Question

A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured the change in the cylinder’s height. The cylinder had a radius of 2 cm, and an initial height of 10 cm. This group of students would like to know what kind of material their cylinder was. Based on their data shown below, and the Young’s modulus values shown, which of the materials below could the cylinder be made from? Select all that apply.

Answer 1

The material that the cylinder is made from is Butyl Rubber.

What is Young's modulus?

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

Area of the cylinder

A = πr²

$A = \pi \times (0.02)^2 = 0.00126 \ m^2$

Young's modulus of the cylinder

$E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae}$

Where;

• e is extension

When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

$E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa$

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

$E = \frac{(50\times 9.8) \times 0.1}{0.00126 \times 0.0227} \\\\E = 1.7 \times 10^6 \ N/m^2\\\\E = 0.0017 \times 10^9 \ N/m^2\\\\E = 0.0017 \ GPa\\\\E \approx 0.002 \ GPa$

The Young's modulus is between 0.001 GPa  to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

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Answer 2

Based on the calculated Young’s modulus of the material, the material the cylinder could most likely be made of is butyl rubber which has a Young’s modulus of 0.001 to 0.002 GPa.

What is Young’s modulus?

Young's modulus, Y is a measure of a solid's ability to resist deformation in its length under stress or strain.

Mathematically,

• Young’s modulus Y = stress/strain

where:

• stress = Force/area
• strain = extension/original length

Y = Fl/Ae

• Area of the cylinder = πr^2

where;

π = 3.14

r = 2 cm = 0.02

Area of the cylinder = 3.14 × 0.02^2 = 0.00125 m^2

When a mass 5 kg was placed on the cylinder:

F = 5 × 10 = 50 N

l = 10 cm = 0.1 m

e = 10 - 7.73 = 0.39 cm = 0.0039 m

A = 0.00125 m^2

Using Y = Fl/Ae

Y = (50 × 0.1)/(0.00125 × 0.0039)

Y = 1025641.026 N/m^2

1 N/m^2 = 1 × 10^-9 GPa

Y = 0.001025 Gpa

When a mass 50 kg was placed on the cylinder:

F = 50 × 10 = 500 N

l = 10 cm = 0.1 m

e = 10 - 7.73 = 2.27 cm = 0.0227 m

A = 0.00125 m^2

Using Y = Fl/Ae

Y = (500 × 0.1)/(0.00125 × 0.0227)

Y = 1762114.537 N/m^2

1 N/m^2 = 1 × 10^-9 GPa

Y = 0.00176 Gpa

The Young’s modulus of the material is between 0.001025 Gpa and 0.00176 Gpa.

Therefore, the material the cylinder could most likely be made of is butyl rubber which has a Young’s modulus of 0.001 to 0.002 GPa.

Learn more about Young’s modulus at: brainly.com/question/6864866