A thin uniform rod of mass M and length L is hinged an end and released from rest in the horizontal position- The tension a point located a distance L-3 from the hinged point- when the rod becomes vertical- will be-22 Mg-27-11 Mg-13-6 Mg-11-2 Mg

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Toppr Admin · Accepted Answer

2mg0-mg-xD7-0-12Iw2-mg-x2212-L2-12-mL23-w2-mgL2w-x221A-3gLT-mg-mv2r-m-g-rw2-m-g-2Lg-xD7-3g2-2m3-g-2g-2mg

A thin uniform rod of mass $M$ and length $L$ is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance $L / 3$ from the hinged point, when the rod becomes vertical, will be

$22 M g / 27$
$11 M g / 13$
$6 M g / 11$
$2 M g$

$2 m g$

$0 + m g \times 0 = \frac{1}{2} I w^{2} + m g (- \frac{L}{2})$

$\frac{1}{2} (\frac{m L^{2}}{3}) w^{2} = \frac{m g L}{2}$

$w = \sqrt{\frac{3 g}{L}}$

$T = m g + \frac{m v^{2}}{r}$

$= m [g + r w^{2}]$

$= m [g + \frac{2 L}{g} \times \frac{3 g}{2}]$

$= \frac{2 m}{3} (g + 2 g)$

$= 2 m g$

A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance L/3 from the hinged point, when the rod becomes vertical, will be22 Mg/2711 Mg/136 Mg/112 Mg

2mg0+mg×0=12Iw2+mg(−L2)12(mL23)w2=mgL2w=√3gLT=mg+mv2r=m[g+rw2]=m[g+2Lg×3g2]=2m3(g+2g)=2mg

A thin uniform rod of mass $M$ and length $L$ is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance $L / 3$ from the hinged point, when the rod becomes vertical, will be

$22 M g / 27$
$11 M g / 13$
$6 M g / 11$
$2 M g$

$2 m g$

$0 + m g \times 0 = \frac{1}{2} I w^{2} + m g (- \frac{L}{2})$

$\frac{1}{2} (\frac{m L^{2}}{3}) w^{2} = \frac{m g L}{2}$

$w = \sqrt{\frac{3 g}{L}}$

$T = m g + \frac{m v^{2}}{r}$

$= m [g + r w^{2}]$

$= m [g + \frac{2 L}{g} \times \frac{3 g}{2}]$

$= \frac{2 m}{3} (g + 2 g)$

$= 2 m g$