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Homework 3 Solution Problem 1: A speed governor consists of a block of mass m that slides within a smooth groove in a turntable that rotates about its center point at angular speed Ξ©. The identical opposing springs, whose stiffness is k, are precompressed. Consequently the springs maintain their contact with the block regardless of the displacement s. The system lies in the horizontal plane. Derive an expression for the normal force exerted by the groove wall on the block and the differential equation governing s as a function of time in the case where Ξ© is an arbitrary function of time. Then determine the natural frequency of vibratory motion of the block within the groove for the case in which Ξ© is constant, and explain how that result can be used to monitor when Ξ© exceeds a critical value. Solution: Attach a moving coordinate system to the disk, with the moving x axis aligned with the groove. The origin of the moving coordinate system is at the center of the disk. To solve this problem we must first find the total acceleration of the block. We will write this acceleration in the moving coordinate system so that the normal force will be easy to identify. There is only one angular velocity in this problem πΜ1 = πΜ πΜ 1 = Ξ©Μ π1 = Ξ© π1 = π π = Ξ©πΜ π = πΜ 1 πΜ1 + π1 (π1 × πΜ1 ) = Ξ©ΜπΜ The vector from the origin to the block is π π¬ = {β } 0 And the velocity and acceleration of the block relative to the moving coordinate system is π Μ π¬Μ π = {0} 0 π Μ π¬Μ π = {0} 0 Since the origin of the moving coordinate system does not translate, we have π«Μ0 = π Thus, the total acceleration is π π 0 0 0 0 π Μ π Μ π = {0} + { 0 } × {β} + 2 { 0 } × {0} + { 0 } × ({ 0 } × {β}) 0 0 Ξ© Ξ© Ξ© Ξ©Μ 0 0 π 0 ββ π Μ 2 β Μ π = {0} + Ξ© { π } + {2π Μ Ξ©} β Ξ© { } 0 0 0 0 π Μ β Ξ©Μβ β Ξ©2 π π = {Ξ©Μπ + 2π Μ Ξ© β Ξ©2 β} 0 Use the πΜ component to find the normal force on the sidewall of the groove: πΉπ = π (Ξ©Μ π + 2π Μ Ξ© β Ξ©2 β) The equation of motion along the groove is β πΉπ‘ = β2ππ = π (π Μ β Ξ©Μ β β Ξ©2 π ) Rearrange this into standard form 2 ππ Μ + (2π β πΞ© ) π = πΞ©Μ β If the angular velocity of the disk is constant, this has the form of a mass/spring system ππ₯Μ + ππ₯ = 0 And the natural frequency is 2π β πΞ©2 2π β ππ = = β β Ξ©2 π π If Ξ©2 > 2π π then the natural frequency will be imaginary, which implies an unstable system. In reality, the mass will cease to vibrate back and forth, and will remain pinned against one end of the groove. If a limit switch were placed at the end of the groove, it could be used to indicate when the disk had reached a critical speed. Problem 2: The disk spins about its axis CD at πΜ as the system rotates about the vertical axis at πΜ. The angle of elevation of the arm supporting the disk is Ξ². The rates πΜ , πΜ , π½Μ are constant. The length from A to B and from B to the axis of rotation of the disk is L, and the radius of the disk is R. Determine the velocity and acceleration of point E, which is the lowest point on the perimeter of the disk. Solution: There are several ways we could go about solving this problem. In class, we have typically attached the moving coordinate system to the object with the most βmotionβ β in this case, the disk. We will try a slightly different approach this time by attaching the moving coordinate system to the outer arm, with its origin at B. The moving x axis is aligned with