## 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 ${ L }_{ 0 }$.

## Step-by-step

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

${ ({ h }_{ Oz }) }_{ 1 }={ ({ h }_{ Oz }) }_{ 2 }$     (1)

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

${ \overrightarrow { h } }_{ O }=\overrightarrow { r } \times m\overrightarrow { \upsilon } =r{ \widehat { u } }_{ r }\times m({ \upsilon }_{ r }{ \widehat { u } }_{ r }+{ \upsilon }_{ \theta }{ \widehat { u } }_{ \theta })mr{ \upsilon }_{ \theta }\widehat { k }$     (2)

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

${ ({ h }_{ Oz }) }_{ 1 }=m{ r }_{ 1 }{ \upsilon }_{ 1 }=m{ r }_{ 0 }{ \upsilon }_{ 0 }\quad and\quad { ({ h }_{ Oz }) }_{ 2 }=m{ r }_{ 2 }{ \upsilon }_{ 2 }=m{ L }_{ 0 }{ \upsilon }_{ 2 }$     (3)

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

${ \upsilon }_{ 2 }=({ { r }_{ 0 } }/{ { L }_{ 0 } }){ \upsilon }_{ 0 }$     (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

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

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

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

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

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

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