Biomechanics 3

A quarterback is chased from the pocket on a crucial 4th down play. He has a mass of 85kg and is running toward the first down marker at 9m/s. The opposing linebacker (mass=100kg) meets him head-on with a velocity of -8m/s. The linebacker wraps up the quarterback during their collision. What is their resulting velocity after the collision?

-.19

A 155 kg football player is running toward another player at 13 m/s. How much average force (in N) needs to be applied over 2.0 seconds to bring him to a stop?

-1,007.5 Negative # F = m x a F = m x (change in velocity / change in time)

If the mass of the foot is 1.16kg and the horizontal and vertical acceleration of the foot are -1.35 m/s2 and 7.65 m/s2, respectively, what is the foot's horizontal joint-reaction force (in N)?

-1.57

Using static analysis solve for the torque of the ball (in Nm).

-101.13 T(ball) = F(ball) x l(ball) = (-250)(0.5cos(36)) = -101.13 Nm KEY b. 𝑇𝑏𝑎𝑙𝑙 = 𝐹𝑏𝑎𝑙𝑙𝑙𝑏𝑎𝑙𝑙 = (−250)(0.5) cos(36) = −101.13 𝑁�

Your team recorded the lower extremity motion and force in a walking study. Given the foot acceleration and force information shown in the diagram, find the horizontal joint-reaction force at the ankle: Fax

-106.9 𝑅𝑥 = 𝑚𝑎𝑥 = 𝑇𝑅𝑥 = 𝑅𝑥𝑟𝑥 =

Consider the free body diagram for the elbow joint for a biceps curl exercise. Up is positive. If d1 = 0.045 m, d2 = 0.16 m; d3 = 0.5 m, using static analysis solve for the moment at the elbow (in Nm).

-2.59 T(elb) = F x L(moment) = (-20)(0.16 x cos(36deg)) = -2.59 Nm Key 𝑇𝑒𝑙𝑏𝑜𝑤 = 𝐹𝑎𝑟𝑚𝑙𝑎𝑟𝑚 = (−20)(0.16) cos(36) = −2.59 𝑁𝑚

During a Bulldogs baseball game, the catcher (mass = 90kg) was analyzed to determine the effects of catching fast balls. He caught a 45m/s pitch over a 0.3 second time period. The mass of the ball is 0.15kg. How much force was created by the impact of the ball against the glove?

-22.5

An ice-skater is spinning counter-clockwise and changes her arm configuration during the routine. At a given instant her angular acceleration is 2 rad/s2 and her angular velocity is 2.5 rad/s. If her current moment of inertia is 4.8 kg.m2, what is the rate (kg.m2/s) at which her moment of inertia is changing? Assume ice is almost frictionless and applies no torque on the skater.

-3.84 The skater is spinning counter-clockwise, so the torque, moment of inertia, angular velocity are all positive. Her angular acceleration is also positive, so angular velocity is increasing. 𝑎 = 𝜔𝑓 − 𝜔𝑖 / ∆𝑡 𝜔𝑓 = 𝑎∆𝑡 + 𝜔i By law of conservation of angular momentum, 𝐼𝑓𝜔𝑓 = 𝐼𝑖𝜔𝑖 , so 𝐼𝑓(𝑎∆𝑡 + 𝜔𝑖 ) = 𝐼𝑖𝜔𝑖 , re-arrange the equation, we obtain (𝐼𝑓 − 𝐼𝑖)/∆𝑡 = −𝑎𝐼𝑓 /𝜔𝑖 = - (2)(4.8) / 2.5 = -3.84

An ice-skater is spinning counter-clockwise and changes her arm configuration during the routine. At a given instant her angular acceleration is 1.1 rad/s2 and her angular velocity is 3.6 rad/s. If her current moment of inertia is 5.2 kg.m2, what is the rate (kg.m2/s) at which her moment of inertia is changing? Assume ice is almost frictionless and applies no torque on the skater.

-3.8Kg x m^2 T = dL / dt - L = Iw T = d(Iw) / dt T = I (dw /dt) + w (dI /dt) T = 0 I (dw/dt) = - w (dI /dt) (dI /dt) = -I/w x (dw/dt) Dw/dt = angular acceleration (dI /dt) = - (I x angular acceleration / w) = - ( 5.2 x 1.1 / 3.6) = - 1.589 KEY What's known: ai = 1.5 rad/s2 𝜔i = 2 rad/s If = 5.1 kgm2 The skater is spinning counter-clockwise, so the torque, moment of inertia, angular velocity are all positive. Her angular acceleration is also positive, so angular velocity is increasing. 𝑎 = 𝜔𝑓 − 𝜔𝑖 ∆𝑡 𝜔𝑓 = 𝑎∆𝑡 + 𝜔𝑖 By law of conservation of angular momentum, 𝐼𝑓𝜔𝑓 = 𝐼𝑖𝜔𝑖 , so 𝐼𝑓(𝑎∆𝑡 + 𝜔𝑖 ) = 𝐼𝑖𝜔𝑖 , re-arrange the equation, we obtain 𝐼𝑓 − 𝐼𝑖 ∆𝑡 = −𝑎𝐼𝑓 𝜔𝑖 = −(1.5)(5.1) 2 = −3.8𝑘𝑔𝑚2 /�

During a basketball throw-in, the player's triceps contract to propel her forearm into launching the ball (mball= 0.45 kg). The player's forearm has a mass of approximately 4.5 kg, the center of mass of the forearm is located 0.17m from the elbow joint center, and the distance along the forearm from the elbow joint center to the ball is approximately 0.35m (shown in diagram). As the forearm begins to rotate, the radius of gyration is .20m. Previous muscle studies in basketball shooting have indicated that, on average, the triceps act a point 0.013 m behind the elbow joint center and produce a force of 750 N. What will be the angular acceleration of the player's forearm as she performs the throw (in rad/s2)?

