,m_(2)=10kg,omega=15` rad/s r=0.4m
`r_(1)=0.225m`
`:.r_(2)=r_(1)-r=0.4-0.225=0`
`.175`
Now, for masses `m_(1)` and `_m_(2)`
`m_(2)r_(2)omega^(2)ltm_(1)r_(1)omega^(2)`
and friction on `m_(1)` will be inward (towards centre)
`:.f+m_(2)r_(2)omega^(2)=m_(1)r_(1)omega^(2)`
`f=m_(1)r_(1)omega^(2)-m_(2)r_(2)omega^(2)`
`=(m_(1)r_(1)-m_(2)r_(2))omega^(2)`
`=(15xx0.225-10xx0.175)xx(15)^(2)`
`:.f=365.625N`
`f=365.62N`
(Rouonding of of 2 decimal places)