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Worksheet 3.6 Ripcord Cars with Energy

In this worksheet you will model the motion of a ripcord car. If you don’t know what a ripcord car is, this video should help.
Figure 3.6.1. What is a ripcord car? The coolest toy ever!
This worksheet is based on the SIMIODE modeling scenario [7], but modified to use energy transfer as a modeling concept.
A ripcord toy (call it a "car") uses a handle with a toothed extension cord (the "t-stick") to spin a wheel. When placed on the ground the wheel initially spins exerting a force that accelerates the car. During the spinning on the ground the speed of the wheel decreases, which causes a loss of rotational kinetic energy in the wheel. Assuming energy is conserved (little energy is dissipated as heat or lost to air resistance), the rotational energy lost by the wheel is converted to translational kinetic energy of the car. This transfer of energy continues until pure rolling is achieved. That is, letting \(v\) be the velocity of the car, \(R\) be the radius of the wheel, and \(\omega\) by the angular velocity of the wheel, the energy transfer from the rotation of the wheel to the translation of the car stops when
\begin{equation*} v = \omega R. \end{equation*}
Throughout the remaining problems we will assume the following:
  • The total mass of the car (including the driving wheel) is \(m_{\text{car}} = 0.5\) kg and the mass of the spinning wheel is \(m_{\text{wheel}} = 0.2\) kg.
  • When we pull the ripcord, we impart a total energy of \(E_0\) J as rotational kinetic energy for the driving wheel of the car.
  • There are two forces exerted on the car prior to the moment when pure rolling is achieved.
    1. A constant propulsive force \(F\) exerted by the spinning wheel.
    2. A resistive force proportional to the translational velocity of the car.
    After pure rolling is achieved only the resistive force is present.
The data that has been collected (from a film at 30 frames per second) on our car is the following:
Table 3.6.2. Distance Traveled by Ripcord Toy
Frame (at 30 fps) Distance (in cm)
0 0
4 3
8 10
12 21
16 34
20 50
24 68
28 85
32 103
36 120
40 136
44 153
48 168
52 184
56 199
60 214
64 228
68 243
72 257
For reference, the formula for translational kinetic energy is \(KE_{\text{trans}} = \frac{1}{2}mv^2\) and the formula for rotational kinetic energy is \(KE_{\text{rot}} = \frac{1}{2}R^2\omega^2\text{.}\)

1.

Derive an equation involving masses and energy of the wheel and car that holds when pure rolling occurs, i.e. when \(v = \omega R\text{.}\) (This will be useful when developing the formula for a flow rate.)

2.

Show that the derivative (with respect to time) of \(KE_{\text{trans}}\) is \(mF_{\text{total}}v\) where \(m\) is the mass of the car, \(F_{\text{total}}\) is the sum of all forces on the car, and \(v\) is the velocity of the car.

3.

Create and Insight to model the motion of the car. Here are some hints:
  • Use velocity, position, kinetic energy of the wheel, and kinetic energy of the car as stocks.
  • Since energy is conserved, inflow to the kinetic energy of the car equals the outflow from the kinetic energy of the wheel.
  • Energy transfer occurs until the equation from Worksheet Exercise 3.6.1 holds. After this time, you can ignore the energy stocks.
  • There should be two flows for the velocity stock.

4.

Estimate \(E_0\text{,}\) \(F\text{,}\) and \(k\) to match the data in Table 3.6.2. Observe that the data has input of frames rather than time. You will need to account for that.

5.

As an alternative approach, without using energy, estimate the time when velocity stops increasing from Table 3.6.2. Use this as a parameter so that only two variables need to be estimated. Which method yields a smaller SSE from the data?