-4.06 T(cm) = I(cm) x a (Icm / Ia) = (T(ball) + T(fa) + T(triceps)) / (Icm) = (1.54) + (7.48) +(-9.75) /0.18 A = -4.06 KEY According to Newton's 2nd Law, ∑ 𝑇𝐶𝑀 = 𝐼𝐶𝑀𝑎 Torques for the whole arm: ∑ 𝑇𝐶𝑀 = 𝑇𝑏𝑎𝑙𝑙 + 𝑇𝑓𝑜𝑟𝑒𝑎𝑟𝑚 + 𝑇𝑡𝑟𝑖𝑐𝑒𝑝𝑠 ∑𝑇𝐶𝑀 = 𝐼𝐶𝑀𝑎 = 𝑇𝑏𝑎𝑙𝑙 + 𝑇𝑓𝑜𝑟𝑒𝑎𝑟𝑚 + 𝑇𝑡𝑟𝑖𝑐𝑒𝑝𝑠 (0.18)𝑎 = 1.55 + 7.5 + (−9.75) 𝑎 = −3.89 𝑟𝑎𝑑/𝑠 2

Find the vertical joint-reaction force at the ankle: Fay

-517.5 𝑅𝑦 = 𝑚(𝑎𝑦 + 𝑔) 𝑇𝑅𝑦 = 𝑅𝑦𝑟y

Find the net moment (torque) around the ankle.

-80

What is the static friction of the desk (in N)?

-87.5 FsMAX ≤ UsN. Normal force N = Fy - Wdesk = (100)sin(330) - 75 = -125 N So Fs = UsN = (0.7) * 125 = 87.5 N

During a basketball throw-in, the player's triceps contract to propel her forearm into launching the ball (mball= 0.45 kg). The player's forearm has a mass of approximately 4.5 kg, the center of mass of the forearm is located 0.17m from the elbow joint center, and the distance along the forearm from the elbow joint center to the ball is approximately 0.35m (shown in diagram). As the forearm begins to rotate, the radius of gyration is .20m. Previous muscle studies in basketball shooting have indicated that, on average, the triceps act a point 0.013 m behind the elbow joint center and produce a force of 750 N. The torque of the triceps is __________ N.m.

-9.75 T(tri) = F(tri) x d(tri) = (-750) x (0.013) = -9.75 KEY 𝑇𝑡𝑟𝑖𝑐𝑒𝑝𝑠 = 𝐹𝑡𝑟𝑖𝑐𝑒𝑝𝑠𝑑𝑡𝑟𝑖𝑐𝑒𝑝𝑠 = (−750)(0.013) = −9.75 𝑁�

What would be the strain energy stored in this compression?

.02 SE = (1/2)K x (change in distance)^2 = (1/2) 100 x (0.02)^2 = 50 x 0.0004 = 0.02 Elastic energy: SE = 1 2 𝑘∆𝑥 2 = 1 2 (100)(0.02) 2 = 0.02 𝐽

A jumping person (m = 57 kg; vi = 0) has an average GRFy of 750 N for 0.52 seconds.

.15

What is the height Kate jumped for this trial?

.55 ∆𝐾𝐸 = ∆𝑃𝐸 1 2 𝑚𝑣𝑓 2 − 1 2 𝑚𝑣𝑖 2 = 𝑚𝑔∆ℎ 1 2 (73.4)(3.56) 2 − 0 = (73.4)(9.81)∆ℎ ∆ℎ = 0.65 m

A 0.8 kg basketball is dropped from a height of 2.1 m. Video analysis is used to determine that the ball bounced to a height of 0.80 m. What is the coefficient of restitution between the floor and the ball?

.703 ***Different numbers) Slide 96 on Exam 3 ppt E = (relative speed after collision / relative speed before collision) = (V1f - V2f)/(V1i - V2i) Vi^2 = 2g(2.3) = sqrt (45.08) = 6.71 Vf^2 = 2g(0.98) = sqrt (19.208) = 4.38 E = 4.38 / 6.71 = 0.65 𝑒 = √ 𝑏𝑜𝑢𝑛𝑐𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑑𝑟𝑜𝑝 ℎ𝑒𝑖𝑔ℎ𝑡 = √ 0.95 /1.5 = 0.80

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 1.7 m/s2, the vertical acceleration ay= 8.2 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -25 rad/s2. The mass of the foot is 1.3 kg The moment from horizontal joint reaction force is _______ N.m.

