A uniform rod of length 2l and mass m is suspended from one end by inextensible string and other end lies on smooth ground. The angle made by rod with vertical is θ=sin−113. If N1 and N2 represents the contact force from ground on rod just before and just after cutting string then find the ratio of N1/N2.

N12l=mgl N1=mg2 Mg−N2=macm N2l2sinθ=ml212α acm=l2α sinθ Mg−N2m= 312N2sin2θ2m mg−N2=3N2 N2=mg4

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A uniform rod of length 2 $l$ and mass m is suspended from one end by inextensible string and other end lies on smooth ground. The angle made by rod with vertical is $θ = \sin^{- 1} (\frac{1}{\sqrt{3}})$ . If $N_{1}$ and $N_{2}$ represents the contact force from ground on rod just before and just after cutting string then find the ratio of $N_{1} / N_{2} .$

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Detailed Solution

$N_{1} 2 l = m g l N_{1} = \frac{m g}{2} M g - N_{2} = m a_{c m} N_{2} \frac{l}{2} \sin θ = \frac{m l^{2}}{12} α a_{c m} = \frac{l}{2} α \sin θ \frac{M g - N_{2}}{m} = \frac{^{3} 12 N_{2} \sin^{2} θ}{2 m} m g - N_{2} = 3 N_{2} N_{2} = \frac{m g}{4}$