THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 438 Assignment # 4:
 
MORE MECHANICS & METABOLISM
 
SOLUTIONS:
 
Thu. 15 Feb. 2007 - finish by Thu. 1 Mar.

Please hand in one assignment per group and list the names & Email addresses of all group members at the top of each sheet. During Reading Break you can gather information from the literature and attempt each problem on your own; then when you return you can pool your information and ideas with the rest of your team and your final solutions should emerge quickly. As always, if you think some necessary information is missing, make a reasonable assumption. But always write down what that assumption is. Remember to estimate your uncertainty in any measured quantity, and don't forget to specify all units. If possible, justify your input. For original comments you may score bonus points!

1. THE ENERGY LOSS OF A HUNT:
   1. Search the literature to collect data about a fast hunter such as wild dog, leopard, or lion.¹ Find mass M, top speed U_t, length of leg L, length of step while foot is on the ground S, period $T = 2\pi/\omega$ (either from T = 1/n where n is the number of steps per second, or from the distance $\lambda$ between two imprints of the feet on the ground: $T = \lambda/U_t$, duration $\tau$ of the hunt, or the range $R = U_t \tau$. Treat the motion of the rear leg like simple harmonic motion where the position of the foot relative to the vertical is $x = A \sin(\omega t)$, where A = S/2 is the amplitude, and $U = \omega A \cos(\omega t)$ is the velocity of the foot. The maximum foot velocity is $U_{\rm max} = \omega A$, the instantaneous acceleration of the foot is $a = \omega^2 A \sin(\omega t)$, and the maximum foot acceleration is $a_{\rm max} = \omega^2 A$.  ANSWER: 

      
      Cheetah, Acinonyx jubatus. (Fig. from Hildebrand, 1962)²

      Body mass: M = 40 kg (Marker and Dickman, 2003).³
      Average top speed: U_t = 14.8 m/s (Hildebrand, 1962).
      Limb Length: L = 0.79 m (Marker and Dickman, 2003).
      Average period: T = 0.31 s (Hildebrand, 1962).
      Angular deflection of the limb: $\phi = 30^\circ$ (Ahlborn, 2004).
      Distance covered during hunt: $X_{\rm hunt} = 60$ m.
      Duration of hunt: $\tau = 5$ s.
      The parameters that control the speed U are stride length $\lambda$ and stride frequency f (Alexander, 2003):⁴ $U = \lambda \times f$. Therefore, we can calculate a stride length for a given speed and stride frequency: f = 1/T = 1/0.31 s = 3.3 Hz and (at top speed) $\lambda = U_t/f$ = (14.8 m/s)/(3.3 Hz) or .
      To get the length S of a "step" requires some knowledge of the leg motion. If we treat this as the swing of a straight rod of length L through an arc subtending $2 \phi = 60^\circ$ at constant $\omega$ (ignoring the fact that this would be a pretty bumpy ride), we can set $U = L \omega$ and pretend that the leg just keeps swinging around in a circle at the same $\omega$ (like in a "Roadrunner" cartoon), in which case the step length S is the same fraction of the total stride length $\lambda$ as $60^\circ$ is of $360^\circ$ - namely 1/6. This immediately gives (at top speed) a step length $S = \lambda/6$ or . Alternatively, we can ignore the details of how the leg might bend during the step and simply assume that it begins with the leg stretched out straight at a $30^\circ$ angle in front (so that the shoulder is $L \sin 30^\circ = 0.79/2$ m behind the foot at the moment the foot first touches the ground) and ends with the leg stretched out straight at the same angle behind the shoulder (so that the shoulder is 0.79/2 m ahead of the foot as it leaves the ground) for a total step length of , consistent with the estimate above.

