Parte 5 - Cinemática Plana de Corpos Rígidos-2023-2

1 Parte 5 - Cinemática Plana de Corpos Rígidos Conceito: Geometria apenas, sem considerar a causa do movimento. Corpo Rígido: é um sistema de partículas para o qual as distâncias entre elas permanecem inalteradas. Tipos de Movimento Plano: Translação, rotação em torno de um eixo fixo e movimento plano geral. Vide Figura 5/1 abaixo. 2 5.1 Rotação em Torno de Um Eixo Fixo e Método do Movimento Absoluto 3 Lista de Exercícios 11: Problema resolvido 5/2, Problema Resolvido 5/3, Problema 5/17, Problema 5/28. SAMPLE PROBLEM 5/8 Crank CB oscillates about O through a limited arc, causing crank OA to os- cillate about O. When the linkage passes the position shown with CB horizontal and OA vertical, the angular velocity of CB is 2 rad/s counterclockwise. For this instant, determine the angular velocities of OA and AB. Solution 1 (Vector). The relative-velocity equation vA = vB + vAB is rewritten as ωOA × rCA = aCA × rpB + aAB × rAB where ωOA = ωOA k aCA = 2k rad/s aAB = ωABk rpA = 100j mm rpB = 751i mm rAB = −175i + 50j mm Substitution gives ωOA k × 100j = 2k × (−751i) + ωABk × (−175i + 50j) −100ωOAi = −150j − 175ωABi + 50ωABj Matching coefficients of the respective i- and j-terms gives −100ωOA − 175ωAB = 0 25/6 + (−7ωAB) = 0 The solutions of which are ωAB = −6/7 rad/s Ans. and ωOA = −3/7 rad/s Ans. Solution II (Scalar-Geometric). Solution by the scalar geometry of the vec- tor triangle is particularly simple here since vA and vB are at right angles for this special position of the linkages. First, we compute vAB which is [v = ru] vAB = 0.075(2) = 0.150 m/s and represent it and its correct direction as shown. The vector vAB must be per- pendicular to AB, and the angle of between vAB and vA is also the angle made by AB with the horizontal direction. This angle is given by tan θ = 100 − 50 = ? 250 − 75 = ? The horizontal vector vA completes the triangle for which we have vAB = vA cos θ = 0.150/cos θ tan θ = vA tan θ = 0.150(2/7) = 0.30/7 m/s The angular velocities become [ω = vr] ωAB = vAB = 0.150 cos θ AB 0.250 cos θ 0.075 = 6/7 rad/s CW Ans. [ω = vr] ωOA = vA = 0.30 1 OA 7 0.100 = 3/7 rad/s CW Ans. Helpful Hints 1 We are using here the first of Eqs. 5/3 and Eq. 5/6. 2 The minus signs in the answers indi- cate that the vectors ωOA and ωOA are in the negative k-direction. Hence, the angular velocities are clockwise. 3 Always make certain that the se- quence of vectors in the vector polygon agrees with the quality of vectors specified by the vector equation. 6 5.2 Método do Movimento Relativo para Eixos Transladados. Análise de Velocidades Conceito: Movimento Plano Geral = Translação Pura + Rotação em Torno de Um Eixo Fixo (Figura 5/5 (a)) XY: sistema inercial A Figura 5/5 (a) mostra que, tomando dois pontos A e B de um corpo rígido, pode se escrever a seguinte equação, ∆𝐫A = ∆𝐫B + ∆𝐫A B ∆𝐫A: deslocamento absoluto do ponto A, ∆𝐫B: deslocamento absoluto do ponto B e ∆𝐫A B: deslocamento relativo do ponto A com relação ao ponto B. Dividindo a equação acima por ∆t e fazendo o limite quando ∆t tende a zero, resulta, lim ∆𝑡→0 ∆𝐫𝐴 ∆𝑡 = lim ∆𝑡→0 ∆𝐫𝐵 ∆𝑡 + lim ∆𝑡→0 ∆𝐫𝐴 𝐵 ∆𝑡 Ou, 𝐯A = 𝐯B + 𝐯A B 7 𝐯A: velocidade absoluta do ponto A, 𝐯B: velocidade absoluta do ponto B e 𝐯A B: velocidade relativa do ponto A com relação ao ponto B. Conceito: O movimento relativo do ponto A com relação ao ponto B é a rotação do ponto A com relação a um eixo fixo em B. Vide Figura 5/5 (b) Portanto, 𝐯A B = 𝛚 × 𝐫 𝛚: rotação absoluta do corpo rígido que contém os dois pontos A e B e 𝐫: posição relativa do ponto A com relação ao ponto B. Portanto, a equação de velocidades fica como, 𝐯A = 𝐯B + 𝐯A B = 𝐯B + 𝛚 × 𝐫 Vide Figura 5/6 8 Lista de Exercícios 12: Problema resolvido 5/7, Problema Resolvido 5/8, Problema 5/64, Problema 5/74, Problema 5/119 e Problema 5/120. 