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The 4kg block in the figure is attached to the vertical rod by means of two strings. When the system rotates about the axis of the rod, the two strings are extended as indicated in figure. Angular velocity of system in order that the tension in upper string is 60 N is $ω$ . The value of $ω^{2}$ is )(g=10 m/s²)

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Detailed Solution

The forces acting on block P of mass is 4kg are shown in the figure $θ$ is the angle made by strings for equilibrium in the vertical direction
$\begin{array}{l} T_{1} \cos θ = T_{2} \cos θ + m g \\ (T_{1} - T_{2}) \cos θ = m g \\ \cos θ = \frac{1}{1.25} = \frac{4}{5} [∵ \cos θ = \frac{O A}{A P} = \frac{1}{1.25}] \\ ∴ T_{1} - T_{2} = \frac{m g}{\cos θ} = \frac{5 m g}{4} \end{array}$
$T_{1} \sin θ + T_{2} \sin θ = m r ω^{2}$

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The 4kg block in the figure is attached to the vertical rod by means of two strings. When the system rotates about the axis of the rod, the two strings are extended as indicated in figure. Angular velocity of system in order that the tension in upper string is 60 N is ω. The value of ω2 is )(g=10 m/s2)