the arm and the moving z axis is parallel to the axis of rotation of the disk. To get from the fixed coordinate system to the moving coordinate system requires one rotation about the Y axis. cos π½ ππ¦ = [ 0 sin π½ cos π½ πΌΜπ₯π¦π§ = [ 0 sin π½ 0 1 0 β sin π½ 1 cos π½ 0 ] {0} = { 0 } cos π½ 0 sin π½ 0 β sin π½ 1 0 ] 0 cos π½ cos π½ Μπ₯π¦π§ = [ 0 πΎ sin π½ 0 1 0 β sin π½ 0 β sin π½ 0 ] {0} = { 0 } cos π½ 1 cos π½ There are two angular velocities associated with this system π1 = πΜ π2 = π½Μ Μ πΜ1 = πΎ πΜ2 = πΜ πΜ 1 = 0 πΜ 2 = 0 The total angular velocity of the outer arm is βπΜ sin π½ π = { π½Μ } πΜ cos π½ The angular acceleration of the outer arm is π = πΜ 1 πΜ1 + π1 (π1 × πΜ1 ) + πΜ 2 πΜ2 + π2 (π2 × πΜ2 ) βπ½Μ πΜ cos π½ π={ } 0 βπ½Μ πΜ sin π½ The vector from the origin of the moving coordinate system to the point E is π1 = 0 π2 = π πΏ+π π¬={ 0 } 0 From the viewpoint of an observer in the moving coordinate system, the velocity of point E is purely tangential arising from the rotation of the disk. 0 π¬Μ π = {π πΜ } 0 and the relative acceleration of point E is entirely centripetal βπ πΜ 2 π¬Μ π = { 0 } 0 The acceleration at point B is also purely centripetal βπΏπΜ 2 cos π½ π«Μ0 = βπΏπ πΌ = { } 0 βπΏπΜ 2 sin π½ Μ2 Μ Some of the necessary cross-products are β(πΏ + π )π½Μ 2 β (πΏ + π )πΜ 2 cos2 π½ π × (π × π¬) = { } β(πΏ + π )π½ΜπΜ sin π½ 2 Μ β(πΏ + π )π cos π½ sin π½ 0 π × π¬ = {β(πΏ + π )π½Μ πΜ sin π½} 0 β2π πΜ πΜ cos π½ 2π × π¬Μ π = { } 0 β2π πΜ πΜ sin π½ Thus, the total acceleration at point E is βπ πΜ 2 β πΏπΜ 2 cos π½ β (πΏ + π )π½Μ 2 β (πΏ + π )πΜ 2 cos2 π½ β 2π πΜ πΜ cos π½ ππΈ = { } β2(πΏ + π )π½Μ πΜ sin π½ 2 2 Μ Μ Μ βπΏπ sin π½ β (πΏ + π )π cos π½ sin π½ β 2π πΜ π sin π½ Problem 3: Collar C slides relative to the curved rod at a constant speed u. The rotation rate about bearing axis AB is constant at Ξ©. Determine the acceleration of the collar in terms of the angle ΞΈ. Also derive expressions for the dynamic forces exerted on the collar by the rod and the tangential force required to hold u constant. Gravity may be neglected. Solution: Attach a moving coordinate system to the center of the ring with the βy axis aligned with the collar, as shown in the figure above. There is one rotation needed to switch from the fixed XYZ system to the moving xyz system cos π π π§ = [β sin π 0 sin π cos π 0 0 0] 1 cos π πΌΜπ₯π¦π§ = {β sin π } 0 sin π π½Μπ₯π¦π§ = {cos π } 0 There are two angular velocities associated with this system π1 = Ξ© π2 = πΜ πΜ1 = πΌΜ πΜ2 = πΜ πΜ 1 = 0 πΜ 2 = 0 But the rate of change of ΞΈ can be written in terms of the velocity of the collar πΜ = π’ π Thus, the total angular velocity is Ξ© cos π π = {βΞ© sin π} π’/π π1 = 0 π2 = π And the angular acceleration is π=β π’Ξ© sin π {cos π} π 0 The vector from the origin of the moving coordinate system to the collar is 0 π¬ = {βπ } 0 π¬Μ π = π π¬Μ π = π The acceleration of the origin of the moving coordinate system is entirely centripetal sin π π«Μ0 = βπ Ξ© π½ = βπ Ξ© {cos π} 0 2Μ 2 A few of the necessary cross products are π Ξ©2 sin π cos π π × (π × π¬) = {π Ξ©2 cos 2 π + π’2 /π } π’Ξ© sin π 0 π×π¬= { 0 } π’Ξ© sin π And the total acceleration of the collar is βπ Ξ©2 sin π (1 β cos π) π = {βπ Ξ©2 cos π (1 β cos π) + π’2 /π } 2π’Ξ© sin π The tangential force is the mass multiplied by the x component of the acceleration. The other two components (multiplied by mass) give the force that is exerted by the ring on the collar.