Chapter 4 - Work, Power and Energy 1. How much kinetic energy does a 2 kg discus have if its velocity is 20 m/s? 3. At the top of a giant swing on the gymnastics high bar, Candy's velocity is 1 m/s, and she is 3.5m high. If Candy's mass is 50 kg, what is her total mechanical energy at this instant? 5. An archer draws his compound bow and shoots an arrow. The 23 g arrow leaves the bow with a velocity of 88 m/s. The power stroke of the bow is 57 cm; that is, the bowstring exerts force on the arrow through a displacement of 57 cm. The peak draw weight of the bow is 312 N. (This is the maximum force that the archer has to exert on the bowstring.) a. How much kinetic energy does the arrow have after release? b. How much work does the bowstring do on the arrow? c. What average force does the bowstring exert on the arrow? 8. Which bowling ball has more energy, a 5 kg ball rolling at 4 m/s or a 6 kg ball rolling at 3 m/s? 12. In a vertical jump-and-reach test, 60 kg Nellie jumps 60 cm, while 90 kg Ginger jumps 45 cm. Assuming both jumps took the same amount of time, which jumper was more powerful? 13. Zoe is pole-vaulting. At the end of her approach run, she has a horizontal velocity of 8 m/s and her center of gravity is 1.0 m high. If Zoe's mass is 50 kg, estimate how high she should be able to vault if her kinetic and potential energies are all converted to potential energy. Chapter 5 - Torques and Moments of Force 1. The Achilles tendon inserts on the calcaneus (the heel bone) at a distance of 8 cm from the axis of the ankle joint. If the force generated by the muscles attached to the Achilles tendon is 3000 N, and the moment arm of this force about the ankle joint axis is 5 cm, what torque is created by these muscles about the ankle joint? 3. Katherine is attempting a biceps curl with a 10kg / 100 N dumbbell. She is standing upright, and her forearm is horizontal. The moment arm of the dumbbell about Katherine's elbow joint is 30 cm. a. What torque is created by the dumbbell about the elbow joint? b. What torque is created by Katherine's elbow flexor muscles to hold the dumbbell in this position? Ignore the weight of her arm and hand. c. If the elbow flexor muscles contract with a force of 1000 N and hold the barbell in static equilibrium, what is the moment arm of the elbow flexor muscles about the elbow joint? 7. A pole-vaulter is holding a vaulting pole parallel to the ground. The pole is 5 m long. The vaulter grips the pole with his right hand 10 cm from the top end of the pole and with his left hand 1 m from the top end of the pole. Although the pole is quite light (its mass is only 2.5 kg), the forces that the vaulter must exert on the pole to maintain it in this position are quite large. How large are they? (Assume that the vaulter exerts only verticalup or downforces on the horizontal pole and that the center of gravity of the pole is located at the center of its length.) 8. The patellar tendon attaches to the tibia at the tibial tuberosity. The tibial tuberosity is 7 cm from the center of the knee joint (a). The patellar tendon pulls at an angle of 35 to a line passing through the tibial tuberosity and the center of the knee joint. If the patellar tendon produces an extension torque around the knee joint equal to 400 Nm, how large is the force in the patellar tendon? Chapter 6 - Angular Kinematics 1. Fred examines the range of motion of Oscar's knee joint during his rehabilitation from a knee injury. At full extension, the angle between the lower leg and thigh is 178. At full flexion, the angle between the lower leg and thigh is 82. During the test, Oscar's thigh was held in a fixed position and only the lower leg moved. What was the angular displacement of Oscar's leg from full extension to full flexion? Express your answer in (a) degrees and (b) radians. 