Question

A rock of mass 0.340 kg is spun horizontally at the end of a wire that has a diameter of 1.00 mm. If the wire gets stretched by 2.00 mm when the rock moves at a speed of 19.0 m/s, what is the Young's modulus of the wire?

Answer 1

Answer:

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Explanation:

First we will find the centripetal force acting on the wire as follows:

F = mv²/r

where,

F = Force = ?

m = mass of rock = 0.34 kg

v = speed = 19 m/s

r = length of wire

Therefore,

F = (0.34)(19)²/r

F = 122.74/r  

now, we find cross-sectional area of wire:

A = πd²/4

where,

A = Area = ?

d = diameter of wire = 1 mm = 0.001 m

Therefore,

A = π(0.001)²/4

A = 7.85 x 10⁻⁷ m²

Now, we calculate the stress on wire:

Stress = F/A

Stress = (122.74/r)/(7.85 x 10⁻⁷)

Stress = 1.56 x 10⁸/r

Now, we calculate strain:

Strain = Δr/r

where,

Δr = stretch in length = 2 mm = 0.002 m

Therefore,

Strain = 0.002/r

now, for Young's modulus (Y):

Y = Stress/Strain

Y = (1.56 x 10⁸/r)/(0.002/r)

Y = 78.13 x 10⁹ Pa = 78.13 GPa