Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes and angle of 37° with the vertical.A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released.

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Solution

Let the length of the rod be l.
Mass of the rod be m.
Let the angular velocity of the rod be ω when it makes an angle of 37° with the vertical.

On applying the law of conservation of energy, we get:

$\frac{1}{2} I ω^{2} - 0 = m g \frac{l}{2} (\cos 37 ° - \cos 60 °) \Rightarrow \frac{1}{2} \times \frac{m l^{2} ω^{2}}{3} = m g \frac{l}{2} (\frac{4}{5} - \frac{1}{2}) \Rightarrow ω^{2} = \frac{9 g}{10 l}$

Let the angular acceleration of the rod be α when it makes an angle of 37° with the vertical.
$Using τ = I α, we get : I α = m g \frac{l}{2} \sin 37 ° \Rightarrow \frac{m l^{2}}{3} α = m g \frac{l}{2} \times \frac{3}{5} \Rightarrow α = 0.9 (\frac{g}{l})$

Force on the particle of mass dm at the tip of the rod
$F_{c} = centrifugal force = (d m) ω^{2} l = (d m) \frac{9 g}{10 l} l \Rightarrow F_{c} = 0.9 g (d m) F_{t} = tangential force = (d m) α l \Rightarrow F_{t} = 0.9 g (d m)$

So, total force on the particle of mass dm at the tip of the rod will be the resultant of F_cand F_t.

$∴ F = \sqrt{({F_{c}}^{2} + {F_{t}}^{2})} = 0.9 \sqrt{2} g (d m)$

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A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60 $^{\circ}$ and then released. Find the magnitude of the force acting on a particle of mass 'dm' at the tip of the rod, when rod makes an angle of 37 $^{\circ}$ with the vertical.