Question 10.46: The connecting rod of an oil engine weighs 800 N, the distan...

\text { Given: } m=\frac{800}{9.81}=81.55 kg , l=1 m , D=120 mm , d =75 mm .

f_{n 1}=\frac{100}{200}=0.5 cps .

f_{n 2}=\frac{100}{170} cps .

\text { Let } a_{1}=\text { distance of centre of gravity from the top of small end bearing } .

a_{2}=\text { distance of centre of gravity from the top of big end bearing } .

For a simple pendulum, natural frequency of rod when suspended from small end bearing is given by,

f_{n 1}=\frac{1}{2 \pi}\left(\frac{g}{l_{e 1}}\right) .

0.5=\frac{1}{2 \pi} \sqrt{\frac{9.81}{l_{e 1}}} .

l_{e l }=0.994 m .

=\frac{K^{2}+a_{1}^{2}}{a_{1}}           (1).

When suspended from the big end,

f_{n 2}=\frac{1}{2 \pi}\left(\frac{g}{l_{e 2}}\right) .

\frac{100}{170}=\frac{1}{2 \pi} \sqrt{\frac{9.81}{l_{e 2}}} .

l_{e 2}=0.718 m .

=\frac{K^{2}+a_{2}^{2}}{a^{2}} .

or      K^{2}=a_{2}\left(0.718-a_{2}\right)          (2).

From Eqs. (1) and (2), we have

a_{1}\left(0.994-a_{1}\right)=a_{2}\left(0.718-a_{2}\right)         (3).

Also    a_{1}+a_{2}=l+\frac{1}{2}(D+ d ) .    (see Fig.10.35)

=1.0+\frac{1}{2}(0.120+0.075) .

=1.0975 m.

or    a_{2}=1.0975-a_{1}          (10.50).

Substituting in Eq. (3), we have

0.994 a_{1}-a_{1}^{2}=0.718\left(1.0975-a_{1}\right)-\left(1.0975-a_{1}\right)^{2} .

=0.788-0.718 a_{1}-1.204-a_{1}^{2}+2.195 a_{1} .

0.483 a_{1}=0.416 .

a_{1}=0.861 m .

K^{2}=0.861(0.994-0.861)=0.1145 m ^{2} .

I = mK ^{2}=81.55 \times 0.1145=9.34 kg m ^{2} .

Distance of centre of gravity from small end centre

=a_{1}-\frac{d}{2} .

= 861 – 37.5=823.5 mm.