0.15 Fx = Rx = m(ax) = (1.3)(1.7) = 2.21 Trx = R(x)r(x) = (2.21) x (0.07) = 0.15 Nm

During a basketball throw-in, the player's triceps contract to propel her forearm into launching the ball (mball= 0.45 kg). The player's forearm has a mass of approximately 4.5 kg, the center of mass of the forearm is located 0.17m from the elbow joint center, and the distance along the forearm from the elbow joint center to the ball is approximately 0.35m (shown in diagram). As the forearm begins to rotate, the radius of gyration is .20m. Previous muscle studies in basketball shooting have indicated that, on average, the triceps act a point 0.013 m behind the elbow joint center and produce a force of 750 N. The moment of inertia ICM of the forearm is _______ kg.m2

0.18 SUMTcm = Icm^a T = Tball + T(forearm) + T (triceps) Icm = M(fa) K^2 = (4.5)(0.20)^2 = 0.18 KEY 𝐼𝐶𝑀 = 𝑚𝑓𝑜𝑟𝑒𝑎𝑟𝑚𝑘𝑓𝑜𝑟𝑒𝑎𝑟𝑚 2 = (4.5)(0.20) 2 = 0.18 𝑘𝑔𝑚2

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 2.1 m/s2, the vertical acceleration ay= 8.7 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -28 rad/s2. The mass of the foot is 1.5 kg The moment from horizontal joint reaction force is _______ N.m.

0.22 𝑅𝑥 = 𝑚𝑎𝑥 = (1.5)(2.1) = 3.15 𝑇𝑅𝑥 = 𝑅𝑥r(x) = (3.15)(0.07) = 0.22

What is the moment of inertia of an object about an axis if there are three point masses (m1 = 1.35 kg, m2 = 1.60 kg, m3 = 2.20 kg) located at r1 = 0.07 m, r2 = 0.17 m, and r3 = 0.30 m distance from the axis of rotation (in kgm2)?

0.2509 I = m x r^2 = (m1 x r1^2) + (m2 x r2^2) + (m3 x r3^2) 𝐼 = ∑𝑚𝑖𝑟1 2 𝑁 𝑖=1 = 𝑚1𝑟1 2 + 𝑚2𝑟2 2 + 𝑚3𝑟3 2 = (1.5)(0.10) 2 + (1.75)(0.20) 2 + (2)(0.35) 2 = 0.015 + 0.07 + 0.245 = 0.33𝑘𝑔𝑚2

A subject weighing 800 N performs a simple heel raise. The weight of the body is assumed to pass through the heads of the metatarsals of the standing foot. The center of motion of the ankle joint is marked with the X. The distance from the metatarsal heads to the ankle rotation is 19.1 cm (this is the moment arm distance when the heel is near the ground). When the heel is near the ground, the moment arm of the force through the Achilles tendon is 8.1 cm. When the patient performs a calf raise, he lifts his heel to an angle (between the foot and the ground) of 58º (assume the body weight still passes through the metatarsal heads at a distance of 19.1cm, but now with an associated angle due to the foot lifting). The moment arm of the force through the Achilles tendon in the raised position is 4.2 cm. Disregard the weight of the foot. What is the force (F) transmitted through the Achilles tendon when the patient's just lifts his foot off the ground (his foot is near the ground as in figure A) (in N)?

1,886.42 Static analysis: E(M) = F(m) x r(m) - F(mh) x r(mta) F(m) = (F(m) x r(mta) /r(m) F(m) = (800 x 0.191) /0.081 = 1886 N

What is the average impact force exerted by the glove on the baseball?

1,898.4

What is the force transmitted through the Achilles tendon when the patient performs a calf raise by lifting his heel from the floor?

1,927.90 Fm = (F(mt) x r(mt)) / r(m) Fm = (800 x 0.191/cos(58 deg)) /0.042 = 1929 N *** More Force***

What is the net moment (torque) at the ankle joint during the swing phase of walking? Given the following information: Foot mass =1.16 kg ax=-1.35 m/s2 ay=7.65 m/s2 ICM=0.0096 kg-m2 αz= -14.66 rad/s2

1.17

If an object achieved an angular acceleration of 32 rad/s2 due to a torque of 38 Nm, what was the object's moment of inertia (in kg.m2?

1.5 T = I x angular velocity (AV) I = T / AV KEY 𝑇 = 𝐼𝑎 𝐼 = 𝑇 / 𝑎 = 45 / 30 = 1.5𝑘𝑔𝑚2

During a basketball throw-in, the player's triceps contract to propel her forearm into launching the ball (mball= 0.45 kg). The player's forearm has a mass of approximately 4.5 kg, the center of mass of the forearm is located 0.17m from the elbow joint center, and the distance along the forearm from the elbow joint center to the ball is approximately 0.35m (shown in diagram). As the forearm begins to rotate, the radius of gyration is .20m. Previous muscle studies in basketball shooting have indicated that, on average, the triceps act a point 0.013 m behind the elbow joint center and produce a force of 750 N. The torque of the ball is __________ N.m.