4. When Josh begins his discus throwing motion, he spins with an angular velocity of 5 rad/s. Just before he releases the discus, Josh's angular velocity is 25 rad/s. If the time from the beginning of the throw to just before release is 1 s, what is Josh's average angular acceleration? 5. The tendon from Lissa's knee extensor muscles attaches to the tibia bone 1.5 in. (4 cm) below the center of her knee joint, and her foot is 15 in. (38 cm) away from her knee joint. What arc length does Lissa's foot move through when her knee extensor muscles contract and their point of insertion on the tibia moves through an arc length of 2 in. (5 cm)? 6. During Charlie's golf drive, the angular velocity of his club is zero at the top of the backswing and 20 rad/s at the bottom of the downswing just before ball impact. The downswing lasts 0.20 s, and the distance from the club head to the axis of rotation is 2.0 m at the bottom of the downswing. a. What is the average angular acceleration of the club during the downswing? b. What is the linear velocity of the club head just before impact with the ball? 9. Julie runs around the curve of a track in lane 1 while Monica runs around the curve in lane 8. The radius of lane 8 is twice as big as the radius of lane 1. If Julie has to run 50 m to get fully around the curve in lane 1, how far does Monica have to run to get fully around the curve in lane 8? 10. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the fist moves through an arc length of 100 cm. What is the average speed of the fist during the hook? Chapter 7 - Angular Kinetics 1. A tennis racket has a mass of 0.350 kg. The radius of gyration of the racket about its longitudinal (twist) axis is 7.2 cm, and its radius of gyration about its swing axis is 20 cm. a. What is the moment of inertia of this racket about its longitudinal axis? 18,14 kgcm2 b. What is the moment of inertia of this racket about its swing axis? 140kg kgcm2 2. The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2. What is the radius of gyration of this club head? 4. The moment of inertia of the lower leg and foot about an axis through the knee joint is 0.20 kgm2. What is the moment of inertia of the leg and foot about the knee joint if a 0.50 kg shoe is worn on the foot? Assume that the shoe's mass is all concentrated in one point 45 cm from the knee joint. 7. 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? 9. Doug is driving a golf ball off the tee. His downswing takes 0.50 s from the top of the 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 kgm2. What average torque does Doug exert on the golf club during the downswing? Chapter 4 - Work, Power and Energy Q1. KE=1/2mv2 M=2kg V=20m/s KE= *2kg*202m/s=400kg m2/s2 Or 400J Q3 Total Mechanical energy = potential energy + kinetic energy =m*g*h+1/2mv2 M=50kg G=9.81m/s H=3.5m V=1m/s ME=(50kg*9.18m/s*3.5m)+(1/2*50kg*12) ME=1716.75J+25J Me=1741.75J Q5 a) kinetic energy of arrow KE= mv2 m=23g=0.023kg v=88m/s KE=1/2 * 0.023kg* 882m/s KE=89.056J b) Work done by bow string It is the kinetic energy change=KE change=1/2m (v12-v02) M=0.023kg V1=88m/s V0=0m/s (since the initial position of arrow is at rest) W=1/2 *0.023kg*(88-0)2m/s W=89.056J c) Average force W=Fav*d