Show FULL work please A 4-ft long (b = 2 ft) uniform slender bar weighing 12 lb
ID: 2240797 • Letter: S
Question
Show FULL work please
A 4-ft long (b = 2 ft) uniform slender bar weighing 12 lb is mounted in a right-angle frame of negligible mass. The bar and frame rotate in the vertical plane about a fixed axis at O. The mass moment of inertia of the bar about an axis through its mass center is IG = mL2/12 . If the bar is released from rest in the vertical position (theta = 0degree), determine for this position The force (magnitude and direction) that the pin O exerts on the frame. The angular acceleration alpha of the bar. The horizontal and vertical components of acceleration of the mass center of the bar.Explanation / Answer
whan the bar is vertical the reaction at point O is opposing the bar weight.
F(O) = m*g =12*0.4536*9.81 =53.4 N
direction is vertical upward.
The height in the right triangle is h=b*tan(45) =2*0.3*tan(45) =0.6 m
you can use the equation Torque(= F*h)
and T = I*epsilon but you need to compute the inertia moment of the bar about the point O.
I = M*L^2/12 +M*h^2 (This is the Steiner theorem)
I = 12*0.4536*(1.2)^2/12 +12*0.4536*0.6^2 = 2.612 kg*m^2
T = I*epsilon
epsilon = T/I = F*h/I = 53.4*0.6/2.612 =12.26 rad/sec^2
Initially the bar is vertical. The linear acceleration is tangent to the bar trajectory, it means it is vertical.
ax =0 m/s^2
ay = epsilon*h = 12.26*0.6 =7.356 m/s^2
ay is less than g (=9.8 m/s^2) since a part of the gravitational force in the center of mass is used to rotate the bar about the point O.