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
Join The Discussion
Comments ( 1 )
In considering friction of a V-thread, the virtual coefficient of friction (μ1) is given by
A. μ1 = μsinβ
B. μ1 = μcosβ
C. $${\mu _1} = \frac{\mu }{{\sin \beta }}$$
D. $${\mu _1} = \frac{\mu }{{\cos \beta }}$$
The lower pairs are _________ pairs.
A. Self-closed
B. Force-closed
C. Friction closed
D. None of these
In a coupling rod of a locomotive, each of the four pairs is a ________ pair.
A. Sliding
B. Turning
C. Rolling
D. Screw
A kinematic chain is known as a mechanism when
A. None of the links is fixed
B. One of the links is fixed
C. Two of the links are fixed
D. None of these
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).