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

mg sqrt7

mg sqrt10/5

mg/5

mg/sqrt5

Gosztillaw93

Beginner2023-03-15Added 4 answers

Taking torque equation about hinge H,

${\tau}_{net}=I\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow mgcos{37}^{\circ}\times \frac{l}{2}=\frac{m{l}^{2}}{3}\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =\frac{6g}{5l}$

[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,

$mg\mathrm{cos}{37}^{\circ}-{N}_{1}=m{a}_{t}\phantom{\rule{0ex}{0ex}}\Rightarrow mg\times \frac{4}{5}-{N}_{1}=m\times \frac{3g}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {N}_{1}=\frac{mg}{5}$

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

$N=\sqrt{{N}_{1}^{2}+{N}_{2}^{2}}=\sqrt{(\frac{mg}{5}{)}^{2}+(\frac{3mg}{5}{)}^{2}}=\frac{mg\sqrt{10}}{5}$

${\tau}_{net}=I\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow mgcos{37}^{\circ}\times \frac{l}{2}=\frac{m{l}^{2}}{3}\alpha \phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =\frac{6g}{5l}$

[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,

$mg\mathrm{cos}{37}^{\circ}-{N}_{1}=m{a}_{t}\phantom{\rule{0ex}{0ex}}\Rightarrow mg\times \frac{4}{5}-{N}_{1}=m\times \frac{3g}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {N}_{1}=\frac{mg}{5}$

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

$N=\sqrt{{N}_{1}^{2}+{N}_{2}^{2}}=\sqrt{(\frac{mg}{5}{)}^{2}+(\frac{3mg}{5}{)}^{2}}=\frac{mg\sqrt{10}}{5}$