Physics, asked by 118me0020, 5 months ago

a cylindrical roller bearing with bore diameter of 40mm is subjected to a radical forces of 25 kn.the coefficient of friction is 0.0012 and the speed of rotation is 1440rpm.calculate the power lost in friction

Answers

Answered by knjroopa

Explanation:

Given a cylindrical roller bearing with bore diameter of 40 mm is subjected to a radial forces of 25 kn. The coefficient of friction is 0.0012 and the speed of rotation is 1440 rpm.calculate the power lost in friction

So bore diameter = 40 mm
So radius will be d/2 = 20 mm
Radial force Pr = 25 kn = 25,000 N
Force of friction F = μ x pr
0.0012 x 25000
F = 30 N
So radius of bore r = 20 mm
Now frictional moment = 30 x 20
= 600 Nmm
= 0.6 Nm
Given Speed = 1440 rpm
Now angular speed ω = 1440 x 2π / 60
= 150.72 rad/sec
Due to friction in bearing there is a loss of power.
Therefore power lost in friction will be
150.72 x 0.6
90.432 Watts

Reference link will be

https://brainly.in/question/32933314

https://brainly.in/question/2691041

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