A torsional system with discs of moment of inertia $${I_1}$$ and $${I_2}$$ as shown in the below figure, is gear driven such that the ratio of speed of shaft B to shaft A is 'G'. Neglecting the inertia of gears, the equivalent inertia of disc on shaft B at the speed of shaft A is equal to
A. $${\text{G}} \times {I_2}$$
B. $${{\text{G}}^2} \times {I_2}$$
C. $$\frac{{{I_2}}}{{\text{G}}}$$
D. $$\frac{{{I_2}}}{{{{\text{G}}^2}}}$$
Answer: Option B
Solution (By Examveda Team)
This Question Belongs to Mechanical Engineering >> Theory Of Machine
Equivalent inertia of disc B = m.i. of disc B/ (gear ratio)2.
gear ratio=n1/n2
m.i. of B=I2
So, equi. inertia of B=I2/(n1/n2) = I2*G^2 (besause G=n2/n1).