written 5.8 years ago by |
Time of oscillation when suspended from small end =175sec and from big and =1.6sec. Determine m – I about the axis passing there mass centre (ii) The dynamically equivalent system of 2 mass one located at small and centre.
Solution: L=L1+L2=800mm
tp1=1.75 sec/cycle.
tp1=2π√k2+(L′1)2g.L′1
1.75=2π√k2(L1+0.03)2g(L1+0.03)
k2=0.701L1+0.0237–L21
tp2=2π√k2+(L′2)2gL′2
Where,
L′1=L1+0.03
L′2=L2+0.04
But L2=0.8–L1
So L′2=0.8–L1+0.04
L′2=0.84–L1
{For big end}
1.6=2π√k2+(0.84−L1)29.81(0.84–L1)
k2=1.044L1–0.1717–L21
0.701L1+0.0237–L21=1.044L1–0.1717–L21
L′1=0.569 m
L′1=0.569+0.030
L′1=0.599
L2=0.800–0.569
L2=0.2303
Substituting value of L1
K=0.31m
∴ I_G = mK^2
=40 (0.31)^2
= 43.9 \ kgm^2
2]
k^2 = l_1 . l_2
(0.314)^2 = 0.569 . l_2
l_2 = 0.173 \ m
m_2 = \frac{mL_1}{L_2} = \frac{40\times 0.5694}{0.569 + 0.173} = 30.69 \ kg
m_1 = \frac{mL_2}{L_1 + L_2} = 9.31kg