Problemset6%20key%202009F - Biomechanics IPHY 4540 Problem...

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Biomechanics IPHY 4540 Problem Set #6 Key 1. When running, your body (center of mass) is about 3.5 cm higher at the instant that you toe-off than at the instant that you land (on your other heel). Aerial time is 0.18 s. a. What is your vertical take-off velocity? you can set this problem up several ways. you can declare that the intial take off point is r yi =0 and r yf = -0.035m or you can: r yi =0.035m, r yf = 0m, a y = -9.8m/s 2 , t= 0.18 sec solve for v yi r yf –r yi = v yi t =0.5at 2 0 – 0.035 = v yi • (0.18) + 0.5 • -9.8 • (0.18) 2 v yi = +0.68m/s b. How much height do you gain after toe-off? again there are multiple ways to solve this, one way is to solve for time first, then solve for r y . at the top of the trajectory, vyf = 0 m/s. v yi = +0.68 m/sec, a = -9.8m/s 2 , solve for t using: v yf = v yi + at 0 = 0.68 + -9.8t; t= 0.07 sec. can re-assign what the starting point r yi is: r yi = 0, v yi = + 0.68m/s, a = -9.8 m/s 2 , & t = 0.07sec, solve for r yf . r yf –r yi = v yi t + 0.5 at 2 r yf = 0.024m so from take off you go up 2.4cm, then fall back to starting point and land 3.5cm below that. 2. A. Draw a graph of elbow angle versus time during a pull-up (both up and down parts of the exercise). Begin at the bottom of the motion. The up and down phases each take 1.0 seconds. The motion is smooth and steady. Indicate approximate magnitudes of the angle on your vertical axis. Define the elbow angle as the angle between the upper and lower arm segments. this graph should be theta vs. time. theta should start at 180 degrees (or π radians), decrease to a small angle at 1 second, then increase back to 180degrees. The shape of the graph should not be like a V. "smooth and steady" means that accelerations are small, not jerky. the inflection points should be rounded off. “steady” could be interpreted as having a nice constant angular velocity during the movement before and after the turnaround. so RK’s ideal answer is curved inflection points with lines of constant slope connecting those curves. i.e. constant angular velocity during the middle of the way up and middle of way down. but, this is a matter of interpretation. does "smooth and steady" mean minimize the peak angular accelerations “or “do your best to have long durations of zero alpha”? Smooth and steady could also be interpreted as being a sinusoidal shape.
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This note was uploaded on 02/15/2010 for the course IPHY 4540 at Colorado.

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Problemset6%20key%202009F - Biomechanics IPHY 4540 Problem...

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