Problem Statement
A 120mm circular disc rotates on a table separated by an oil film of 2mm thickness. Find the viscosity of the oil if the torque required to rotate the disc at 60 rpm is 4×10-4 Nm. Assume the velocity gradient in the oil film to be linear.
- Radius of disc (R) = 120/2 = 60mm = 0.06m
- Thickness (h) = 2mm = 2×10-3m
- Torque (T) = 4×10-4 Nm
- N = 60 rpm
- Viscosity of oil (μ) = ?
Solution
Given:
- Radius of disc (R) = 0.06m
- Thickness (h) = 2×10-3m
- Torque (T) = 4×10-4 Nm
- N = 60 rpm
Calculations:
Step 1: Angular velocity (ω):
ω = 2Nπ / 60
ω = 2×60×π / 60 = 6.28 rad/s
Step 2: Torque equation:
T = ∫0R 2πμω/h × r³ dr
Solving:
T = (πμωR4) / 2h
Substitute values:
4×10-4 = (π × μ × 6.28 × (0.06)4) / (2 × 2×10-3)
μ = 0.0062 Ns/m2
Result:
The viscosity of the oil is 0.0062 Ns/m2.
Explanation
This solution can be understood as follows:
- Torque and Viscosity Relationship: The torque required to rotate a disc in a viscous medium is proportional to the viscosity of the medium and the rate of rotation. The viscosity acts as the resistance against motion due to the shearing of the oil layers.
- Angular Velocity Calculation: The angular velocity (ω) is calculated from the given rotational speed (N). The formula ω = 2πN / 60 converts rpm to rad/s, giving us the rate of rotation in standard units.
- Torque Equation Derivation: The torque T is related to the viscosity μ, angular velocity ω, and the geometry of the disc and oil layer. The integral accounts for the distribution of shear forces across the radius of the disc.
- Final Viscosity Calculation: Substituting the known values (radius, thickness, torque, and angular velocity) into the derived formula allows us to solve for the viscosity. The result represents the oil’s resistance to flow under the given conditions.
- Physical Interpretation: A higher viscosity implies greater resistance to motion, requiring more torque for the same rotational speed. This calculation provides a quantitative measure of the oil’s properties.
