Question

A steel ball (density = 7860kg/m3) with a diameter of 7.3 cm is tied to an aluminum wire 80 cm long and 2.8 mm in diameter. The ball is whirled about in a vertical circle with a tangential speed of 8.0 m/s at the top of the circle and 9.4 m/s at the bottom of the circle. (Young Modulus of the wire is 6.9*10^10 N/m^2)

Part A) Find the amount of stretch in the wire at the top of the circle.

Part B) Find the amount of stretch in the wire at the bottom of the circle.

Answer 1

volume of the ball=(4/3)*pi*r^3

=(4/3)*pi*(0.073)^3

=1.63*10^-3 m^3

so mass of the ball = rho*V

=1.63*10^-3*7860

=12.8 kg

Youngs modulus = stress/strain

a)Y=F*L/(A*x)

where x is the stretch in the wire.so,

F at the top = mv^2/r - mg

=12.8*8^2/0.8 - 12.8*9.81

=898.432 N

A=pi*(0.5*2.8*10^-3)^2

=6.15*10^-6 m^2

so,

6.9*10^10 = 898.432*0.8/(x*6.15*10^-6)

or x=1.69*10^-3 m

b)here F = mv^2/r+mg

=12.8*9.4^2/0.8 + 12.8*9.81

=1539.33 N

so using the above formula again,

6.9*10^10 = 1539.33*0.8/(x*6.15*10^-6)

or x=2.9*10^-3 m