A mass of 10kg is suspended from the end of a steel rod of length 4 metres and radius 1mm What is the extension of the rod beyond the original length of y =200 X10-9 N/m²
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To find the extension of the rod beyond its original length, we can use Hooke's Law, which states that the extension of a material is directly proportional to the applied force.
Hooke's Law can be written as:
F = k * ΔL
Where:
F is the applied force (in Newtons),
k is the spring constant (in N/m),
ΔL is the change in length (in meters).
In this case, the rod is acting like a spring with a spring constant (k) given by:
k = (π * r^2 * E) / L
Where:
r is the radius of the rod (in meters),
E is the Young's modulus (in N/m^2 or Pascal),
L is the original length of the rod (in meters).
Given:
Mass of the object (m) = 10 kg
Length of the rod (L) = 4 m
Radius of the rod (r) = 0.001 m
Young's modulus (E) = 200 x 10^9 N/m²
First, let's find the spring constant (k):
k = (π * (0.001)^2 * (200 x 10^9)) / 4
= (π * 0.000001 * 200 x 10^9) / 4
= (π * 200 x 10^3) / 4
≈ 157079.6327 N/m
Now, let's calculate the extension of the rod (ΔL):
F = m * g
= 10 * 9.8
= 98 N
ΔL = F / k
= 98 / 157079.6327
≈ 0.0006232 m
So, the extension of the rod beyond its original length is approximately 0.0006232 meters (or 0.6232 mm).
To calculate the extension of the rod, we need to use Hooke's Law, which states that the extension of an object is directly proportional to the force applied to it, as well as its length and a constant known as the modulus of elasticity.
In this case, the force applied to the rod is the weight of the mass, given by the equation:
F = m * g
where F is the force, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass is 10 kg, the force F is:
F = 10 kg * 9.8 m/s² = 98 N
The extension of the rod is also influenced by the length of the rod and the modulus of elasticity.
The length of the rod is given as 4 meters.
The modulus of elasticity, denoted by 'y', is given as 200 x 10^-9 N/m².
The formula for the extension of a rod is:
ΔL = (F * L) / (A * y)
where ΔL is the extension, F is the force, L is the length, A is the cross-sectional area, and y is the modulus of elasticity.
Now, let's calculate the cross-sectional area of the rod.
The radius of the rod is given as 1 mm, which is equal to 0.001 meters.
The cross-sectional area can be calculated using the formula:
A = π * r²
where A is the cross-sectional area and r is the radius.
Substituting the values, we get:
A = π * (0.001 m)² = 0.00000314 m²
Now, we can substitute all the known values into the formula for the extension:
ΔL = (98 N * 4 m) / (0.00000314 m² * 200 x 10^-9 N/m²)
Simplifying:
ΔL ≈ 124585.987 m
Therefore, the extension of the rod beyond its original length is approximately 124,586 meters.