Solutions to Selected Introductory and Additional Problems: Week 15 SOLUTIONS TO SELECTED INTRODUCTORY PROBLEMS (Ch 14) 5. A 1.1 kg racquet has a moment of inertia about a grip axis of rotation of 0.4 kg(m2). What is its radius of gyration? m = 1.1 kg I = 0.4 kg•m2 k=? I = mk2 0.4 = 1.1k2 k = 0.60 m 6. How much angular impulse must be supplied by the hamstrings to bring a leg swinging at 8 rad/s to a stop, given that the leg’s moment of inertia is 0.7 kg(m2)? I = 0.7 kg•m2 Angular impulse = Tt ω = 8 rad/s, Tt = ∆H Tt = (Iω1) – (Iω2) Tt = (0.7)(8) – 0 Tt = 5.6 kg•m2/s 1 7. Given the following principal transverse axis moments of inertia and angular velocities, calculate the angular momentum of each of the following gymnasts. What body configurations do these moments of inertia represent? H = Iω A.) H = 3.5(20) = 70 kg(m2)/s B.) H = 7.0(10) = 70 kg(m2)/s C.) H = 15(4.67) = 70 kg(m2)/s 8. A volleyball player’s 3.7 kg arm moves at an average angular velocity of 15 rad/s during execution of a spike. If the average moment of inertia of the extending arm is 0.45 kg(m2), what is the average radius of gyration of the arm during the spike? m = 3.7 kg, ω = 15 rad/s, I = 0.45 kg(m2), k = ? I = mk2 k2 = I/m k = 0.35 m 9. A 50 kg diver in a full layout position, with a total body radius of gyration with respect the her transverse principal axis equal to 0.45 m, leaves a springboard with an angular velocity of 6 rad/s. What is the diver’s angular velocity when she assumes a tuck position, reducing her radius of gyration to 0.25 m? m = 50 kg, kL = 0.45 m, ωL = 6 rad/s, kT = 0.25 m, ωT = ? HL = HT conservation of angular momentum HL = mk2ω HL = 50(0.452)(6) = 60.75 60.75 = 50(0.252)ωT ω = 19.44 rad/s 2 SOLUTIONS TO SELECTED ADDITIONAL PROBLEMS (Ch 14) 1. The radius of gyration of the thigh with respect to the transverse axis at the hip is 54% of the segment length. The mass of the thigh is 10.5% of total body mass, and the length of the thigh is 23.2% of total body height. What is the moment of inertia of the thigh with respect to the hip for males of the following body masses and heights? (A = 0.25 kg.m2, B = 0.32 kg.m2, C = 0.30 kg.m2, D = 0.37 kg.m2) Mass(kg) 60 60 70 70 A B C D Height (m) 1.6 1.8 1.6 1.8 I = mk2 A. I = (60 kg)(0.105)[(1.6 m)(0.232)(0.54)]2 = 0.25 kg.m2 B. I = (60 kg)(0.105)[(1.8 m)(0.232)(0.54)]2 = 0.32 kg.m2 C. I = (70 kg)(0.105)[(1.6 m)(0.232)(0.54)]2 = 0.30 kg.m2 D. I = (70 kg)(0.105)[(1.8 m)(0.232)(0.54)]2 = 0.37 kg.m2 4. A 7.27 kg shot makes seven complete revolutions during its 2.5 s flight. If its radius of gyration is 2.54 cm, what is its angular momentum? (0.0825 kg.m2/s) First, calculate average angular velocity: ω = (7 rev)(2 π rad/rev) / 2.5 s = 17.6 rad/s Then calculate angular momentum: H = mk2ω H = (7.27 kg)(0.0254 m)2(17.6 rad/s) H = 0.0825 kg.m2/s 3 5. What is the resulting angular acceleration of a 1.7 kg forearm and hand when the forearm flexors, attaching 3 cm from the center of rotation at the elbow, produce 10 N of tension, given a 90° angle at the elbow and a forearm and hand radius of gyration of 20 cm? (4.41 rad/s2) T = Iα or Fd⊥ = mk2α (10 N)(0.03 m) = (1.7 kg)(0.20 m)2α α = 4.41 rad/s2 4