A vertical shaft 150 mm in diameter rotating at 200 rpm rests on a flat foot step bearing. The shaft carries a vertical load of 20 kN. Considering uniform pressure distribution and coefficient of friction as 0.05, what will be the frictional torque on the bearing?

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This question was previously asked in

MPPGCL JE ME 19 March 2019 Official Paper

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Answer 1

Concept:

For Uniform Pressure Distribution, Total Frictional Torque is given by

\(T = \frac{2}{3}\mu WR\)

where µ = Coefficient of Friction, W = Applied Load, R = Radius of Shaft

Calculation:

Given:

D = 150 mm = 0.150 m, N = 200 rpm, W = 20 kN, µ = 0.05

For Uniform Pressure Distribution, Total Frictional Torque is given by

\(T =\frac{2}{3} \times 0.05 \times 20000 \times 0.075\)

T = 50 Nm

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