Answer: viscosity of the oil becomes μ = 0.2003 pa.s
Explanation:
from the given question, we have that
torque T = 4Nm
height H = 450mm = 0.45m
rotational speed ω = 30rpm = 30×2π/60 = 3.14 rad/sec
also to calculate for the linear velocity of the intermediate cylinder,
V = Rω
where R is the radius of the cylinder.
now we substitute values of R and ω, we have
V = Rω
V = 0.15 × 3.14 = 0.471 m/s
calculating the drag force on the two side of the cylinder;
Fd = 2μA (Δv/Δy)
Fd = 2μA V/h
where A = 2πRH
Fd = 2μ(2πRH) V/h
h = thickness, μ = viscosity of oil, H = height of cylinder, and V = mean velocity of cylinder
substituting values we have,
Fd = 2μ * (2π*0.15*0.45)* 0.471/0.003
Fd = 133.10μN
from the torque equation, we can calculate the viscosity
thus, T = Fd*R
we already have Fd = 133.10μN, substituting it into the above expression we have
T = 133.10μ * 0.15 where T is 4Nm,
∴ μ = 4/ 133.1*0.15
μ = 0.2003 pa.s
i hope this helps, cheers