therefore Fav=W/d W=89.056 D=57cm=0.57m Fav=89.056J/0.57m=156.24N Fav=156.24N Q8 Say first ball with 5kg and 4m/s KE=1/2 mv2= *5kg*42m/s=40J For the second ball with 6kg and 3m/s KE=1/2 mv2= *6kg*32m/s=27J From the two, the first ball of 5kg rolling at 4m/s has more energy Q12. WE have to find the potential energy for both First for Nellie PE=m*g*h M=60kg G=9.81m/s H=60cm=0.60m PE=60kg*9.81m/s*0.45m=353.16J Second for Ginger PE=m*g*h M=90kg G=9.81m/s H=45cm=0.45m PE=90kg*9.81m/s*0.45m=397.305J Ginger was more powerful than Nellie Q 13. KE+PErunning=PEjump KE=1/2*m*v2 PE=m*g*hruning PEjump=m*g*hruning *m*82+m*9.8*1=m*g*Hjump Cancelling the m all through *82+9.8=9.8Hjump 41.8=9.8Hjump Hjump=41.8/9.8=4.265m Height he can jump is 4.265m Chapter 5 - Torques and Moments of Force Q1. The 5cm is the moment/effort arm Torque, T=Fr F=3000N r=5cm=0.05m T=3000N*0.05m=150Nm Q3. a) torque created by elbow joint; Torque=F*r=100N*0.3m T= 30Nm b) the torque applied=30Nm now the torque created by the elbow flexor muscle to hold the weight in this position T>30Nm c) moment of flexon muscles=1000N moment arm=r moment = force * moment arm for static equilibrium 30=1000*r r=30/1000=0.03m r=3cm Q7. Fnet = 0 Fr+Fl - Fg = 0 Fr+Fl = Fg = 2.5*9.8 = 24.5 N from top end net torque = 0 Fr*0.1 + Fl*1 - Fg*2.5 = 0 Fr*0.1 + Fl*1 = Fg*2.5 0.1*Fr + (24.5-Fr) = 24.5*2.5 Fr = -40.83 N Fl = 65.33 N Q8. The distance of tibial tuberosity from the center of the knee joint s r=7cm=0.07m Angle=350 Extension torque around the knee joint T=400Nm The force in the patellar tendon F=? But T=Fr sin 35 Therefore F=T/[r sin 35] = 9962.55 N Chapter 6 - Angular Kinematics Q1. a) 17882 = 96 degrees b) 96 degrees = 96*2*pi/360 = 1.675 radians Q4. Average angular acceleration =f-i/t =? f=25m/s i=5m/s t=1s =(25-5)m/s /1s= 20m/s2 Average angular accelaration is 20m/s2 Q5. Set up a ratio in that Arc/distance=arc/distance 2/1.5=x/15 X=2*15/1.5=20 Arc length of tibia =20in Q6. a) Average angular acceleration =f-i/t =? f=20rad/s i=0rad/s t=0.20s =(20-0)rad/s /0.2s= 20rad/s2 b) Vt=r =20 r=2 =20*2=40rad/s Q9. Julie's track 2 * r=50 Monica's track 2 * r1=D Comparing the two, D/50=r1/r Since the radius r1 is twice r we replace with 2r D/50=2r/r D=50*2=100 Monica has to run 100m Q10. Average speed, Vt=r For the arc of 5cm during hook, v=75cm/s Therefore Vt=r for the first hook should be Vt=r for r=100 Since angular velocity for both hook is same, then we compare V100=100cm= 75cm/s=5cm 75*100/5=V100 V100=1500cm/s Chapter 7 - Angular Kinetics Q1. a) Moment of inertia of axis Ia=mK2 M=0.350kg K is radius of gyration=7.2cm Ia=0.35*7.22=18.144kg.cm2 b) Moment of inertia of axis Ia=mK2 M=0.350kg K is radius of gyration=20cm Ia=0.35*202=140kg.cm2 Q2. Moment of inertia of axis Ia=mK2 Where K is radius =? Ia=5000gcm2 M=200g K2=Ia/m=5000gcm2/200g K2=25cm2 K=sqrt 25cm2 K=5cm Q4. Moment of inertia of body rotating about an axis is I=I 0+mr2 I0=0.20kg.m2 M=0.5kg R=45cm=0.45m I=0.20kg.m2+(0.50kg*0.452m2) I=0.20+0.10125 I=0.30125kg m2 Q7. Torque, T=I But I=mr2 M=2kg Radius of gyration=12cm But radius/sqrt2=radius of gyration 12=r/sqrt2 R=12*sqrt2=16.971cm=0.16971m T=I T=100N 100=2*0.69712/2* = 100*2/2*0.69712 =3472.04rad/s2 angular speed /t =3472.04/0.2 =17360.22rad/s Q9. Time t=0.5s Angular velocity at impact =30rad/s Moment of inertia of club about the grip, I=0.22kg m 2 Average torque=(I* -0)/t We subtract 0 since initial condtion is zero velocity Torque = ((0.22*30)-0)/0.5 Torque = 13.2Nm The average torque on the club during downswing is 13.2Nm