Refer to Fig 10.38.
\text { Given: } m=60 kg , l=900 mm , d_{1}=80 mm , d_{2}=100 mm , T_{1}=1.85 s , T_{2}=1.70 s .
(a) For an equivalent simple pendulum,
T_{1}=2 \pi \sqrt{\frac{l_{e 1}}{g}} .
1.85=2 \pi \sqrt{\frac{l_{e 1}}{9.81}} .
l_{e 1}=0.85 m .
=\frac{K^{2}+a_{1}^{2}}{a_{1}} .
K^{2}=a_{1}\left(0.85-a_{1}\right) (1).
T_{2}=2 \pi \sqrt{\frac{l_{e 2}}{g}} .
1.70=2 \pi \sqrt{\frac{l_{e 2}}{9.81}} .
l_{e 2}=0.718 m .
=\frac{K^{2}+a_{2}^{2}}{a_{2}} .
K^{2}=a_{2}\left(0.718-a_{2}\right) (2).
From Eqs. (1) and (2)
a_{1}\left(0.85-a_{1}\right)=a_{2}\left(0.718-a_{2}\right) .
Now a_{1}+a_{2}=0.900+\frac{1}{2}(80+100) 10^{-3}=0.990 m .
or a_{2}=0.990-a_{1} .
∴ a_{1}\left(0.85-a_{1}\right)=\left(0.99-a_{1}\right)\left(0.718-0.99+a_{1}\right) .
0.85 a_{1}-a_{1}^{2}=\left(0.99-a_{1}\right)\left(a_{1}-0.272\right) .
=0.99 a_{1}-a_{1}^{2}-0.26928+0.272 a_{1} .
=1.262 a_{1}-a_{1}^{2}-0.26928 .
0.412 a_{1}=0.26928 .
a_{1}=0.6536 m .
K^{2}=0.6536(0.85-0.6536)=0.1284 m ^{2} .
K=0.3583 m.
(b) I=m K^{2}=60 \times 0.1284=7.704 kg m ^{2} .
\text { (c) Distance of centre of gravity from small end centre, } a=a_{1}-\frac{d_{1}}{2} .
=0.6536-\frac{0.08}{2}=0.6136 m .
\text { Let } m_{a}=\text { mass placed at small end centre } .
m_{b}=\text { second mass } .
b=\text { distance of } m_{b} \text { from centre of gravity } .
For dynamically equivalent system,
a b=K^{2} .
b=\frac{K^{2}}{a}=\frac{0.1284}{0.6136}=0.2092 m
Now m_{a}=\frac{b m}{a+b}=\frac{0.2092 \times 60}{0.6136+0.2092}=15.25 kg .
m_{b}=m-m_{a}=60-15.25=44.75 kg .
