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Derive the equation for the correction couple to be applied to make two mass systems dynamically equivalent.

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Derive the equation for the correction couple to be applied to make two mass systems dynamically equivalent.

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written 4.8 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 = $T_1 - T$

$T = m k_1^2 \ \alpha – m K_G^2 \ \alpha$

$= m \ \alpha [k_1^2 – K_G^2]$

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