Question

Controlling force of a spring controlled governor, $F_C=2800r-76$ , where $F_c$ is in Newton and r is in metre. If mass of ball is 5 kg, extreme radii of the balls are 100 mm and 175 mm, the coefficient of insensitivity of governor at extreme rad ius ( 175 mm) if friction in governor is 5 N at each ball, is

• A. 1.19 %
• B. 0.094 %
• C. 9.22 %
• D. 3.29 %

Answer 1

The correct option is A 1.19 %

$F_c=2800r-76$

m = 5 kg,

$r_1=175mm=0.175m,$

At maximum speed,

$F_{c1}=2800\times 0.175-76=414N$

$414=mr{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{414}{5\times 0.175}}=21.75rad/sec$

(mean speed without friction)

With an equivalent frictionalforce 5N at the balls ,the controlling force at the balls,the controlling force at maximum and minimum equilibrium speed is

$F_{c1}^{\prime }=414+5=419N$

$F_{c2}^{\prime }=414-5=409N$

$({\omega }_1^{\prime }{)}^2=\frac{419}{m{r}_1}=\frac{419}{5\times 0.175}=478.85$

${\omega }_1^{\prime }=21.88rad/s$

$({\omega }_1^{\prime }{)}^2=\frac{409}{m{r}_1}=\frac{409}{5\times 0.175}=467.428$

${\omega }_2^{\prime }=21.62rad/s$

Coefficient of insensitiveness

$=\frac{({\omega }_1^{\prime }-{\omega }_2^{\prime })}{\omega }$

$=\frac{(21.88-21.62)}{21.75}=0.0119$

$=1.19\%$