A materials scientist is testing a cylindrical sample of lightweight material. T

Question

A materials scientist is testing a cylindrical sample of lightweight material. The cylinder has a radius r = 0.50 cm and an unknown but reasonably large length L. Unfortunately for length measurements, the scientist only has a device that measures very small distances. She does, however, have a device that measures time precisely. She decides to determine the sample’s length by using the cylinder as a rod or string in a pendulum. She attaches one end of the cylinder to a pivot joint and attaches a mass m = 50 kg to the other end. Attaching the mass causes an increase in the cylinder’s length by L = 0.50 mm. Setting this pendulum swinging, she finds that the period of oscillation is T = 2.457 s.

Answer the following questions assuming that:

• The mass of the cylinderical sample is negligible in comparison to the 50 kg attached to the end of it.

• The length of the cylindrical sample is constant throughout the pendulum motion.

(a) What is the unstretched length L of the cylinder?

(b) If the cylinder behaves elastically, what is the Young’s modulus of the material?

(c) How much elastic potential energy is stored in the cylinder due to the change in length L?

Explanation / Answer

(a) We know that the time period
T = 2Pi(l/g)1/2
therefore
l = 1.522 m
Therfore unstretched length
L = l - 0.5*10-3 = 1.5216 m
(b) w = 2Pi/T= (s/m)1/2
s = 326.97 N/m
S = AE/l = (Pi*0.52)*10-4E/(1.5216)
E = 6.33475*106 N/m2
(c) Potential energy = (1/2)sX2
= (1/2)326.97*0.52*10-6 = 40.87*10-6 J

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