Atorque of 4 nm is required to rotate the intermediate cylinder at 30 radians/min. calculate the viscosity of the oil. all cylinders are 450 mm long. neglect end effects.

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

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