   2. Assuming that the maximum acceleration of the foot is equal to the maximum acceleration of the body, find the acceleration time t₁ needed to reach top speed U_t = a t₁.  ANSWER:  The maximum acceleration of the foot (in the context of simple harmonic motion) is $a_{\rm max} = A \omega^2$ where A is the amplitude, $A = L \sin \phi$ = (0.79 m $)(\sin 30^\circ)$ $\Rightarrow$ $a_{\rm max} = A \omega^2$ = (0.79 m)(3.3 Hz)² = 8.6 m/s². Of course, this maximum horizontal acceleration occurs at the end of the range of motion (e.g., for the "Road Runner" picture, when the leg is horizontal) and at mid-stride the horizontal acceleration in this model would be zero; so we are bound to be overestimating the maximum acceleration by something like a factor of two. The problem tells us to assume that the acceleration of the foot is equal to the acceleration of the body, $a_{\rm max} = a$, so the time to reach top speed is t₁ = U_t/a = (14.8 m/s)/(8.6 m/s²) or .⁵

   3. Determine the speed-up distance s₁ (similar to the free fall distance $s = {1\over2} g t^2$).  ANSWER:  The speed-up distance is similar to free fall distance, but instead of gravity we use our calculated acceleration: $s_1 = {1\over2} a t_1^2$ = (0.5)(8.6 m/s²)(1.7 s)² or .

   4. Calculate the kinetic energy K which the animal attains at top speed.  ANSWER:  This is straightforward: $K = {1\over2} M U_t^2$ = (0.5)(40 kg)(14.8 m/s)² or .⁶

   5. Assuming that K was obtained by the application of an average force F acting during the time t₁ or over the distance s₁, the energy equation reads K = F x s₁, from which F can be found.  ANSWER:  Given K = 4380 J and s₁ = 7.3 m, we use F = K/s₁ to get .

   6. Find the total energy expense of the hunt, $E_{\rm tot} = F \times X_{\rm hunt}$ and express it in mass of body fat burned, taking into account a reasonable inefficiency of converting the body fat into muscle fuel.⁷ ANSWER:  $E_{\rm tot} = F X_{\rm hunt}$ = (600 N)(60 m) or . Such a short sprint would not normally be reflected in the conversion of fat to fuel, but rather governed by anærobic metabolic pathways. The muscle biochemistry and histology of cheetah locmotor muscles show a high capacity for glycolysis and indicate that they are well equipped for anærobic metabolism.⁸Nevertheless, the energy has to eventually come from some form of "chemical fuel", and if the cheetah had no other source than its own body fat, it would have to "burn" a mass $\Delta M = E_{\rm tot} / \eta h$ where $\eta \approx 25$% is the efficiency of utilization for work and h = 32 x 10⁶ J/kg is the specific energy content of fat. This gives $\Delta M = 0.45 \times 10^{-2}$ kg or . (The cheetah needs to be successful at least once in every few hundred tries or she will get pretty skinny!)

2. LANDING ON YOUR FEET: Two young people jump down from a height of h = 2.0 m onto the forest floor which has a mud hole and a rocky flat. The boy (M_b = 70 kg) lands with stiff legs on the muddy ground, which "gives" so that he depresses the ground by $\Delta y = 5$ cm when being decelerated from the impact velocity U₀ to rest.
   1. Determine the impact speed U₀.  ANSWER:  You can use either the constant acceleration equation U₀² = 2 g h or the conservation of energy ${1\over2} M_b U_0^2 = M_b g h$ to get .

   2. Calculate the deceleration (negative acceleration) of the the boy's center of mass during landing, and determine the average force on the soles of his shoes during impact.  ANSWER:  We have little choice but to assume the mud exerts a constant force F_b during the boy's impact, absorbing his entire kinetic energy $K = {1\over2} M_b U_0^2$ as work $W = F_b \Delta y$ done on the mud. Thus $F_b = M_b U_0^2/2\Delta y$ = (70 kg)(6.26 m/s )²/(2 x 0.05 m) or or 40 times his weight. Ouch!

   3. The girl (M_g = 65 kg) lands on the rock, but in order to reduce the impact force she lands with soft knees and moves her center of gravity relative to her feet by $\Delta y = 0.5$ m as she lands. What is the average impact force on the soles of her feet during the landing?  ANSWER:  The velocity on impact U₀ is the same for both. For the young woman we use the same equation, except that $\Delta y$ is 10 times bigger and her mass M_g is less, so the average force is more than ten times smaller: $F_g = M_g U_0^2/2\Delta y$ = (65 kg)(6.26 m/s )²/(2 x 0.5 m) or or 4 times her weight.