1.54 Tball = W(ball) x d(ball) = M(ball)g x d(ball) = (0.45)(9.81) x (0.35) = 1.54 Nm KEY b. 𝑇𝑏𝑎𝑙𝑙 = 𝑊𝑏𝑎𝑙𝑙𝑑𝑏𝑎𝑙𝑙 = 𝑚𝑏𝑎𝑙𝑙𝑔𝑑𝑏𝑎𝑙𝑙 = (0.45)(9.81)(0.35) = 1.55 𝑁m

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 1.7 m/s2, the vertical acceleration ay= 8.2 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -25 rad/s2. The mass of the foot is 1.3 kg The moment from vertical joint reaction force is _______ N.m.

1.64N x m 𝑅𝑦 = 𝑚(𝑎𝑦 + 𝑔) = (1.3)(8.2 + 9.81) = 23.41𝑁 𝑇𝑅𝑦 = 𝑅𝑦𝑟𝑦 = (23.41)(0.07) = 1.64𝑁 ∙ m

A jumping person (m = 57 kg; vi = 0) has an average GRFy of 750 N for 0.52 seconds. What's their vertical takeoff velocity?

1.74

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 1.7 m/s2, the vertical acceleration ay= 8.2 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -25 rad/s2. The mass of the foot is 1.3 kg Given the acceleration about the center-of-mass and other computed forces, the muscle torque acts on the ankle joint is _______ N.m.

1.77 T(ankle) = sum of T(a) + T(rx) + T(ry) Tank = (Icm)(a) Tank = (0.001)(-25) + (0.155 + 1.638) = 1.77 Nm *** dorsiflexion c. ∑ 𝑇𝐶𝑀 = 𝐼𝐶𝑀𝑎𝑧 𝑇𝑎𝑛𝑘𝑙𝑒 = ∑𝑇𝐶𝑀 + 𝑇𝑅𝑥 + 𝑇𝑅𝑦 = 𝐼𝐶𝑀𝑎𝑧 + 𝑇𝑅𝑥 + 𝑇𝑅𝑦 𝑇𝑎𝑛𝑘𝑙𝑒 = (0.001 × (−25)) + 0.15 + 1.64 = 1.77𝑁 ∙ m

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 2.1 m/s2, the vertical acceleration ay= 8.7 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -28 rad/s2. The mass of the foot is 1.5 kg The moment from vertical joint reaction force is _______ N.m.

1.94 𝑅𝑦 = 𝑚(𝑎𝑦 + 𝑔) = (1.5)(8.7 + 9.8) = 27.75 N 𝑇𝑅𝑦 = 𝑅𝑦𝑟𝑦 = (27.75) x (0.07) = 1.94 N x m

Compute I in the figure. ICM is the moment of inertia about an axis passing through the center of mass of the object.

12.1 Iprox = Icm + mr2 = 12 + (2.5 x 0.1^2) = 12.1 KEY 𝐼𝑝𝑟𝑜𝑥 = 𝐼𝐶𝑀 + 𝑚𝑑 2 = 12 + (2.5)(0.2) 2 = 12.1𝑘𝑔𝑚2

As an expert biomechanist, you've just been hired by Schuttto supervise a team designing a safer helmet to reduce concussions in NFL football players. The team presents you with two models which simulate a player's head hitting the inside of the helmet. During both simulations, the helmet is hit with the same external force and the model of the head is assumed to have a mass of 3.6 kg. The first simulation shows that the exterior of Helmet 1 buffers some of the force so that the player's head is moving at 5 m/s immediately after application of the external force. The inside of Helmet 1 is cushioned so that the cushion gives for .15 seconds before the head comes to rest. What is the force transferred to the player's head in the first simulation?

120 F = m x a F = m x (change in V / change in T) = 3.6 x (5/ 0.15) = 120 KEY mhead = 3.6 kg Helmet 1: v1 = 5 m/s, v2 = 0 (the player's head comes to rest), t = .15 seconds Helmet 2: v1 = 5.3 m/s, v2 = 0, t = .22 seconds In order to know which helmet's safer, it makes sense that we would want to know how much force is being transferred to the player's head. So, what do we know? For each simulation, we know velocities at two different time points, we know a force is being transferred to the head, we know the time of impact, and we know the mass of the head. Since we know mass and velocity at two time points, it leads us in the direction of linear momentum and, since we know the time and we're curious about the force, let's look at impulse. Impulse = Ft = ΔM = Δmv= m2v2-m1v1 For Helmet 1,Ft = m2v2-m1v1F(.15 s) = (3.6 kg)(0 m/s) -(3.6 kg)(5 m/s) F = -120 N

How much work does the catcher do on the baseball during the catch (J)?

151.9

The displacement of the baseball due to the deformation of the catcher's glove and the movement of the catcher's hand is 8 cm in a horizontal direction from the instant it first makes contact with the glove until it stop. How much kinetic energy does the baseball possess just before it strikes the glove (J)?

151.9

What would be the force (in N) exerted by a spring with a stiffness of 100 N/m that was compressed 0.02 m?

2 F = K(change in distance) F = 100 x (0.02) = 2 Elastic force: F = k(delta)x = 100(0.02) = 2 N

Using static analysis solve for the net muscle force, Fm (in N).

