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

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Gosztillaw93 · Accepted Answer

Taking torque equation about hinge H,            τ              n      e      t        =  I      α      ⇒  m  g  c  o  s      37    ∘    ×      l    2    =            m              l        2              3        α      ⇒      α    =            6      g              5      l       [torque due to hinge reaction is zero] Let the hinge force components be N1 and N2       a    t    =      l    2        α    =            3      g        5   Force equation perpendicular to the rod,  m  g  cos  ⁡      37    ∘    −      N    1    =  m      a    t      ⇒  m  g  ×      4    5    −      N    1    =  m  ×            3      g        5      ⇒      N    1    =            m      g        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    =  m  g  sin  ⁡      37    ∘    =            3      m      g        5  Hinge force (N) is  N  =            N      1      2        +          N      2      2        =      (                  m        g            5              )      2        +    (                  3        m        g            5              )      2        =            m      g              10              5

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

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