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Investigate the relationship between speeds recorded and the distance travelled in this kinematic scenario.

Board and Spool

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

1) Suppose that the spool is rotating with an angular speed $\omega$ about a point A. The spool is a rigid body, this means that the angular speed is the same for all points. Write equations for points B and C:

$$\omega_C = \frac{v_C}{R},      \omega_B = \frac{v_B}{R + r}$$

but $\omega_B = \omega_C = \omega$ and $v_B = u$ because the board does not slip on the spool. Thus, $v_C = \frac{R}{R+r}u$. This means that the speed at which the man is approching the spool is $v = u - v_C = u -\frac{R}{R+r}u =\frac{r}{R+r}u$. This means that the time needed for the man to reach the spool is $$t = \frac{l}{v} = \frac{l(R+r)}{ru}\;.$$

 2) The man will travel $s = ut = \frac{l(R+r)}{r}$.

3) If $r = R$ then $s = 2l$.

4) Plug numbers to the equations but do not forget to change units, $t = 6.48\mathrm{s}$ and $s = 6.48\mathrm{m}$.