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Question

A uniform stick of length L and mass M hinged at one end is released from rest at an angle θ0 with the vertical. Show that when the angle with the vertical is θ the hinge exerts a force Fr along the stick and Ft perpendicular to the stick given by
Fr=12Mg(5cosθ3cosθ0) and Ft=14Mgsinθ

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Solution

(i) C is the centre of mass of the rod. Let ω the angular speed of rod about point O at angle θ From
conservation
of mechanical energy,MgL2(cosθcosθ0)=12(ML23)ω2
ω2=3gL(cosθcosθ0)....(i)
Now, FrMgcosθ=M(L2)ω2....(ii)
From Eqs. (i) and (ii), we get
Fr=12Mg(5cosθ3cosθ0) Hence proved.
(ii) Angular acceleration of rod at this instant,
α=τI=MgL2sinθML23
=32gsinθL
Tangential acceleration of COM,
at=(α)(L2)=34gsinθ.....(iii)
Now, Ft+Mgsinθ=Mat....(iv)
From Eqs. (iii) and (iv), we get
Ft=14Mgsinθ
Hence proved.
Here negative sign implies that direction of Ft is opposite to the component Mgsinθ
236593_219195_ans_8e94027ac66e42ec9011a5e6050fe238.png

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