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Derive the equation for the correction couple to be applied to make two mass systems dynamically equivalent.
1 Answer
written 5.6 years ago by |
Solution:
Let,
$L_3$ = Distance of ass placed at D from G,
$I_1$ = New mass moment of inertia of 2 masses.
$k_1$ = New radius of duration.
$\alpha$ = Angular acceleration of the body.
$I =$ Mass moment of inertia of dynamically equivalent system.
$K_G$ = Radius of gyration of a dynamically equivalent system.
$\rightarrow$ Torque required to accelerate body,
$T = I. \alpha$
$= m. (KG)^2 . \alpha$
$\rightarrow$ Torque required to accelerate two mass system placed arbitariy.
$T_1 = I_1 \ \alpha$
$= m.(k_1)^2 . \alpha$
$\therefore T =$ Correction couple = $T_1 - T$
$T = m k_1^2 \ \alpha – m K_G^2 \ \alpha$
$= m \ \alpha [k_1^2 – K_G^2]$