A 1.03 kg mass is attached to a spring of force constant 10.6 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.525 m at a rate of 2.20 revolutions per second?
To find out how much the spring will stretch, we need to analyze the forces acting on the system.
The first force is the centripetal force, which keeps the mass moving in a circular path. This force is given by Fc = m * (v^2) / r, where m is the mass, v is the velocity, and r is the radius.
The second force is the force exerted by the spring. This force is given by Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as Fs = -k * x, where k is the force constant of the spring and x is the displacement.
In this case, the displacement of the spring is equal to the difference between the stretched length and the equilibrium length. Since the spring is initially at equilibrium, the stretched length is equal to the displacement.
Given:
Mass (m) = 1.03 kg
Force constant (k) = 10.6 N/cm = 10.6 N / 0.01 m
Radius (r) = 0.525 m
Number of revolutions per second = 2.20 rev/s
First, let's calculate the velocity:
The circumference of the circular path can be calculated as C = 2 * π * r.
The velocity (v) can be calculated as v = C * n, where n is the number of revolutions per second.
Therefore, v = (2 * π * r) * n.
Substituting the given values:
v = (2 * 3.14159 * 0.525) * 2.20 m/s
Now, let's calculate the centripetal force (Fc):
Fc = m * (v^2) / r
Substituting the given values:
Fc = 1.03 kg * ((2 * 3.14159 * 0.525 * 2.20)^2) / 0.525 m
Next, let's calculate the spring force (Fs) using Hooke's Law:
Fs = - k * x
Since the mass is in equilibrium, the spring force (Fs) will be equal in magnitude and opposite in direction to the centripetal force (Fc).
Therefore, Fs = Fc.
Now we can solve for the displacement (x):
Fs = - k * x
|x| =
Taking the magnitudes and substituting the calculated values:
|x| = (1.03 kg * ((2 * 3.14159 * 0.525 * 2.20)^2) / 0.525 m) / (10.6 N / 0.01 m)
Simplifying the equation and calculating the result:
|x| = 3.84199 m
Therefore, the spring will stretch by approximately 3.84199 meters.