Krazy Katt, te mecanical cat Te cat rigting relex is a cat's innate ability to orient itsel as it alls in order to land on its eet. Te rigting relex begins to appear at 3 4 weeks o age, and is perected at 7 weeks. Te minimum eigt required or tis to occur in most cats would be around 30 cm. You are part o an engineering team trying to reproduce tis in a mecanical cat. Your prototype, aptly named Katt, wen dropped motionless, upside down rom.5 m, can rigt itsel beore it its te ground by rotating its. Katt s is a solid cylinder 30 cm in lengt and 15 cm in diameter, wit a mass o 3.0 kg. Attaced to te center o one end o te is Katt's, a 0-cm long rod wic extends out perpendicular to Katt's and as only 10% te mass o te. Te will rotate due to a small electric motor in te. Your task is to determine te energy demand (power) o te motor so you can order te piece. 1
Understanding te model, identiying te pysical principles to be used, identiying te target. Cats ability to turn teir bodies in te air is based on te same principle tat allows a diver to turn in te air: i we neglect air resistance (wic is small), te only external orce acting on bot bodies is teir own weigt. Tis orce acts on te center o mass o te, and tus produces zero torque. Wen tere is no net external torque acting on a system, total angular momentum is conserved. Note te word external : tere may be important internal orces between dierent parts o te system, and tese orces may produce torque on te parts, and yet te total angular momentum will be conserved. Tis is indeed te case in tis problem. Tis is te model o Katt and its. Since tis is suc a simple model, let us identiy te top o Katt s back wit a red line. n te normal position, tis is te part tat sould be on top. m L M R Katt is to be dropped upside down rom a eigt =.5 m and sould turn itsel around beore toucing te ground.
ω ω Tat means te cylinder needs to rotate 180 degrees. Te way to accomplis tis is by rotating te in te opposite direction te rigt amount. Te rigt amount is determined by conservation o angular momentum. (Note tat te rigt amount could a priori be any angle, so te position o te at te end o te all in te igure is completely arbitrary) Te rotation involves te production o kinetic energy by te engine. We ave a limited amount o time to produce tis energy (te drop time), and tus te engine needs to be able to produce some minimum power. Tis minimum required power is te goal o our problem. Setup Let us start wit te time limitation. An object dropped rom eigt is given by reaces te ground in a time t given by: 1 gt t g During tis time, te must rotate π radians. te accelerates wit a constant angular acceleration 1, we ave: 1 te acceleration is constant, te cat will reac te loor in te rigt position but wit some angular speed (and tus migt skid upon landing, not a very gracious perormance). A better option, just sligtly more complicated, would be to assume tat te accelerates during te irst al o te time and decelerates during te oter al. Tis results in a sligtly more demanding engine: te must acquire a iger speed and in only al o te time. n tis solution, we will stick to te constantly accelerated system model or simplicity. 3
1 t g Tus te required acceleration o te is: g [Equation 1] And te inal angular speed o te will be g t [Equation ] Now we need to link tis to te rotation o te. Bot parts (cylinder and ) rotate about te axis o te cylinder. Te moments o inertia o te cylinder and te or rotations about tat axis are: 1 1 MR ml 3 (rod about axis troug an end) At any given time, i te and te are rotating wit angular speeds ω and ω, te total angular momentum o te system is: L Wen te model is released, noting is rotating, so te angular momentum is zero. Conservation o angular momentum means tat tis needs to be true at any moment during te all, tus: 0 Tis means tat te two angular speeds are related troug te ratio o te moments o inertia : [Equation 3] Note tat equation 3 makes it very obvious tat ω, L (and α) are not te magnitude o tese quantities but, strictly speaking, te z component o tese quantities, were z is te direction perpendicular to te plane o rotation. n oter words, tey ave a sign! 4
We are inally ready to compute te required power output o te engine. Te instantaneous power is given by: P Te acceleration o te system as been assumed constant, and tus te torques are also constant. Tereore, te igest power demand appens wen te speeds are te largest (i.e., at te end). Newton s tird law requires, and tereore P And we can use equation 3 to write everyting in terms o te pysical quantities: P 1 Te torque on te can be obtained using equation 1 in And, as argued above, we are interested in te maximum speed, so te angular speed o te is given by equation. Putting all tis togeter gives: g P 3 g 1 Numerical values Wit M = 3.0 kg, R = 0.075 m, m = 0.30 kg and L = 0.0 m, =0.0084 kg m ; =0.0040 kg m ; Wit =.5 m,.1 5
3 9.8 m/s P 0.0084 kg m 1.1.8 W.5 m Cecks To make sure tat te system can work, let s compute te required maximum speeds: Equation : Equation 3: g 1 8.8 s 1.4 turns per second 0.47 s.9 turns per second 1 Bot values seem easible. Various qualitative cecks on te inal expression: te eigt increases, te required power decreases n a world wit a larger g, te cat would ave less time to lip around, so power would be larger Power increases or iger values o te ratio between te moment o inertia o te and tat o te, i.e., or smaller values o te moment o inertia o te. ndeed, i te was very sort or very ligt, we intuitively expect tat te lip will be arder. Final note: Cats also use teir lexible spines to twist teir bodies during a all. n a more realistic model, we could 1) allow our cat s to adopt a V sape; and ) break te V into two cylinders tat can rotate relative to one anoter. Tis cat s model is tus made o tree rotating parts. Tis can be appreciated in tis video: ttp://www.youtube.com/watc?v=rajyivxuiw 6