2,305 (D3 x cos(36) x 250) + (d2 x cos(36) x 20) - (Fm x d1) = ((250)(0.5) (cos (36)) + ((0.16)(20)(cos(36)) - (0.045) Fm = 0 = (101.137 + 2.589 ) / 0.045 Fm = 2305.022 KEY In static analysis, ∑ 𝑇 = 0, so 𝑇𝑚 + 𝑇𝑒𝑙𝑏𝑜𝑤 + 𝑇𝑏𝑎𝑙𝑙 = 0 𝐹𝑚𝑑1 = −𝐹𝑎𝑟𝑚𝑙𝑎𝑟𝑚 − 𝐹𝑏𝑎𝑙𝑙𝑙𝑏𝑎𝑙𝑙 So 𝐹𝑚 = −𝐹𝑎𝑟𝑚𝑙𝑎𝑟𝑚−𝐹𝑏𝑎𝑙𝑙𝑙𝑏𝑎𝑙𝑙 𝑑1 = −(−20)(0.16)cos(36)−(−250)(0.5) cos(36) 0.045 = 2305�

Kate is taking part in jumping drills at the start of the University of Georgia basketball season to measure her maximum vertical jump height. She is performing non-counter movement jumps from a force platform which enables her vertical ground reaction forces to be recorded. She has a mass of 73 kg. For a particular jump, the trainer recorded an average vertical ground reaction force (FG,y) of 2130 N applied over a duration of 185ms. What is the average power produced by the legs to increase the body's kinetic energy (in W)?

2,533.01 So ∆𝐾𝐸 = 1 2 𝑚𝑣𝑓 2 − 1 2 𝑚𝑣𝑖 2 = 1 2 (73.4)(3.56)^2 = 465.12 𝑃 = 𝑊 ∆𝑡 = ∆𝐾𝐸 ∆𝑡 = 465.12 0.185 = 2514.2 𝑊

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 2.1 m/s2, the vertical acceleration ay= 8.7 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -28 rad/s2. The mass of the foot is 1.5 kg Given the acceleration about the center-of-mass and other computed forces, the muscle torque acts on the ankle joint is _______ N.m.

2.13 N x m ∑ 𝑇𝐶𝑀 = 𝐼𝐶𝑀𝑎𝑧 𝑇𝑎𝑛𝑘𝑙𝑒 = ∑𝑇𝐶𝑀 + 𝑇𝑅𝑥 + 𝑇𝑅𝑦 = 𝐼𝐶𝑀𝑎𝑧 + 𝑇𝑅𝑥 + 𝑇𝑅𝑦 𝑇𝑎𝑛𝑘𝑙𝑒 = (0.001 x (-28)) + 0.22 + 1.94 = 2.13 N x m

During the support phase of running, if the vertical ground reaction force was 1,000 N for a runner (mass = 80 kg), what would be the acceleration of the center of mass?

2.7 m/s2 Sum Fy = GRF - W ma(y) = 1000N - mg (80) ay = 1000- (80) x (9.8) a = 2.7 m/s2 acceleration of the center of mass is 2.7 m/s2

If the mass of the foot is 1.16kg and the horizontal and vertical acceleration of the foot are -1.35 m/s2 and 7.65 m/s2, respectively, what is the foot's vertial joint-reaction force (in N)?

20.3

A football fan is trying to find her seat. Her seat is in the 90th row. The difference in height between the seats is 35cm. The fan weighs 800N. How much work does she do by climbing to the 90th row (in J)?

25,299 W = F x d W = 800 x 31.50 W = 25200 a. W = Fd F = 800 N d = 90(0.35) = 31.5 m W = 800(31.5) = 25200 J b. P = W/(delta)t P = 25200 / 90 = 280 W

What is the drag force (in N) Harry experiences at this speed?

27.2

How much power is generated if she climbs to her seat in 90 seconds (in W)?

280 P = W / change in time P = 25200 / (90 sec) P = 280 W

An ultimate Frisbee player (body mass of 54 kg) sprints at 7.1 m/s and makes a quick turn to her right. In the stance phase during this turn, her vertical acceleration is 5.0 m/s2 and her horizontal deceleration is -2.0 m/s2. If the static coefficient of friction between her foot and the turf is 0.8, how small of a radius (m) can she have on her turn not to slip? Assume that air resistance is negligible.

4.32 KEY What's known: Body mass m = 54kg Tangential velocity v = 7.1m/s Vertical acceleration az = 5.0 m/s2 Horizontal acceleration ax = -2.0 m/s2 Coefficient of static friction u = 0.8 The question asks the smallest radius allowed by the friction force. The key relationship here is: FsMAX ≤ UsFN = Ff The friction force is provided by the centripetal force, horizonal GRF, and normal force (vertical GRF). Centripetal force: 𝐹𝑐 = 𝑚𝑎𝑐 = 𝑚 ( 𝑣𝑇 2 𝑟 ) = 54 ( 7.1 2 𝑟 ) Normal force: 𝐹𝑁 = 𝐹𝑔𝑟𝑓,𝑧 = 𝑚(𝑎𝑧 + 𝑔) = 54(5 + 9.81) = 799.74𝑁 Horizontal GRF: 𝐹𝑔𝑟𝑓,𝑥 = 𝑚𝑎𝑥 = 54(−2.0) = −108𝑁 Friction force is the resultant force from centripetal and normal force: 𝐹𝑓 2 = 𝐹𝐶 2 + 𝐹𝑔𝑟𝑓,𝑥 2 = (54 ( 7.1 2 𝑟 )) 2 + (−108)^2 UsFN = Ff ((0.8)(799.74))^2 = (54 ( 7.1 2 𝑟 )) 2 + (−108)^2 Solve for r, we have 𝑟 = (54)(7.1) 2 630.61 = 4.3𝑚

What is the COM of the arm in this diagram (x value)?

