The symmetric mechanism shown in the figure rotates about the vertical axis at c

Question

The symmetric mechanism shown in the figure rotates about the vertical axis at constant angular speed Ohm = 3 rad/s in the direction shown. The vertical shaft is rigidly fixed to horizontally link , which is pinned at one end to link and at other to link . Two identical disks of radius R = 8 cm and mass m = 5 kg are mounted at point A and B on the ends of links and , respectively link. A spring with an unstretched length of 30 cm and spring constant k = 320 N/m connects link to link at the illustrated points. Determine the angular speed omega, and its direction, that is required to maintain an angle theta = 30 degree between the vertical axis and links and . The mass of the shafts and spring is negligible.

Explanation / Answer

This problem deals with gyroscopic precession. Let's start by finding out how much the spring stretches: Each "arm" is leaning at an angle of 30°. The spring is attached 15cm up each linkage. This means that on each linkage the spring extends 15*sin(30) beyond its base, or 7.5cm. Since each arm extends 7.5cm at the spring connection point the sping the total stretch is 15cm. The force the spring puts on each arm is then (.15)(320N/m) = 48N. =================================== Next lets assume the disks are not spinning, only rotating about the central pole at 3rad/sec. Perform an simple force equilibrium on one side of the arms. The goal of this is to determing how much force is needed to keep the "arms" at a 30° angle. Let's begin looking at the right arm, DB: You have 5kg at the end of the linkage. There is a force due to gravity with a magnitude of (5)(9.81) = 49N acting strigt down then you have a force acting to the left at a location of 15cm along the linkage from point D. You finally have a centripital force acting at the center of mass of the disk that is directed to the right. This force has a magnitude of m*v^2/r where v is equal to r*w: F_centripital = (5kg)(.15m + .1m)(3rad/s)^2 = 11.25N. Now lets do a summation of torques about point D. (F_c)(20*cos(30)) + (F_gravity)(20*sin(30)) = (F)(15*cos(30)) (11.25N)(17.3cm) + (49N)(10cm) = F (12.99cm) F = (194.625 + 490)/(12.99) F = 52.7N. So this is how much force is required to keep the linkage BD leaning outward by 30° while rotatin around the central pole at 3rad/sec. Since this is for one arm we need to multiply it by two to get the total force required. F_net = 52.7N * 2 = 105.4N If the spring could provide 105.4N of force then the disks would not have to spin, but since the spring is only providing a force of 48N the gyroscopic precession needs to account for the additional (105.4-48) = 57.4N of force. Lets again look at one arm which needs to provide 1/2 of the 57.4N since there are 2 arms. The torqe needed will be (57.4/2)*(15cos(30)) = 372.8 N-cm or 3.72N-m The equation for gyroscopic precession torque is: T = (theata)*L*sin(alpha) where (theata) is the precession angualr velocity which in this case is 3rad/sec; L is the rotational inertia of the disk; and (alpha) is the angle between (theata) and L which is 30° in this case. The equation for rotationl inertia is: L = I*w where I is the moment of inertia (mr^2/2) for a disk and w is the angualr velcotiy Your equaiton then should be: 3.72 = 3 * (5)(.08)^2(.5)* w * sin(30) 3.72 = .024w w = 155 rad/sec I hope this helps!

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