5/64 The circular disk of radius 0.2 m is released very near the horizontal surface with a velocity of its center vo = 0.7 m/s to the right and a clockwise angular velocity ω = 2 rad/s. Determine the velocities of points A and P of the disk. Describe the motion upon contact with the ground. 5/74 For an interval of its motion the piston rod of the hydraulic cylinder has a velocity vA = 4 ft/sec as shown. At a certain instant θ = β = 60°. For this instant determine the angular velocity ωBC of link BC. 11 5.3 Método do Movimento Relativo para Eixos Transladados. Análise de Aceleração. Derivando a equação de velocidade 𝐯A = 𝐯B + 𝐯A B = 𝐯B + 𝛚 × 𝐫 com relação ao tempo dá, aA = aB + (aA/B)t + (aA/B)n (aA/B)n = ω × (ω × r) (aA/B)t = α × r SAMPLE PROBLEM 5/13 The wheel of radius r rolls to the left without slipping and, at the instant considered, the center O has a velocity vo and an acceleration ao to the left. Determine the acceleration of points A and C on the wheel for the instant considered. Solution. From our previous analysis of Sample Problem 5/4, we know that the angular velocity and angular acceleration of the wheel are ω = vo/r and α = ao /r The acceleration of A is written in terms of the given acceleration of O. Thus, aA = ao + aA/O = ao + (aA/O)t + (aA/O)n The relative-acceleration terms are viewed as though O were fixed, and for this relative circular motion they have the magnitudes (aA/O)t = rα = r • ao/r = (ao)i = (aA/O)n = rω² = r(vo/r)² = (vo²/r) and the directions shown. Adding the vectors head-to-tail gives aA as shown. In a numerical problem, we may obtain the combination algebraically or graphically. The algebraic expression for the magnitude of aA is found from the square root of the sum of the squares of its components. If we use x- and y-directions, we have aA = sqrt{(aA)x² + (aA)y²} = sqrt{[vo cosθ + (aA/O)t]² + [ao sinθ + (aA/O)n]²} = sqrt{[vo cosθ + r(ao/r)]² + [ao sinθ + (vo²/r)]²} The direction of aA can be computed if desired. The acceleration of the instantaneous center C of zero velocity, considered a point on the wheel, is obtained from the expression aC = ao + aC/O where the components of the relative-acceleration term are (aC/O)t = ra² directed from C to O and (aC/O)n = ra directed to the right because of the counter-clockwise angular acceleration of line CO about O. The terms are added together in the lower diagram and it is seen that aC = ra² Helpful Hints 1 The counterclockwise angular acceleration α of OA determines the positive direction of (aA/O)t. The normal component (aA/O)n is, of course, directed toward the reference center O. 2 If the wheel were rolling to the right with the same velocity vo but still had an acceleration ao to the left, note that the solution for aA would be unchanged. 