Glycerin at 20 8 C fills the space between a hollow sleeve of diameter 12 cm and a fixed coaxial solid rod of diameter 11.8 cm. The outer sleeve is rotated at 120 rev/min. Assuming no temperature change, estimate the torque re-quired, in N · m per meter of rod length, to hold the inner rod fixed.

Complete question:

Glycerin at 20 °C fills the space between a hollow sleeve of diameter 12 cm and a fixed coaxial solid rod of diameter 11.8 cm. The outer sleeve is rotated at 120 rev/min. Assuming no temperature change, estimate the torque required, in N · m per meter of rod length, to hold the inner rod fixed

Answer:

The torque required to hold the inner rod fixed is 25.41 N. m/rod length

Explanation:

Given angular velocity, ω = 120 RPM

angular velocity, ω (rad/s) = (2πN) / (60) = (2π*120) / (60) = 12.566 rad/s

Tangential velocity of the sleeve (V) = ωr

where, r is radius of the sleeve = 12/2 = 6cm = 0.06 m

Tangential velocity of the sleeve (V) = ωr = (12.566 rad/s) * (0.06 m) = 0.75396 m/s

Distance between the cylinders (h) = (12-11.8) / 2 = 0.1 cm = 0.001 m

Shear stress (τ) on glycerin = (μ*V) / (h)

where, μ is the dynamic viscosity of glycerin at 20°C = 1.49 kg/m. s

Shear stress (τ) = (1.49 * 0.75396) / (0.001) = 1123.4 N/m²

Shearing force (F) in tangential direction to the sleeve = Shear stress (τ) * Area of the sleeve

F = τ * A

F = τ * 2π*r*L

Force per unit rod length

F/L = (1123.4*2π*0.06) = 423.567 N/m

To hold the inner rod fixed, the opposite torque by the inner rod must be equal to torque produced by the shearing force.

Torque per unit length (N. m/length) = [force per unit length (N/m) * radius of the sleeve (m) ]

Torque per unit length = 423.567 * 0.06

Torque per unit length = 25.41 N. m/rod length

Therefore, the torque required to hold the inner rod fixed is 25.41 N. m/rod length.