Martha Pacheco

2023-03-14

A uniform rod of mass m and length l can rotate in a vertical plane about a smooth horizontal axis hinged at point H. Find the force exerted by the hinge just after the rod is released
mg sqrt7
mg sqrt10/5
mg/5
mg/sqrt5

Gosztillaw93

Taking torque equation about hinge H,
${\tau }_{net}=I\alpha \phantom{\rule{0ex}{0ex}}⇒mgcos{37}^{\circ }×\frac{l}{2}=\frac{m{l}^{2}}{3}\alpha \phantom{\rule{0ex}{0ex}}⇒\alpha =\frac{6g}{5l}$
[torque due to hinge reaction is zero]
Let the hinge force components be N1 and N2
${a}_{t}=\frac{l}{2}\alpha =\frac{3g}{5}$
Force equation perpendicular to the rod,
$mg\mathrm{cos}{37}^{\circ }-{N}_{1}=m{a}_{t}\phantom{\rule{0ex}{0ex}}⇒mg×\frac{4}{5}-{N}_{1}=m×\frac{3g}{5}\phantom{\rule{0ex}{0ex}}⇒{N}_{1}=\frac{mg}{5}$
Angular velocity (ω)=0 at the time of release and net force along the rod is zero since there is no acceleration towards the center,
${N}_{2}=mg\mathrm{sin}{37}^{\circ }=\frac{3mg}{5}$
Hinge force (N) is
$N=\sqrt{{N}_{1}^{2}+{N}_{2}^{2}}=\sqrt{\left(\frac{mg}{5}{\right)}^{2}+\left(\frac{3mg}{5}{\right)}^{2}}=\frac{mg\sqrt{10}}{5}$

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