4.34

What would be the centripetal force of an object (m = 3 kg) rotating about a radius of 0.5 m at 1.8 rad/s?

4.86 Fc = mw^2r KEY Centripetal acceleration is 𝑎𝐶 = 𝑣𝑇 2 𝑟 = 𝜔 2 𝑟 U sing Newton's 1st Law again, 𝐹𝐶 = 𝑚𝑎𝐶 = 𝑚𝜔2 𝑟 = 2(1.5) 2 (0.5) = 2.25𝑁

What is the centripetal force (in N) Harry experiences at this speed?

433.3

If a 200 N force (F) was applied for 4s (∆t) to a 20-kg (m) object initially moving at 6 m/s (vi), what would be the object's final velocity (vf)?

46 m/s F = m x a F = m x (Vf - Vi / delta t ) 200 = 20kg (Vf - 6 / 4) Vf = 46 m/s

What is the COM of the arm in this diagram (y value)?

5.47

A person (mg = 800 N) walks 1000 m on a 45° uphill slope. How much mechanical work (in J) is required to lift the object up the hill?

565,680

A 3kg mass is accelerating at 2m/s2 in a straight line. How much force is pushing the mass?

6 F = m x a

Harry (80 kg) is riding a bike (8.5 kg) on wet concrete. He wants to go around a circular path with a radius of 10.5 m but does not want to fall. The coefficient of static friction between the bicycle tire and the wet concrete is 0.5. Assume that the both tires provides the propulsive force to overcome air resistance and that the vertical and lateral ground reaction forces are both equally distributed between the two tires. The frontal surface area of the rider and bike is 0.54 m2, the drag coefficient of the rider and bike is 0.82, and the density of air is 1.2 kg.m3.

7.17

During a basketball throw-in, the player's triceps contract to propel her forearm into launching the ball (mball= 0.45 kg). The player's forearm has a mass of approximately 4.5 kg, the center of mass of the forearm is located 0.17m from the elbow joint center, and the distance along the forearm from the elbow joint center to the ball is approximately 0.35m (shown in diagram). As the forearm begins to rotate, the radius of gyration is .20m. Previous muscle studies in basketball shooting have indicated that, on average, the triceps act a point 0.013 m behind the elbow joint center and produce a force of 750 N. The torque of the forearm is __________ N.m.

7.5 T(fa) = W(fa) x r(fa) = (m)(g)(r) = (4.5)(9.81)(0.17) = 7.5 KEY 𝑇𝑓𝑜𝑟𝑒𝑎𝑟𝑚 = 𝑊𝑓𝑜𝑟𝑒𝑎𝑟𝑚𝑟𝑓𝑜𝑟𝑒𝑎𝑟𝑚 = 𝑚𝑓𝑜𝑟𝑒𝑎𝑟𝑚𝑔𝑟𝑓𝑜𝑟𝑒𝑎𝑟𝑚_𝐶𝑂𝑀 = (4.5)(9.81)(0.17) = 7.50 𝑁m

A non-Biomechanics student is trying to move a desk. He pushes on the desk with a force of 100N at an angle of 330°. The desk weighs 75N and the coefficient of static friction is 0.70.

86.6 CaH = Cos (angle) = h / adj 100 x cos (330) = 86.60

The second simulation shows that the exterior of Helmet 2 buffers some of the force so that the player's head is moving at 5.3 m/s immediately after application of the external force. After the force reaches the head, the inside of Helmet 2 is cushioned so that the cushion gives for .22 seconds before the head comes to rest. What is the force transferred to the player's head in the second simulation?

86.7 F = m x a F = 3.6 x (5.3 / 0.22) = 86.73 KEY For Helmet 2,Ft = m2v2-m1v1F(.22 s) = (3.6 kg)(0 m/s) -(3.6 kg)(5.3 m/s) F = -86.73 N

For any set amount of applied elbow flexor muscle force, the largest torque is produced when the muscular line of pull is at an angle of _____ from the ulna.

90 deg

If all of the following objects are at rest, which will exhibit the highest potential energy?

A .15 kg baseball stuck above a speaker 50 meters above the ground

Which of the following best describes inertia?

A measure of an objects resistance to a change in motion

During the stance phase of human walking, which lower limb joint usually has the greatest muscle power?

Ankle

During diving, when the diver has left the board (vertical line hash mark in the figure), what could happen to the local and remote angular momenta for different body segments?

Both local and remote angular momenta can change for all segments

Consider a situation when only two torques are acting about the elbow: torque due to the biceps brachii and torque due to resistance (i.e., weight). During an elbow flexion exercise, if the biceps brachii generate 250 N of force with a moment arm of 0.05 m, what type of contraction would occur if the resistance force was 62 N (moment arm = 0.2 m)?