3 We note that the acceleration of the instantaneous center of zero velocity is independent of α and is directed toward the center of the SAMPLE PROBLEM 5/14 The linkage of Sample Problem 5/8 is repeated here. Crank CB has a con- stant counterclockwise angular velocity of 2 rad/s in the position shown during a short interval of its motion. Determine the angular acceleration of links AB and OA for this position. Solve by using vector algebra. Solution. We first solve for the velocities which were obtained in Sample Problem 5/8. They are ωAB = –6t rad/s and ωOA = –3/7 rad/s where the counterclockwise direction ( + k-direction) is taken as positive. The ac- celeration equation is aB = aA + a(B/A)t + a(B/A)n where, from Eqs. 5/3 and 5/6b, we may write aA = ɔOA x rA/O + aOA x (ɔOA x rA/O) = ɔOA x [ 100j + 13 k x ( –32 k x 100j) = – 100ɔOAi – 100(ɔOA) mm/s a(B/A)n = ɔAC x rB/A + ɔAC x (ɔAC x rB) = 0 + 2k x (2k x [ – 75i]) = 300i mm/s a(B/A)t = ɔAB x (ωAB x rB/A) = 56 ij x ( – 56 j) x ( – 75i + 50j)] = -57 (171.5 – 50)i mm/s a(B/A)n = ɔAB x r(B/A) = ɔAB x (–75i + 50j) = – 50 ɔABi – 175 ɔABj mm/s a(B/A)t = ɔAB x r(B/A)/ We now substitute these results into the relative-acceleration equation and equate separately the coefficients of the i-terms and the coefficients of the j-terms to give -100ɔOA = 429 – 50ɔAB - 18.37 = – 36.7 – 175ɔAB The solutions are aAB = – 0.1050 rad/s2 and aOA = – 4.34 rad/s² Since the unit vector k points out from the paper in the positive z-direction, we see that the angular accelerations of AB and OA are both clockwise (negative). It is recommended that the student sketch each of the acceleration vectors in its proper geometric relationship according to the relative-acceleration equa- tion to help clarify the meaning of the solution. 5/121 The center O of the wheel is mounted on the sliding block, which has an acceleration aO = 8 m/s2 to the right. At the instant when 𝜃 = 45° and 𝜃’ = 3 rad/s and 𝜃’’= -8 rad/s2. For this instant determine the magni- tudes of the accelerations of points A and B. 5/146 The mechanism of Prob. 5/75 is repeated here. Each of the sliding bars A and B engages its respec- tive rim of the two riveted wheels without slipping. If, in addition to the information shown, bar A has an acceleration of 2 m/s² to the right and there is no acceleration of bar B, calculate the magnitude of the acceleration of P for the instant depicted. vA = 0.8 m/s vB = 0.6 m/s 5/147 The four-bar linkage of Prob. 5/88 is repeated here. If the angular velocity and angular acceleration of drive link OA are 10 rad/s and 5 rad/s² respectively, both counterclockwise, determine the angular accelerations of bars AB and BC for the instant represented. Problem 5/147 5/153 The elements of a power hacksaw are shown in the figure. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counterclockwise speed of 60 rev/min, determine the acceleration of the blade for the position where 𝜃 = 90°, and find the corre- sponding angular acceleration of the link AB. Problem 5/153 5/155 An oil pumping rig is shown in the figure. The flexi- ble pump rod D is fastened to the sector at E and is always vertical as it enters the fitting below D. The link AB causes the beam BCE to oscillate as the weighted crank OA revolves. If OA has a constant clockwise speed of 1 rev every 3 s, determine the acceleration of the pump rod D when the beam and the crank OA are both in the horizontal position shown. 17 5.4 Método do Movimento Relativo para Eixos Girantes. Análises de Velocidade e Aceleração. Desenvolvimento das equações no quadro. 