Measuring viscosity with torsion pendulum

[Rosengrip]

Homework Statement

We have a round disc with radius of 10 cm and mass of 1 kg, suspended on a wire and lying on a thin (1 mm) layer of oil. We torsionally nudge it and we get some info out of it:

At angular velocity of 10 s$$^{-1}$$ the amplitude is maximal.
At angular velocity of 12 s$$^{-1}$$ the amplitude is 1/2 of maximal amplitude (is reduced by half)

What's the viscosity of oil?

Homework Equations

Torsion pendulum motion equation:
$$\phi$$=$$\phi$$0 * e^-(beta*t)(cos(w*t) + phase)
beta - coefficient of damping
w- angular velocity
F = -c*v (force, opposing the oscillation of pendulum)

c - coefficient of damping
v - velocity

I was thinking along the lines that first I need to find a dampening cofficient (but I need a force for that) and then I got really lost (dampening increases with angular velocity, it's a bit confusing).

Any point in the right direction would be greatly appreciated.

 

[jbsteele]

The Attempt at a SolutionThe damping coefficient (beta) can be calculated by using the following equation:β = F/(mv) where F is the force opposing the oscillation of the pendulum, m is the mass of the disc, and v is the velocity. Since the force is dependent on the viscosity of the oil, we can use the following equation to calculate the viscosity of the oil:μ = F/(Av) where A is the area of the disc and v is the velocity. Using the given information, we can then calculate the viscosity of the oil as follows:At angular velocity of 10 s^{-1}:F = mv*β μ = F/(π*(10 cm)2*10s^{-1}) μ = 1 kg*10s^{-1}*β/(π*100 cm2*10s^{-1}) μ = β/(10π cm2/s) At angular velocity of 12 s^{-1}:F = mv*β/2 μ = F/(π*(10 cm)2*12s^{-1}) μ = 1 kg*12s^{-1}*β/(2*π*100 cm2*12s^{-1}) μ = β/(20π cm2/s) Since the damping coefficient (β) is the same for both angular velocities, we can equate the two equations to solve for β:10π cm2/s = 20π cm2/s β = 10π cm2/s Therefore, the viscosity of the oil is 10π cm2/s.