BIOL/PHYS 438

● Logistics● Ch. 5: “Animals in Motion” wrapup

– CentriXXXal forces, Terminal velocity, . . .● Ch. 6: “Locomotion” wrapup

– Swim, Fly, Walk . . .

Logistics

Assignment 1: Solutions now online!

Assignment 2: Solutions online soon!

Assignment 3: Solutions online soon!

Assignment 4: due Thursday

Jess will be away next week (5-9 March)Hopefully your Projects are underway by now . . .

Circular Motion review

Fvr

•

in order to cause the centripetal acceleration a = v

2/r = r ω 2 implicit in uniform

circular motion.

•F'

Clarification: centripetal vs. “centrifugal” forces:In order for a particle of mass m to move at a constant speed v in a circle of radius r about a center C, it must be subjected to a

m

C

centripetal (“toward the center”) force of magnitude F = m v

2/r = m

r ω 2 (where ω ≡ d/dt )

This implies an equal and opposite centrifugal force F' acting on the center C (not on the particle!).

Predator vs. Prey: Turning Radius

If both animals have the same coefficient of friction µ with the ground, then the largest centripetal force they can exert is µMg, causing the same centripetal acceleration a = u

2/r = µg for both animals. If the prey has a smaller turning radius, it is because it is slower than the predator, not smaller.

Figure 5.7a in the textbook suggests that the prey can turn more sharply than the predator (and thus escape) because it is smaller. This is not the case.

Lesson: slow down for “broken field running” !

Crane StanceA karate master stands in position A. How can she get to position C in one move without bending her knees or sliding her feet on the floor ?

A C

Answer: first she quickly lifts her arms, raising her centre of gravity to the desired level and giving it a small upward momentum; then she abruptly reverses the motion and lowers her arms, lifting the rest of her body so that she can move her legs inward freely.

A CBExtra normal forces Normal forces = 0

Exam question (?)

Could a heron on the airless Moon get off the ground by flapping its wings (while holding its breath)? Explain!

Allometrics: X = a M

X a

Heartbeat Quota:

Nhb = 4.0211.89(365246060) M –

0.05

= 1.5109 M –

0.05

more or less independent of M.

Terminal Velocity -- when Mg = Fdrag

Recr ≈ 100 (external flow) vs. ≈ 2300 (pipes)

Re = 2 u R / = 2 u R /

Sphere: 4/3 R 3 g = Fdrag = 6 R u + ½ R 2 CD u 2 FStokes

Fhydrodynamic

For Re > 100, CD ~ 1; for Re < 100, Fh « Fs so it does no harm to set CD = 1. Then we can solve a quadratic equation for u:

u = (6 /R )

(√1 + (R 3 g /27 2

)

− 1 )

Raindrop Terminal Velocityu = (6 /R ) (√1 + (R 3 g /27

2 ) − 1 )

with g = 9.81, = 103 and = 1.8 10−5 gives

u = (1.08 10−7 /R ) (√1 + 1.12 1015 R 3 − 1 ):

rainmist

fog

cloud

Jet PropulsionBernoulli: Δp = ½ u

2 mass flow: J = r

2 u

Thrust: F = r 2

Δp = ½ r 2 u

2 = J u = J 2

/ r 2

There's a factor of 2 discrepancy . . . .

High Jumpers

Allometrics: X = a M

X a

Cost of TransportE

nerg

y pe

r Dis

tanc

e pe

r Bod

y M

ass

Fish Can't Swim? (Gray's paradox)

Kicking off the Eddies(what goes around, comes around)

Allometrics: X = a M

X a

SwimmersSpeed Usw ≈ 0.5 M 0.19 Length L large ≈ 1.5 M 0.43

FliersSpeed U ≈ 15 M 1/6 Metabolic velocity Um ≈ 40 M –1/4

RunnersEndurance running speed Uer ≈ Cr M

– 0.023 ≈

const.

Limits