5/160 The disk rotates about a fixed axis through O with angular velocity ω = 5 rad/sec and angular acceleration α = 3 rad/sec² in the directions shown at a certain instant. The small sphere A moves in the circular slot, and at the same instant, β = 30°, β̇ = 2 rad/sec, and β̈ = -4 rad/sec². Determine the absolute velocity and acceleration of A at this instant. Problem 5/160 5/163 An experimental vehicle A travels with constant speed v relative to the earth along a north–south track. Determine the Coriolis acceleration a_cor as a function of the latitude θ. Assume an earth-fixed rotating frame Bxyz and a spherical earth. If the vehicle speed is v = 500 km/h, determine the magnitude of the Coriolis acceleration at (a) the equator and (b) the north pole. Problem 5/163 5/176 For the instant represented, link CB is rotating counterclockwise at a constant rate N = 4 rad/s, and its pin A causes a clockwise rotation of the slotted member ODE. Determine the angular velocity ω and angular acceleration α of ODE for this instant. Problem 5/176 5/182 The crank OA revolves clockwise with a constant angular velocity of 10 rad/s within a limited arc of its motion. For the position θ = 30° determine the angular velocity of the slotted link CB and the acceleration of A as measured relative to the slot in CB. Problem 5/182 5/183 The Geneva wheel of Prob. 5/56 is shown again here. Determine the angular acceleration α_C of wheel C for the instant when θ = 20°. Wheel A has a constant clockwise angular velocity of 2 rad/s. Problem 5/183 5/191 The pin A in the bell crank AOD is guided by the flanges of the collar B, which slides with a constant velocity vB of 3 ft/sec along the fixed shaft for an interval of motion. For the position θ = 30° determine the acceleration of the plunger CE, whose upper end is positioned by the radial slot in the bell crank. Problem 5/191 25 Parte 6 - Cinética Plana de Corpos Rígidos Equações do movimento SAMPLE PROBLEM 6/5 A metal hoop with a radius r = 6 in. is released from rest on the 20° incline. If the coefficients of static and kinetic friction are μs = 0.15 and μk = 0.12, determine the angular acceleration α of the hoop and the time t for the hoop to move a distance of 10 ft down the incline. Solution. The free-body diagram shows the unspecified weight mg, the normal force N, and the friction force F acting on the hoop at the contact point C with the incline. The kinetic diagram shows the resultant force maG through G in the direction of motion and the couple Iα. The counterclockwise angular acceleration requires a counterclockwise moment about G, so F must be up the incline. Assume that the hoop rolls without slipping, so that a = rα. Application of the components of Eq. 6/1 with x- and y-axes assigned gives [∑Fₓ = m aₓ ] mg sin 20° − F = maG [∑Fᵧ = m aₐ ] N − mg cos 20° = 0 [∑MG = I α ] Fr = mra Elimination of F between the first and third equations and substitution of the kinematic assumption a = rα give a = rg sin θ / 2 = (32.2 / 2) (0.342) = 5.51 ft/sec² Alternatively, with our assumption of a = rα for pure rolling, a moment sum about C by Eq. 6/2 gives α directly. Thus, [∑MC = IG α + maCₐ ] mg r sin 20° = IG α + ⨀ ma ⨀ r = IG α + m a ⨀ r sin 20° To check our assumption of no slipping, we calculate F and N and compare F with its limiting value. From the above equations, F = mg sin 20° − ⨀ m a ⨀ g sin 20° = 0.1710mg N = mg cos 20° = 0.940mg But the maximum possible friction force is Fmax = μsN Fmax = 0.15(0.940mg) = 0.1410mg Because our calculated value of 0.1710mg exceeds the limiting value of 0.1410mg, we conclude that our assumption of pure rolling was wrong. Therefore, the hoop slips as it rolls. The friction force then becomes the kinetic value [F = μkN ] F = 0.12(0.940mg) = 0.1128mg The motion equations now give [∑Fₓ = m aₓ ] mg sin 20° − 0.1128mg = ma ã = 