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Two blocks of...

Question

Two blocks of mass $m_{1}$ = 10kg and $m_{2}$ = 5kg connected to each other by a massless inextensible string of length 0.3m are placed along a diameter of a turntable. The coefficient of friction between the table and $m_{1}$ is 0.5 while there is no friction between $m_{2}$ and the table. The table is rotating with an angular velocity of 10rad/sec about a vertical axis passing through its center. The masses are placed along the diameter of the table on either side of the center O such that $m_{1}$ is at a distance of 0.124 m from O. the masses are observed to be at rest with respect to an observer on the turntable.
(i) Calculate the frictional force on $m_{1}$
(ii) What should be the minimum angular speed of the turntable so that the masses will slip from this position?

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Solution

$m_{1} = 10 k g, m_{2} = 5 k g, l = 0.3 m, l = l + l_{1} + l_{2}, μ_{1} =$ coefficient of friction between table and $m_{1} = 0.5, w = 10 r a d / s, l_{1} = 0.124 m$
$(i)$ Centrifugal force acting on $m_{1}$ :
$F_{1} = m_{1} w^{2} l_{1}$
Centrifugal force acting on $m_{2}$
$F_{2} = m_{2} w^{2} (l - l_{1})$
The frictional force on $m_{1}$ is equal to the difference of these:
$f = | F_{2} - F_{1} | = w^{2} | m_{2} (l - l_{1}) - m_{1} l_{1} | = {(10)}^{2} | 5 \times (0.3 - 0.124) - 10 \times 0.124 | = 36 N$

$(i i)$ If the masses start to slip, $f$ should be equal to kinetic friction:
$f = | F_{2} - F_{1} | = w^{2} | m_{2} (l - l_{1}) - m_{1} l_{1} | = {(10)}^{2} | 5 \times (0.3 - 0.124) - 10 \times 0.124 | = 36 N$

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