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.
