The 8 kg slender bar is released from rest in the horizontal position. When it h
ID: 1852915 • Letter: T
Question
The 8 kg slender bar is released from rest in the horizontal position. When it has fallen to the position shown, what are the x and y components of force exerted on the bar by the pin support A?
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Explanation / Answer
The only force the pin provides is the one creating centripetal acceleration for the bar.
The centripetal force acting on the bar is given by m*^2*r where is the angular velocity of the bar and r is the distance from the pin to the center of mass of the bar which of course is 1m.
To find we use energy consevation. The potential energy lost must equal the KE gained.
mgh = 0.5*I*^2;
I is the moment of inertia about the pin and is given by m*L^2/3.
h is the vertical distance of fall of the center of mass and is given by 1*cos(45)=0.7 m
F= 6*m*g*h/L^2 ( from easy substitution)
= 6*8*10*0.7/4=84N
this force is directed alon the bar and towards the pin from the center of mass
so the component along the X and Y axes respectively becomes :
Fx=Fcos(45)=58.8N
Fy = Fsin(45)=58.8N
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