Four identical point masses m are joined by l | vedclass

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(B) Consider one of the point masses $m$. The distance of each mass from the center of the square is $r = \frac{l}{\sqrt{2}}$.The centripetal force re

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$m{\omega ^2}l/2$

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(B) Consider one of the point masses $m$. The distance of each mass from the center of the square is $r = \frac{l}{\sqrt{2}}$.The centripetal force required for circular motion is provided by the tension $T$ in the two strings connected to the mass.Each string makes an angle of $45^\circ$ with the radial line pointing towards the center.The component of tension from each string towards the center is $T \cos(45^\circ) = \frac{T}{\sqrt{2}}$.Since there are two such strings acting on each mass,the total radial force is $2 	imes \frac{T}{\sqrt{2}} = T\sqrt{2}$.Equating this to the centripetal force $F_c = m{\omega ^2}r$:$T\sqrt{2} = m{\omega ^2} \left( \frac{l}{\sqrt{2}} ight)$$T\sqrt{2} = \frac{m{\omega ^2}l}{\sqrt{2}}$$T = \frac{m{\omega ^2}l}{2}$

Four identical point masses $m$ are joined by light strings of length $l$ and arranged to form a square frame. The center of the table coincides with the center of the arrangement. If the arrangement rotates with a constant angular velocity $\omega$,find the tension in each string.

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