A simple model of orbital motion under a central force can be constructed by considering the motion of a disk D sliding with no friction over a horizontal surface while connected to a fixed point O by a linear elastic cord of constant k and unstretched length L0. Let the mass of D be m = 0.45 kg

Question

A simple model of orbital motion under a central force can be constructed by considering the motion of a disk D sliding with no friction over a horizontal surface while connected to a fixed point O by a linear elastic cord of constant k and unstretched length { L }_{ 0 }. Let the mass of D be m = 0.45 kg and { L }_{ 0 } = 1 m. Suppose that when D is at its maximum distance from O, this distance is { r }_{ 0 } = 1.75m and the corresponding speed of D is { \upsilon }_{ 0 } = 4m/s. Determine the elastic cord constant k such that the minimum distance between D and O is equal to the unstretched length { L }_{ 0 }.

Accepted Answer

The FBD to the right implies that the angular momentum of D about O in the z direction is conserved, i.e.,

[latex]{ ({ h }_{ Oz }) }_{ 1 }={ ({ h }_{ Oz }) }_{ 2 }[/latex] (1)

where the subscripts 1 and 2 denote the instants in which D achieves the maximum and minimum distance from O, respectively. Also,

[latex]{ \overrightarrow { h } }_{ O }=\overrightarrow { r } imes m\overrightarrow { \upsilon } =r{ \widehat { u } }_{ r } imes m({ \upsilon }_{ r }{ \widehat { u } }_{ r }+{ \upsilon }_{ heta }{ \widehat { u } }_{ heta })mr{ \upsilon }_{ heta }\widehat { k }[/latex] (2)

When D is at the maximum or minimum distance from O, [latex]{ \upsilon }_{ r }[/latex] is equal to zero (i.e., D is no longer increasing or decreasing its distance from O). Hence, given that [latex]{ \upsilon }_{ heta } > 0[/latex], we have

[latex]{ ({ h }_{ Oz }) }_{ 1 }=m{ r }_{ 1 }{ \upsilon }_{ 1 }=m{ r }_{ 0 }{ \upsilon }_{ 0 }\quad and\quad { ({ h }_{ Oz }) }_{ 2 }=m{ r }_{ 2 }{ \upsilon }_{ 2 }=m{ L }_{ 0 }{ \upsilon }_{ 2 }[/latex] (3)

where [latex]{ \upsilon }_{ 1 }[/latex] and { \upsilon }_{ 2 }[/latex] are the speeds of D at the instants 1 and 2, respectively, and where we have enforced the fact that [latex]{ r }_{ 2 } = { L }_{ 0 }[/latex]. Substituting Eqs. (3) into Eq. (1) and solving for [latex]{ \upsilon }_{ 2 }[/latex] we have

[latex]{ \upsilon }_{ 2 }=({ { r }_{ 0 } }/{ { L }_{ 0 } }){ \upsilon }_{ 0 }[/latex] (4)

Since the only force doing work on D is the force { F }_{ c }due to due to the elastic cord, applying the work-energy principle, we have

T1 C V1 = T2 C V2 (5)

where

[latex]{ T }_{ 1 }=\frac { 1 }{ 2 } m{ \upsilon }_{ 0 }^{ 2 },{ V }_{ 1 }=\frac { 1 }{ 2 } k{ ({ r }_{ 0 }-{ L }_{ 0 }) }^{ 2 },{ T }_{ 2 }=\frac { 1 }{ 2 } m{ ({ r }_{ 0 }-{ L }_{ 0 }) }{ \upsilon }_{ 0 }^{ 2 },{ V }_{ 2 }=0[/latex] (6)

where we have accounted for the fact that at [latex]{ r }_{ 2 }={ L }_{ 0 }[/latex] and the elastic cord is unstretched. Substituting Eq. (6) into Eq. (5) we have

[latex]{ T }_{ 2 }=\frac { 1 }{ 2 } m{ \upsilon }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } { ({ r }_{ 0 }-{ L }_{ 0 }) }^{ 2 }{ \upsilon }_{ 0 }^{ 2 }\quad[/latex] (7)

where k is the elastic spring constant of the elastic cord. Solving Eq. (7) for k we have

[latex]k=\frac { m{ ({ r }_{ 0 }+{ L }_{ 0 }) }^{ 2 }{ \upsilon }_{ 0 }^{ 2 } }{ { { L }_{ 0 }^{ 2 }({ r }_{ 0 }-{ L }_{ 0 }) } }[/latex]=26.4N/m

where we have used the following numerical data: m = 04.5 kg, [latex]{ { r }_{ 0 } }[/latex] = 1.75 m, [latex]{ L }_{ 0 }[/latex] = 1 m, and [latex]{ \upsilon }_{ 0 }[/latex] = 4m/s.

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