3. DIFFUSION AND METABOLIC RATES OF A NUDIBRANCH: A certain nudibranch of mass M = 0.005 kg living in 10$^\circ$C waters off Vancouver Island carries its gills (mass m) outside its body. Assume that (i) the gills are 10% of the total body mass, (ii) the gills are tree-like structures that have branches with an average diameter of $D = 500 \; \mu$m, filled with fine capillary vessels of diameter $d = 10 \; \mu$m right under the skin of the gills. (See sketch.)

   

   1. Determine the surface area A of the gills that contributes to the diffusion exchange of oxygen.  ANSWER:  Mass of nudibranch: M = 0.005 kg. Mass of gills: m = 0.0005 kg. Gill volume V_g = (0.0005 kg)/(1000 kg/m³) = 5 x 10^-7 m³. Gill `trunk' diameter d_g = 500 x 10^-6 m. Gill capillary diameter d_c = 10 x 10^-6 m. If we treat the entire gill volume as one long cylinder of diameter d_g, its length L_g can be deduced from its volume $V_g = \pi (d_g/2)^2 L_g$: $L_g = V_g / \pi (d_g/2)^2 = 2.55$ m. Then we can calculate its surface area as $A_g = \pi d_g L_g = \pi (500 \times 10^{-6}$ m)(2.55 m) or .

   2. Calculate the diffusion flow rate of oxygen $\dot{N}_{{\rm0}_2}$ [molecules/sec] into the gills. (See Sect. 4.1.3; assume first that the blood returning from the body into the gills is completely deoxygenated.)  ANSWER:  The diffusion flux $\dot{N}_{{\rm0}_2} = A_g D \Delta n/\Delta x$, Where A_g is the area over which diffusion occurs, D is the diffusion coefficient and $\Delta n$ is the change in concentration over a given distance $\Delta x$.
      Here $\dot{N}_{{\rm0}_2} = [(4 \times 10^{-3}$ m ²) (18 x 10^-10 m²/s ) (1.63 x 10²³ molecules O₂/m³ H₂O )] /(10 x 10^-6 m) or .

   3. Use the relation between $\dot{N}$ and $\Gamma$ which is derived in Section 4.3.3 to determine the metabolic rate of the nudibranch.  ANSWER:  $\dot{N}_{{\rm0}_2} = J_{O_2}/M_{O_2} = (3.4 \Gamma)/(M_{O_2} \Delta H)$.
      Thus 1.17 x 10¹⁷ molecules O₂/s $= (3.4 \Gamma)/[(32 \times 1.6 \times 10^{-27}$ kg)(30 MJ/kg)]
      so $\Gamma = [(1.17 \times 10^{17}$ molecules O₂/s ) (32 x 1.6 x 10^-27 kg)(30 MJ/kg)]/3.4 = 5.3 x 10^-8 MJ/s or .

   4. Assuming that for these animals $\Gamma = a M^{3/4}$, determine the metabolic constant a.  ANSWER:  $\Gamma = a M^{0.75}$ $\Rightarrow$ $a = \Gamma / M^{0.75}$ = 0.053/0.005^0.75 or .

   5. Look into the literature to assess if the assumption is reasonable that the blood returning from the body into the gills is completely deoxygenated. If not, how would this change your answers?  ANSWER:  Many references show curves similar to those in Fig. 4.17 on p. 145 of the textbook, indicating that essentially all hæmoglobin oxygen bonding sites are empty when in equilibrium with a reservoir at zero partial pressure of O₂. However, living tissue is unlikely to have zero partial O₂ pressure at any time, so this is not too helpful.

      

      The "Bohr effect" (see above) helps squeeze more O₂ out of hæmoglobin sites by filling them with CO₂ supplied by the same tissues, which helps but does not really answer the question. Figure 4.18(b) on p. 146 suggests that the equilibrium O₂ partial pressure in venous blood returning to the lungs is about 5 kPa, compared with about 12.5 kPa for oxygenated blood leaving the lungs, which means that venous blood is still about 40% oxygenated. Since $\Delta n$ is the O₂ concentration difference between the water (which we assume has the same O₂ partial pressure as air, namely about 20 kPa) and the deoxygenated blood, if the latter has $p_{{\rm O}_2} \approx 5$ kPa instead of zero as assumed above, it will reduce $\dot{N}_{{\rm0}_2}$ by a factor of 15/20 = 0.75, and the other calculations should be adjusted accordingly.