Agymnast does a one-arm handstand. The humerus, which is the...

Agymnast does a one-arm handstand. The humerus, which is the upper arm bone (between the elbow and the shoulder joint), may be approximated as a $0.30$ -m-long cylinder with an outer radius of $1.00 \times 10^{-2} \mathrm{~m}$ and a hollow inner core with a radius of $4.0 \times 10^{-3} \mathrm{~m}$. Excluding the $\mathrm{am}$, the mass of the gymnast is $63 \mathrm{~kg}$. (a) What is the compressional strain of the humerus? (b) By how much is the humerus compressed?

Agymnast does a one-arm handstand. The humerus, which is the upper arm bone (between the elbow and the
shoulder joint), may be approximated as a $0.30$ -m-long cylinder with an outer radius of $1.00 \times 10^{-2} \mathrm{~m}$ and a hollow inner core with a radius of $4.0 \times 10^{-3} \mathrm{~m}$. Excluding the $\mathrm{am}$, the mass of the gymnast is $63 \mathrm{~kg}$.
(a) What is the compressional strain of the humerus?
(b) By how much is the humerus compressed?

Key Concepts

Cross-sectional Geometry of Hollow Cylinders

When dealing with objects like bones or tubes that can be approximated as hollow cylinders, the effective cross-sectional area is calculated as the difference between the areas of the outer and inner circles. This area is crucial for determining the compressive stress when an axial load is applied, as it directly influences the magnitude of stress and subsequent strain.

Compressive Strain

Compressive strain is a dimensionless measure of deformation representing the fractional change in length of an object under compressive loading. It is defined as the change in length divided by the original length, allowing engineers to quantify how much a material compresses under an applied load.

Hooke’s Law and Young’s Modulus

Hooke’s Law states that, for elastic materials, the stress applied to a material is directly proportional to the resulting strain, with Young’s modulus being the proportionality constant. Young’s modulus is a material property that measures its stiffness and is used to calculate the deformation (such as compression) when a force is applied.

Compressive Stress

Compressive stress is the force applied per unit area over the cross-section of an object, measured in units like Pascals (N/m²). It represents how the applied load, such as weight, is distributed across the material’s cross-sectional area and is a key factor in determining the resulting strain via Hooke’s Law.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that aginist does a one -arm handstand, and the omeris is the bone that is supporting his weight, her weight.

00:15 So we're given that the initial length of this bond is approximately 0 .3 meters or 30.

00:31 Centimeters and it is it can be as approximated as a cylinder with an outer radius we're gonna call that our radius of one times 10 to the minus two meters we are also given the the inner radius of four times ten to the minus 3 meters and the mass of the giantess is 63 kilograms.

01:22 And the first question for this problem, it is what is the compressional strain of the humerus? so for part of this problem, we need to find the compressional strain.

01:37 So the compression as the strain is defined as this.

01:40 This is equal to the strain.

01:42 Now this strain is also defined as the force that is being applied over the yon's modulus times the creusceptional area.

02:01 Now, in this case, the force that is being applied to this bound is the weight of the genus.

02:14 So the force is the weight of the genus that we know is defined.

02:22 The weight is defined as the mass of the genus time means the acceleration due to gravity.

02:29 Jans models times the creusceptional area.

02:34 As the problem sets, this bone has an all -inner cord.

02:43 So we have an outer and an inner radius so the cross -ceptional area is going to be something like this if we cut this bond by a cross -ception we will have something like this so the area that we want to obtain is this are in here so to the entire area of the of the bigger of the bigger circle we need to to substrat this tiny area at the center.

03:20 So we are given that information.

03:22 So we know that the area in this case is going to be pi times the radius of the outer circle minus the radius of the inner circle.

03:39 So we are given those information.

03:41 So we just substitute to obtain the area.

03:45 So in this case, is going to be.

03:50 The outer radius is 1 times 10 to the minus 2 meters, this to the square, minus 4 times 10 to the minus 3 meters for the inner radius square...

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