KIN 3309 1st Edition Lecture 20Outline of Last Lecture I. TorqueII. LeversIII. Rotational Analog of Newton’s First LawIV. Moment of InertiaV. Angular MomentumVI. Rotational Analog of Newton’s Second LawVII. Rotational Analog of Newton’s Third LawVIII. Conservation of Angular MomentumIX. OutlineX. Centers of Mass & GravityXI. Total Body Center of MassXII. Linear vs. Angular Kinetic AnalysisXIII. Dynamic AnalysisXIV. Angular ImpulseXV. Angular WorkXVI. Angular PowerXVII. Angular EnergyXVIII. Work-Energy TheoremThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.XIX. QuizOutline of Current Lecture I. How to Solve Biomechanics ProblemsII. Question 1III. Angular and Linear VelocityIV. Question 2V. Question 3VI. Question 4VII. Linear vs. Angular Kinetic AnalysisVIII. Question 5IX. Question 6X. Conservation of Angular MomentumXI. Question 7XII. Angular ImpulseXIII. Question 8XIV. Question 9XV. Question 10Current LectureI. How to Solve Biomechanics Problemsa. Step 1i. Identify clearly what the problem is askingb. Step 2i. Identify the information you are givenc. Step 3i. Identify relationshipsd. Step 4i. Combine the given information and the relationshipsII. Question 1a. During the delivery phase of fastball pitch, the arm internally rotates at the shoulder. The angular velocity of this internal rotation peaks at 120 rad/s. At this instant, the elbow angle is 90 degrees, so the angular velocity of the forearm is also 120 rad/s. The baseball in the pitcher’s hand is 35 cm from this axis of rotation through the shoulder joint. At this instant the linear velocity of the baseball is 45 m/s.i. A) How much of the baseball’s total linear velocity is due to the 120 rad/s angular velocity of the forearm?1. 93%ii. B) What is the centripetal acceleration of the baseball at this instant in m/s21. 5040 m/s2iii. C) How large is the force exerted by the pitcher on the baseball to cause this acceleration? The baseball’s mass is 145 g.1. 731 NIII. Angular and Linear Velocitya. The linear velocity can be determined when the length of the segment (radius) and the angular velocity are knownIV. Question 2a. A person wants to push an 110-kg box across the floor. The coefficient of friction between the box and the floor is 0.3.i. A) What is the normal force experienced by the box as it sits on the floor?1. 1079.1 Nii. B) If the person pushes horizontally on the box, how much force must be applied to move the box?1. F > 323.73 Niii. C) If the person pushes against the box with the force calculated in (b), but at an angle of 20 degrees relative to the horizontal, what is the normal force experienced by the box?1. 1189.82 NV. Question 3a. A 608 N women dives from a 10 m platform. No external forces are applied during diving. (No external forces are applied during diving)i. A) What is her potential and kinetic energy 7 m into the dive?1. KE = 4256 Nm2. PE = 1824 Nmii. B) What is her linear velocity 7 m into the dive?1. v = 11.72 m/sVI. Question 4a. An athlete is doing a knee extension exercise using a 140 N dumbbell strapped toher ankle, 37 cm from the knee joint. The athlete holds her leg so that the horizontal distance from her knee to the dumbbell is 28 cm. The weight of her leg is a 33 N with a moment arm (horizontal distance) of 14 cm from her knee joint.i. A) For this position, what torque is created by the dumbbell about her knee joint axis?1. 43.82 Nmii. B) If the moment arm of the knee extensor muscles is 3 cm about the knee joint axis, what amount of force must these muscles produce to holdthe leg in the position described?1. F = 1460.7 NVII. Linear vs. Angular Kinetic Analysisa. Static Analysisi. Systems at rest or constant velocityb. Dynamic Analysisi. Systems in motionVIII. Question 5a. Quentin is lifting a 10 kg dumbbell by pulling upward on it with a 108.1 N force. What is the acceleration of the dumbbell as a result of this force?i. 1 m/s2IX. Question 6a. Kristen is spinning on the ice at 50 rad/s about her longitudinal axis when she abducts her arms and doubles her radius of gyration about her longitudinal axis from 40 cm to 70 cm. If her angular momentum is conserved, what is her angularvelocity about her longitudinal axis after she increases her radius of gyration?i. 16.33 rad/sX. Conservation of Angular Momentuma. When gravity is the only external force acting on an object, the angular momentum remains constanti. This is because the gravitational force acts through the center of mass of an object (point of rotation) and therefore does not produce a torqueb. This allows divers and gymnasts to manipulate their angular velocities by changing their moments of inertiaXI. Question 7a. The average net torque Justin exerts on a discus about its axis of spin is 100 Nm during a throw. The mass of the discus is 2 kg, and its radius of gyration about the spin axis is 12 cm. If the discus is not spinning at the start of Justin’s throwing action, and the throwing action lasts for 0.20 s, how fast is the discus spinning when Justin releases it?i. 694.44 rad/sXII. Angular Impulsea. Torque applied over a period of timeb. Impulse = change in momentumXIII. Question 8a. Tom’s leg angularly accelerates 52.36 rad/s2 around the hip joint during a roundhouse kick in the transverse plane. The moment of inertia of the leg aroundthe axis of rotation for this kick is 0.75 kg-m2. How large is the torque that produces this acceleration?i. 39.27 NmXIV. Question 9a. Doug is driving a gold ball off the tee. His downswing takes 0.50 s from the top ofthe swing until ball impact. At the top of the swing, the club’s angular velocity is zero; at the instant of ball impact, the club’s angular velocity is 30 rad/s. The swing moment of inertia of the club about the grip is 0.220 kg-m2. What average torque does Doug exert on the golf club during the downswing?i. 13.2 NmXV. Question 10a. Sarah’s twist angular momentum increases from 0 to 50 kg-m2/s in 0.25 s as she initiates a twisting jump on the ice. During the 0.25 s, her moment of inertia about her twist axis is 2.2 kg-m2.i. A) How large is the average torque that produces this change in angular momentum?1. 200 Nmii. B) How fast is Sarah’s twist angular velocity at the end of the 0.25 s?1. 22.7