Question

Red-headed Agama lizards (Agama agama) are able to control the pitch of their bodies when jumping to higher ground or onto vertical surfaces by using their tail to regulate angular momentum. Pitch refers to, for example, the up or down orientation of the nose of an airplane or, in this case, the body and nose of the lizard. The figure below shows the lizard just before it jumps. A change in pitch during the jump would represent a rotation of the forward section of the lizard about an axis perpendicular to the page. When jumping from a horizontal surface, especially one that is smooth, the Agama lizard can attain a significant amount of angular momentum about this axis, associated with a rotation of its body in the counterclockwise direction. To counteract this, the lizard swings its tail in the clockwise direction, stabilizing the pitch of the forward section. To understand this behavior, we construct a structural model in which the lizard consists of two thin colinear, uniform cylindrical sections, a head section of mass 54.5 g and length 13.0 cm and a tail section of mass 11.5 g and length 17.0 cm. Immediately after beginning its jump, the lizard is measured to have an angular momentum of 1.50 x 10 kg. m/s (counterclockwise) about an axis perpendicular to the page and passing through its center of mass. (a) What is the initial angular speed (in rad/s) of the lizard immediately after leaping? 13.5 How is the lizard to be modeled? What is the moment of inertia of each section? Where should the axis of rotation of the lizard be, and what is the angular momentum of a rigid body? rad/s (b) At what angular speed (in rad/s) must the lizard initially rotate its tail to maintain the initial pitch of its forward section throughout the jump? 7.5 * Your answer cannot be understood or graded. More Information rad/s

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Answer 1

> Sol? Mi = 11.5g , Y = 19. M2 = 5405 g 1 %2 = (12+14) - Lizard is made up to two seefim L;=13cm 112717 cm centre of mony X = - * +, M2 X2 ) = (1-5942.) + 54.5)(1964) E Marme) (17.5+5485) [*= 2 12.80-441 = 20- X = 20-886 from left from head Eg =-(200886).+13 = 2o5stem FA Head has moment of inertia I= MC + mogen 1= 154,5x195 65405x42.5pg)?) 12 I= 0767.54173561934 1 = 1123.875 gom? 12

= [= 'lo S0X108 Kg.me = 15000 g Cagays Mow w- _ > 11500bar) = 113-3 46 Vals B rulate ito Tout to maintain the 1.CM The lizard Irudial Pitch of its fooward seefing Ao from Tail (cog) = (20.866 - 12) = 120366 cm Now simlarly for Taal, moment of inerten Izberite + mart. 71 41.50% 192 (0-5 212.3437 = (24*80+ 1758.66) = 2025. sive → w=(t) = ( 1 ) = 17.369 Foto 12