Concentric b/c the torque of resistance is less than the biceps

During figure skating, a skater can manipulate angular velocity despite no external torques being applied to her. Which mechanical principle can explain this phenomenon?

Conservation of angular momentum

Which axis of rotation has the lowest moment of inertia?

D

What kind of movement the foot undergoes?

Dorsiflexion

The mechanical advantage of a third-class lever is always greater than the mechanical advantage of a second-class lever.

False

What type of lever is this?

First Class

Describe the difference among major classes of levers and give an example in the human body for each class.

First class- the effort force and the resistance force are on opposite sides of the fulcrum. Example: patella in knee extention Second Class- , the effort force and the resis- tance force act on the same side of the fulcrum, not many examples but a disputed one its calf raises Third class- The effort force and the resistance force are also on the same side of the fulcrum, arm held in flexion at the elbow

Which of the following statements best describes pressure force?

Force per unit area

Where to place COG to get a quick start in walking/running?

Front edge of the BoS

Which helmet design would you suggest is safer/more effective in reducing concussions if concussions are caused primarily by force application to the head?

Helmet 2 So, since Helmet 2transfers less force than Helmet 1 to the player's head, we can conclude that Helmet 2 appears to be safer and more effective in reducing concussions.

Try pushing a heavy book across a table. Is it easier to get it started moving or keep it moving?

It's easier to keep it moving.

Which term best represents energy resulting from motion?

Kinetic energy

Explain the difference between kinetic energy, potential energy, and elastic energy, and how they convert from one to another. Use equations and examples if needed.

Kinetic energy (KE) is the energy resulting from linear motion. The formula is KE = 1/2(mv^2). Potential energy (PE) is the capacity to do work due to a body's position within the gravitational field. The formula is PE = mgh. Elastic energy (SE) is the capacity to do work due to deformation of a body energy stored when a spring is stretched or compressed. The formula is SE = 1/2(k x change(x)^2)

Eddie imparts side spin on the ball to try and fool the goal keeper by striking it with the inside of his kicking (right) foot (see the figure below). Which direction the ball will move to eventually?

Left The ball spins and moves in the air. The spin is in the same direction of the air drag on the left side of the ball, making the velocity increase and air pressure decreases; whereas on the right side, the spin is in the opposite direction, causing decreased velocity and increased air pressure. As a result, the ball will eventually turn towards the left.

In an experiment on patients with Parkinson's disease, patients' gait data were collected, and joint angles over gait cycle were plotted below. a. Use the kinematics and kinetics knowledge you learned from this class, explain what are the major abnormalities in PD patients walking pattern. b. Relating to the concept of vigor and neural mechanism discussed in this course, what are the possible reasons that cause these deficits?

Looking at the chart in the PD group the cadence mean is higher than the control group. PD patients tend to have a shuffle in their gait pattern which causes their speed (velocity) to be slower, and the length of the stride to be shorter. Look at the graphs the overall power in the hip, knee, and ankle is greater in the control group than in the PD groups. Those with Parkinson's disease have a loss of dopamine in the brain, which leads to the parkinsonian symptoms such as slow gait, stooped posture, shuffling gait, instability, increased risk of falling, as well as difficulty initiating gait. Dopamine is a reward signal that determines movement vigor. Dopamine from the substantia nigra to the neostriatum regulates the likelihood of moving at a certain speed. Those with PD have a higher chance of moving slowly because of the depleted dopamine in their system causing a distortion of speed selection mechanisms. Movement speed is determined by movement energy cost. Movement energy cost is created in response to movement vigor. Therefore, dopamine from the substantia nigra carries a signal for motor motivation. In Parkinson's patients they show a reduction in reduced movement vigor. Vigor interacts with reward in human movement, which with PD patients the vigor is reduced overall in their movement.

Describe magnus effect and give an example in sport.

Magnus effect is a projectile with air drag, which is when a spinning ball is going to change its path. The ball takes a curved path. An example is in soccer the way a soccer ball can curve. For example, the Roberto Carlos goal where his kick called the soccer ball to anti-clockwise spinning ball decreases pressure on its left side causing the ball to curve in that direction. Generation of a sidewise force on a spinning cylindrical or spherical solid immersed in a fluid when there is relative motion between the spinning body and the fluid. Explains the curved path a ball can take while in the air. Occurs only if the object is spinning.

The drag force is a component of fluid resistance that always acts opposite to the motion of the person. This force is not proportional to which of the following:

Mass of the person

Explain succinctly the differences between: moment (torque) and moment of inertia. Use equations, examples and specify units if needed.

Moment (torque) is when a force is applied such that it causes a rotation, the product of that force and the perpendicular distance to its line of action. Its not a force but defined as the tendency of a force to cause a rotation about a specific axis. Its a vector quantity and thus has magnitude and direction. Torque uses the formula T = F x perpendicular Radius. a body's resistance to a change in angular motion Depends on mass and distribution of mass w.r.t. the axis of rotation If ∆T=0, then ∆ω=0, ice skater with arms in versus out

Explain the difference between momentum and impulse and their relationship. Use equations, examples, and specify units if needed.

