Walking with a steady cadence is very important for covering long distances efficiently. How we place our feet, and how quickly we walk, also depends on the roughness of the surface we are walking upon and on the slope of the surface: we walk carefully on slippery surfaces, and take smaller steps whenhiking up a hill. When we are walking at constant speed in a fixed direction, the horizontal and vertical components of the acceleration of our center of mass must be zero. In addition, the sum of torques about the body's center of mass must also be zero. Consider the situation shown in the figure below.ALMAXCMХ СМXCM XCMXCMWe can model the walking gait of a person as a swing of the front leg and torso about the point where the front foot is planted (shown with a red circle in the figure) and a rotation of the trailing leg about the center of mass (CM) of the person. If each leg of this 78.0 kg person is 85.0 cm long and has a massof 13.8 kg, and 0; = 0₁ = 20.0°, what is the angular momentum (in kg • m²/s) of each leg about the center of mass of the person if the sequence of images shown in the figure spans 1.35 s? (Model each leg as a long, thin rod that does not bend at the knee.)(No Response) kg m2/s

Answered: Image /qna-images/answer/a5bb3118-bcbd-41c6-a47a-f743fb08d99c.jpg

Answered: Walking with a steady cadence is very…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter10: Systems Of Particles And Conservation Of Momentum

Section: Chapter Questions

Problem 77PQ

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Three solid, uniform boxes are aligned as in the figure below. Find the x- and y-coordinates (in m) of the center of mass of the three boxes, measured from the bottom left corner of box A. (Consider the three-box system.) HINT 0.200 m 0.280 m 0.120 m y A B C 0.350 m Origin 0.750 kg 1.00 kg 0.650 kg Х ст E m m Уст x
The motion of a human body through space can be modeled as the motion of a particle at the bodys center of mass as we will study in Chapter 9. The components of the displacement of an athletes center of mass from the beginning to the end of a certain jump are described by the equations xj = 0 + (11.2 m/s) (cos 18.5)t 0.360 m = 0.840 m + (11.2 m/s)(sin 18.5)t - 12(9.80 m/s2)t2 where t is in seconds and is the time at which the athlete ends the jump. Identify (a) the athletes position and (b) his vector velocity at the takeoff point. (c) How far did he jump?
To get up on the roof, a person (mass 70.0 kg) places 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to frictionless. The center of ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base.
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In Chapter 9, we will define the center of mass of an object and prove that its motion is described by the particle under constant acceleration model when constant forces act on the object. A gymnast jumps straight up, with her center of mass moving at 2.80 m/s as she leaves the ground. How high above this point is her center of mass (a) 0.100 s, (b) 0.200 s, (c) 0.300 s. and (d) 0.500 s thereafter?
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A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.12a). His motion through space can be modeled precisely as that of a particle at his center of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his hang time), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.12b) with center of mass elevations yi = 1.20 m, ymax = 2.50 m, and yf = 0.700 m. Figure P4.12
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