The object shown is called a speed governor, a mechanical device for the regulation and control of the speed of mechanisms. The system consists of two arms of negligible mass at the end of which are attached two spheres, each of mass m. The upper end of each arm is attached to a fixed collar A. The

Question

The object shown is called a speed governor, a mechanical device for the regulation and control of the speed of mechanisms. The system consists of two arms of negligible mass at the end of which are attached two spheres, each of mass m. The upper end of each arm is attached to a fixed collar A. The system is then made to spin with a given angular speed { \omega }_{ 0 } at a set opening angle { heta }_{ 0 }. Once it is in motion, the opening angle of the governor can be varied by adjusting the position of the collar C (by the application of some force). Let heta represent the generic value of the governor opening angle. If the arms are free to rotate, that is, if no moment is applied to the system about the spin axis after the system is placed in motion, determine the expression of the angular velocity { \omega } of the system as a function of { \omega }_{ 0 }, { heta }_{ 0 }, m, d, and L, where L is the length of each arm and d is the distance of the top hinge point of each arm from the spin axis. Neglect any friction at A and C.

Accepted Answer

The FBD shown implies that the moment of the external forces about the z axis is equal to zero. Because the z axis is fixed we can then say that the angular momentum in the z direction is conserved, i.e.,

[latex]{ h }_{ Oz }(0)={ h }_{ Oz }[/latex] (1)

where [latex]{ h }_{ Oz }(0)[/latex] and [latex]{ h }_{ Oz }[/latex] denote the z component of the angular momentum of the system at the initial time and at a generic subsequent time, respectively. Due to the symmetry of the system, the system’s angular momentum is

[latex]{ h }_{ Oz }=2m(L\sin { heta } +d)\omega (L\sin { heta } +d)[/latex] (2)

where [latex]\omega[/latex] is the angular velocity of the system. Recalling that at the initial time [latex]\omega (0)[/latex]=[latex]{ \omega }_{ 0 }[/latex] and [latex] heta (0)[/latex]=[latex]{ heta }_{ 0 }[/latex], substituting Eq. (2) into Eq. (1) and solving for [latex]\omega[/latex] ,we have

[latex]\omega =\frac { { (L\sin { { heta }_{ 0 } } +d) }^{ 2 } }{ { (L\sin { { heta } } +d) }^{ 2 } } { \omega }_{ 0 }[/latex]

Question : The object shown is called a speed governor, a mechanical de...

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