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Answer: BIO Prevention of Hip Fractures. Falls resulting
Chapter 2, Problem 26E(choose chapter or problem)
Problem 26E
BIO Prevention of Hip Fractures. Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact is about 2.0 m/s. If this can be reduced to 1.3 m/s or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads. (a) If a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what constant acceleration (in m/s2 and in g’s) does the hip undergo to reduce its speed from 2.0 m/s to 1.3 m/s? (b) The acceleration you found in part (a) may seem rather large, but to assess its effects on the hip, calculate how long it lasts.
Questions & Answers
QUESTION:
Problem 26E
BIO Prevention of Hip Fractures. Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact is about 2.0 m/s. If this can be reduced to 1.3 m/s or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads. (a) If a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what constant acceleration (in m/s2 and in g’s) does the hip undergo to reduce its speed from 2.0 m/s to 1.3 m/s? (b) The acceleration you found in part (a) may seem rather large, but to assess its effects on the hip, calculate how long it lasts.
ANSWER:
Solution 26E
Step 1:
We can use Newton’s equations of motion to solve this problem.
a) Here, they have provided that, the initial velocity during the impact is, u = 2 m/s
We have to reduce it up to 1.3 m/s during the impact in order to prevent the fracture. Therefore, the final velocity during impact should be, v = 1.3 m/s
The initial size of the pad is, t1 = 5 cm = 0.05 m
The size of the pad during compression, t2 = 2 cm = 0.02 m
Therefore, we can use the equation, v2 = u2 + 2aS
Here, S = (t1 - t2) which is nothing but the change in thickness of the pad during the compression.
Putting all those values in the equation, we get,
1.32 m2/s2 = 22 m2/s2+ (2 × a × [0.05 m - 0.02 m])
1.69 m2/s2 = 4 m2/s2 + (0.06 m × a )
(1.69 m2/s2 - 4 m2/s2 ) = (0.06 m × a )
- 2.31 m2/s2 = 0.06 a m
Therefore, a = - 2.31 m2/s2 / 0.06 m = - 38.50 m/s2
This is the constant acceleration needed for reducing the velocity from 2 m/s to 1.3 m/s if the pad is compressed from 5 cm thickness to 2 cm thickness.
We know that, g = 9.8 m/s2
Therefore, a = - 38.50 m/s2 / 9.8 m/s2 = - 3.928 g