0.229(32.2) = 7.38 ft/sec² [∑MG = I α ] 0.1128mg(r) = mra α = 0.1128(32.2) / 6/12 = 7.26 rad/sec² The time required for the center G of the hoop to move 10 ft from rest with constant acceleration is [v = v₀ +(a/₂)t] t = √(2x/a) = √(2)(10) / 7.38 = 1.646 sec SAMPLE PROBLEM 6/6 The drum A is given a constant angular acceleration α of 3 rad/sec² and causes the 70-kg spool B to roll on the horizontal surface by means of the connecting cable, which wraps around the inner hub of the spool. The radius of gyration k of the spool about its mass center G is 250 mm, the coefficient of static friction between the spool and the horizontal surface is 0.35. Determine the tension T in the cable and the friction force F exerted by the horizontal surface on the spool. Solution. The free-body diagram and the kinetic diagram of the spool are drawn as shown. The correct direction of the friction force F may be assigned in this problem by observing from both diagrams that with counterclockwise angular acceleration, a moment about point G (and also about point D) must be counterclockwise. A point on the connecting cable has an acceleration ar = rα = 0.25(3) = 0.75 m/s². which is also the horizontal component of the acceleration of point D on the spool. It will be assumed initially that the spool rolls without slipping, in which case it has a counterclockwise angular acceleration α = (aD/g)/G. The acceleration of the mass center G is therefore, ar = (a/φ)(2.5) = 1.125 m/s². With the kinematics determined, we now apply the three equations of motion: [∑Fₓ = m aₓ ] F − T = 70(-1.125) [∑Fᵧ = m aₐ ] N − 709.81 = 0 N = 687 N [∑MG = I α ] (70)(0.450) − 70(0.150) = 700(.250)(2.5) Solving (a) and (b) simultaneously gives F = 75.8 N and T = 154.6 N Ans. To establish the validity of our assumption of no slipping, we see that the surfaces are capable of supporting a maximum friction force Fmax = μsN = 0.306587 = 171.7 N. Since only 75.8 N of friction force is required, we conclude that our assumption of rolling without slipping is valid. If the coefficient of static friction had been 0.1, for example, then the friction force would have been limited to 0.1(687) = 68.7 N, which is less than 75.8 N, and the spool would slip. In this event, the alternate relation a = rα would no longer hold. With (aD) known, the angular acceleration would be α = ((aD)(g))/G. Using this relation along with F = μkN = 68.7 N, we would then resolve the three equations of motion for the unknowns T, F, and α. Alternatively, with point C as a moment centre in the case of pure rolling, we may see Eq. 6/2 and obtain T directly. Thus, [∑MC = IG α + mår] 0.37 = 70(0.25)(2.5) + 70(1.125)(0.45) T = 154.6 N Ans. 6/94 The robotic device of Prob. 6/67 is repeated here. Member AB is rotating about joint A with a counterclockwise angular velocity of 2 rad/s, and this rate is increasing at 4 rad/s². Determine the moment MB exerted by arm AB on arm BC if joint B is held in a locked condition. The mass of arm BC is 4 kg, and the arm may be treated as a uniform slender rod. Problem 6/94 6/105 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in. The piston and piston pin A together weigh 1.80 lb. The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is 3000(2π/60) = 100π rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle θ = 90°. (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Problem 6/105 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS (m = mass of body shown) BODY Circular Cylindrical Shell MASS