Momentum (p) is mass times velocity (p = m x v), and the units are kg x m/s. Impulse is the force x the time interval, and the units are Ns. Impulse is the change of momentum. Creates a force-time curve, you can increase momentum by increasing the force or increase the time interval.

Does he move the desk?

No KEY 86.6N < 87.5N. No, he does not move the desk.

If a collision between two bodies results in a coefficient of restitution of 1.0, this collision can be described as:

Perfectly elastic

Explain the difference between elastic and inelastic collision using the concept of momentum. Use equations and examples if needed.

Perfectly elastic collision is the collision during which the velocity of the system is conserved. Perfectly inelastic (plastic) collision is the collision resulting in the total loss of system velocity. Momentum still conserved, but objects stick together instead of bouncing apart. Football tackles are an example. Everything we do is inelastic but not perfectly. A coefficient of restitution of 1 describes an elastic collision and a value of 0 results in a perfectly inelastic collision where each particle travels at the same speed, example golf ball is <.83, e = alive speed after collison / relative speed before collison

What type of lever is this?

Second class

When walking/riding uphill, do we experience larger or smaller friction compared to that on a flat surface?

Smaller

Which requires the larger muscle force?

Standing on the toes

Use the kinetics knowledge you learned from this class, analyze the following free-body diagram of a person lifting an object, and explain what are the factors for a good lifting strategy to avoid back injury.

The diagram shows the force weight on the head, trunk, arms, and box. The free-body diagram is showing the compression reaction force, shear reaction force, and muscle tension in the L5 vertebrae. It shows how when you lift a box correctly it will help minimize the load put on the back. The best way to lift any box is to squat down with your legs and grab the box and then stand up with your legs. The factor of you squatting down and lifting your legs takes away from having the need to bend your back. You want to lift with your leg muscles, and not your back, keep the load as close to your body as you can and maintain the natural curve in your spine. This helps the force/pressure to be dissipated through the body and not just centered in the back. The bending down and picking up would put force/pressure on your spinal cord. The load on the discs is a combined result of the object weight, the upper body weight, the back muscle forces, and their respective lever arms to the disc center.

Which of the following does not affect the magnitude of static friction force before it reaches to the maximum value?

The mass of the object

During a countermovement jump, there is a period of time where forces acting on the person from the ground are less than body weight (see figure below). Which of the following statements best explains this observation?

Velocity is increasing in the negative direction.

During walking, the order of magnitude of GRF from highest to lowest is

Vertical, Anteroposterior, Mediolateral

Explain the law of conservation of angular momentum. Use an example in sport.

When gravity is the only external force acting on an object, the angular momentum remains constant. This is because the gravitational force acts through the center of mass of an object (point of rotation) and therefore does not produce a torque. In diving, Angular momentum (H) at the instant of takeoff remains constant throughout the dive •When the diver brings in her legs mid-air The radius of gyration (k) is decreased Local moment of inertia (I) is reduced Increasing angular velocity (𝜔) The tighter the tuck, the greater the angular velocity

During running along a straight line at a steady velocity (acceleration = 0), the average braking and propelling impulses are equal (see Figure A below). In Figure B, the person showed

decreased acceleration and increased negative impulse in the breaking phase.

Backspin in a golf ball

helps to keep the ball in the air longer

During the second half of elbow flexion, the net muscle torque will be ______, and the power will be _____.

negative, negative there is a vertical line that goes down the middle of each graph. On the left side is flexion and the right side is extension. So when I asks about the second half of elbow flexion you are looking at the second half of the left section. Then the second graph will show the torque and the bottom graph shows power! And they are both negative in that section

During the first half of elbow extension, the net muscle torque will be ______, and the power will be _____.

negative, positive there is a vertical line that goes down the middle of each graph. On the left side is flexion and the right side is extension. So when I asks about the second half of elbow extension you are looking at the first half of the right section. Then the second graph will show the torque and the bottom graph shows power! And they are both negative in that section

In the middle of the swing phase of a gait cycle in brisk walking, a walker had the horizontal acceleration ax= 2.1 m/s2, the vertical acceleration ay= 8.7 m/s2, the moment of inertia of the foot about its center of mass is ICM=0.001 kg-m2 and the angular acceleration of the foot is 𝛼𝑧= -28 rad/s2. The mass of the foot is 1.5 kg The foot undergoes _________ action.

plantarflexion

Which of the following makes the statement FALSE? Increasing the amount of knee flexion that occurs during landing from a jump ________.

requires less quadriceps muscle fiber lengthening

Describe key factors affecting stability.

the first factor is where the line of gravity falls with respect to the base of support. An object is most stable if the line of gravity is in the geometric center of the base of support. Increasing the area of the base of support generally increases the stability, Increasing the base of support to allow the line of gravity to fall within the base of support, The stability of an object is also inversely proportional to the height of the center of mass. mass of the object increases stability as well.

During running, the dynamic coefficient of friction can be calculated using which formula (Fy= anterior/posterior; Fx = medial/lateral; Fz= vertical force components)?

μ=Fy/Fz KEY Using the general formular for friction Ff = N, = Ff / N. The normal force N is the vertical Ground Reaction Force, Fz, the friction is the anterior/posterior GRF, Fy. So = Ff / N = Fy / Fz The answer is D.