Question 6.TC.1B: The Parametric Swing A child builds up the motion of a swing......
The Parametric Swing
A child builds up the motion of a swing by standing and squatting. The trajectory followed by the center of mass of the child is illustrated in Fig. 6 -2. Let r_{u} be the radial distance from the swing pivot to the child’s center of mass when the child is standing, while r_{d} is the radial distance from the swing pivot to the child’s center of mass when the child is squatting. Let the ratio of r_{d} to r_{u} be2^{\frac{1}{10}}= 1. 072, that is the child moves its center of mass by roughly 7% compared to its average radial distance from the swing pivot.
To keep the analysis simple it is assumed that the swing be mass-less, the swing amplitude is sufficiently small and that the mass of the child resides at its center of mass. It is also assumed that the transitions from squatting to standing (the A to B and the E to F transitions) are fast compared to the swing cycle and can be taken to be instantaneous. It is similarly assumed that the squatting transitions (the C to D and the G to H transitions) can also be regarded as occurring instantaneously.
How many cycles of this maneuver does it take for the child to build up the amplitude (or the maximum angular velocity) of the swing by a factor of two?
(1) The conservation of angular momentum (CAM) from A to B, C to D, E to F and G to H.
{ L}=I{\theta}=m{r}{^{2}}\,\theta, (1)
m = mass of the child,
r = distance of the child’s center of mass to the swing’s pivot P,
θ = the swing’s angular velocity with respect to P.
A to B:
Let \theta_{d} and \theta_{u} are the angular velocity at point A and B respectively,
then according to CAM,
L_{A}=m r_{d}^{2}\,\theta_{d}=L_{B}=m r_{u}^{2}\,\theta_{u}, (2)
so that
\theta_{d}=\frac{r_{u}^{2}}{r_{d}^{2}}\,\theta_{u}\,, (3)
hence each time the swing repeat moving upward (A to B or E to F) its angular speed increases by factor of\left({\frac{r_{d}}{r_{u}}}\right)^{2}.
(2) The Conservation of Mechanical Energy (from B to C)
E_{B}=E_{C}=K+V=\frac{1}{2}m r_{u}^{2}\theta_{B}^{2}-m g r_{u}(1-\cos\theta). (4)
The change of the potential energy (from B to C) is the same as the rotation energy at point B,
m g r_{u}(1-\cos\theta)={\frac{1}{2}}m r_{u}^{2}\,\theta_{u}^{2}. (5)
Using the similar method, we could get the following equation for the transition from D to E,
m g r_{d}(1-\cos\theta)={\frac{1}{2}}m r_{d}^{2}\,\theta_{d}^{2}.From equations (3), (5) and (6) we have
\frac{r_{u}}{r_{d}}\,=\,\left(\frac{r_{u}}{r_{d}}\right)^{2}\left(\frac{\theta_{u}}{\theta_{d^{\prime}}}\right)^{2}\rightarrow\frac{\theta_{d^{\prime}}}{\theta_{u}}\,=\sqrt{\frac{r_{u}}{r_{d}}}\,. (7)
For half a cycle we have \theta_{u ^{\prime}}=\,{\Big(}{\frac{r_{d}}{r_{u}}}{\Big)}^{2}\,\theta_{d^{\prime}}\,=\,{\Big(}{\frac{r_{d}}{r_{u}}}{\Big)}^{\frac{3}{2}}\,\theta_{u}. (8)
For n complete cycles, the growth of angular velocity amplitude as well as the angular amplitude \theta_{A} increases by a factor of \rho_{A,\,n}\,=\,\left(\frac{r_{d}}{r_{u}}\right)^{3 n}.
For \rho_{\mathrm{A},\,n}=2 then with {\frac{r_{d}}{r_{u}}}=2^{\frac{1}{10}} one gets (2^{\frac{1}{10}})^{3n}=2, so n={\frac{10}{3}}.
Alternate Solution
The moment of inertia with respect to the swing pivot
I=M r^{2}. (1)
Since the A to B transition is fast one has by conservation of angular momentum,
I_A{\omega_{A}}=I_B{\omega_{B}}. (2)
The energy at point A is
E_{A}=\frac{1}{2}I_{A}\omega_{A}^{2}. (3)
The energy at point B is
E_{B}=\frac{1}{2}I_{B}\omega_{B}^{2}+M g h\,, (4)
where h = r_{d} – r_{u} is the vertical distance the child’s center of mass moves. The energy at point C (conservation of energy)
E_{\mathrm{c}}=E_{B}={\frac{1}{2}}I_B{\omega_{B}^{2}}+M{\mathrm{g}}h. (5)
As the child squats at the C to D transition, the swing losses energy of the amount Mgh, so
E_{D}=\frac{1}{2}I_{B}\omega_{B}^{2}. (6)
Energy at point E is equal to energy at point D (conservation energy)
E_{E}=E_{D}={\frac{1}{2}}I_{B}\omega_{B}^{2}. (7)
But we have also
E_{E}={\frac{1}{2}}I_{E}\omega_{E}^{2}. (8)
From equation (7) and (8) we have
\omega_{E}^{2}={\frac{I_{B}}{I_{E}}}\omega_{B}^{2}. (9)
Using equation (2) this equation yields
\omega_{E}^{2}=\frac{I_{A}}{I_{B}}\omega_{A}^{2}\,, (10)
where we have used I_{E}=I_{A}.
Using equation (1) one obtains from equation (10)
\omega_{E}^{2}={\frac{r_{d}^{2}}{r_{u}^{2}}}\omega_{A}^{2}. (11)
From this one obtains
{\frac{\mid\omega_{E}\mid}{\mid\omega_{A}\mid}}={\frac{r_{d}}{r_{u}}}. (12)
This ratio gives the fractional increase in the amplitude for one half cycle of the swing motion. The fractional increase in the amplitude after n cycles is thus,
\frac{|\ \omega_{E}\ |_{n}}{|\ \omega_{A}\ |_{0}}=\left(\frac{r_{d}}{r_{u}}\right)^{2n}, (13)
where\left|\omega_{A}\right|_{0} is the initial amplitude and\left | \omega_{E}\right|_{n}is the amplitude after n cycles. Substitute the values,
2=2^{\frac{2n}{10}} (14)
or,
n = 5. (15)
Thus it takes only 5 swing cycles for the amplitude to build up by a factor of two.