CENTER — MASS MOMENTS OF INERTIA Ixx = 1/2 ml² + 1/12 ml² Iyy = 1/2 ml² + 1/12 ml² Izz = ml² Half Cylindrical Shell MASS CENTER x̄ = 2r/π MASS MOMENTS OF INERTIA Ixx = Iyy = 1/2 ml² + 1/12 ml² Iyy = 1/3 ml² Izz = ml² Iz = (1 - 4/π²) ml² Circular Cylinder MASS CENTER — MASS MOMENTS OF INERTIA Ixx = Iyy = 1/4 ml² + 1/12 ml² Izz = 1/2 ml² Semicylinder MASS CENTER x̄ = 4r/3π MASS MOMENTS OF INERTIA Ixx = Iyy = 1/4 ml² + 1/12 ml² Izz = 1/2 ml² Iz = (1 - 1.6/9π²) ml² Rectangular Parallelepiped MASS CENTER — MASS MOMENTS OF INERTIA Ixx = 1/12 m(b² + l²) Iyy = 1/12 m(a² + l²) Izz = 1/12 m(a² + b²) Ixy = 1/3 ml² Iyz = 1/3 m(b² + l²) TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Spherical Shell MASS CENTER — MASS MOMENTS OF INERTIA Izz = 2/3 mr² Hemispherical Shell MASS CENTER x̄ = r/2 MASS MOMENTS OF INERTIA Ixx = Iyy = Izz = 2/3 mr² Iyy = Izz = 5/12 mr² Sphere MASS CENTER — MASS MOMENTS OF INERTIA Izz = 2/5 mr² Hemisphere MASS CENTER x̄ = 3r/8 MASS MOMENTS OF INERTIA Ixx = Iyy = Izz = 3/5 mr² Iyy = Izz = 83/320 mr² Uniform Slender Rod MASS CENTER — MASS MOMENTS OF INERTIA Iyy = 1/12 ml² Iyy.z = 1/3 ml² TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Quarter-Circular Rod Elliptical Cylinder Conical Shell Half Conical Shell Right-Circular Cone MASS CENTER \( \bar{x} = \bar{y} = -\frac{2r}{\pi} \) — \( \bar{z} = \frac{2h}{3} \) \( \bar{x} = \frac{4r}{3\pi} \) \( \bar{z} = \frac{2h}{3} \) \( \bar{z} = \frac{3h}{4} \) MASS MOMENTS OF INERTIA \( I_{xx} = I_{yy} = \frac{1}{2}mr^2 \) \( I_{zz} = mr^2 \) \( I_{xx} = \frac{1}{4}ma^2 + \frac{1}{12}ml^2 \) \( I_{yy} = \frac{1}{4}mb^2 + \frac{1}{12}ml^2 \) \( I_{zz} = \frac{1}{4}m(a^2 + b^2) \) \( I_{yy1} = \frac{1}{4}mb^2 + \frac{1}{3}ml^2 \) \( I_{yy} = \frac{1}{2}mr^2 + \frac{1}{2}mh^2 \) \( I_{yy1} = \frac{1}{2}mr^2 + \frac{1}{2}mh^2 \) \( I_{zz} = \frac{1}{3}mr^2 \) \( I_{yy} = \frac{1}{2}mr^2 + \frac{1}{18}mh^2 \) \( I_{xx} = I_{yy} = \frac{1}{2}mr^2 + \frac{1}{2}mh^2 \) \( I_{xx,1} = I_{yy,1} = \frac{1}{2}mr^2 + \frac{1}{6}mh^2 \) \( I_{zz} = \frac{1}{2}mr^2 \) \( I_{zz1} = \left(\frac{1}{2} - \frac{16}{9\pi^2}\right)mr^2 \) \( I_{yy} = \frac{3}{20}mr^2 + \frac{3}{28}mh^2 \) \( I_{yy1} = \frac{3}{20}mr^2 + \frac{1}{10}mh^2 \) \( I_{zz} = \frac{3}{10}mr^2 \) \( I_{yy} = \frac{3}{20}mr^2 + \frac{3}{80}mh^2 \) TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Half Cone Spheroid Elliptic Paraboloid Rectangular Tetrahedron Half Torus MASS CENTER \( \bar{x} = \frac{r}{7}\pi \) \( \bar{z} = \frac{3h}{4} \) \( \bar{z} = \frac{3c}{8} \) \( \bar{z} = \frac{2c}{3} \) \( \bar{x} = \frac{a}{4} \) \( \bar{y} = \frac{b}{4} \) \( \bar{z} = \frac{c}{4} \) \( \bar{x} = \frac{a^2 + 4R^2}{2\pi R} \) MASS MOMENTS OF INERTIA \( I_{xx} = I_{yy} = \frac{3}{20}mr^2 + \frac{3}{28}mh^2 \) \( I_{xx,1} = I_{yy,1} = \frac{3}{20}mr^2 + \frac{1}{10}mh^2 \) \( I_{zz} = \frac{3}{10}mr^2 \) \( I_{zz1} = \left(\frac{3}{10} - \frac{1}{2\pi}\right)mr^2 \) \( I_{xx} = \frac{1}{5}m(b^2 + c^2) \) \( I_{yy} = \frac{1}{5}m(c^2 + a^2) \) \( I_{zz} = \frac{1}{5}m(a^2 + b^2) \) \( I_{yy1} = \frac{1}{5}m\left(a^2 + \frac{18}{11.9}\right) \) \( I_{yy} = \frac{1}{5}m(b^2 + 1.2a^2) \) \( I_{xx} = \frac{1}{12}m(b^2 + c^2) \) \( I_{yy} = \frac{1}{12}m(a^2 + c^2) \) \( I_{zz} = \frac{1}{12}m(a^2 + b^2) \) \( I_{yy1} = \frac{1}{20}m(a^2 + c^2) \) \( I_{yy1} = \frac{3}{80}m(b^2 + c^2) \) \( I_{zz1} = \frac{3}{80}m(a^2 + b^2) \) \( I_{yy} = \frac{1}{2}mR^2 + \frac{5}{4}ma^2 \) \( I_{xx} = I_{yy} = \left(\frac{1}{2}\right)mR^2 + \frac{3}{4}ma^2 \)