SSC JE Mechanical Engineering 2012 Paper

Question : 170 of 200

Marks: +1, -0

The stiffness of the spring in a Hartnell governor is equal to,
where

S_1

= Spring force exerted on the sleeve at maximum radius of rotation.

S_2

= Spring force exerted on the sleeve at minimum radius of rotation.
h = lift of sleeve

Solution:

The stiffness of spring

= {\difference \of \force \exerted \on \sleeve}/{ \lift \of \sleeve}

={S_1-S_2}/h

where

h={(r_2-r_1).g}/x

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