Question

A rough casting of a rotor, with a mass of 200 kg, is mounted on centres 110 cm apart for machining. The rotor is statically balanced by attaching two masses, one each in the planes A (9 kg) and B (12 kg), which are on opposite sides of the plane containing the mass centre of the rotor, at a distance of 50 cm and 35 cm respectively from it. The angle between the CG of the plane A and that of the plane B is 90${}^{\circ }$. The distance of each mass from the axis of rotation is 37 cm (for the plane A) and 44 cm (for the plane B).

The force on the bearing 'B' when the rotor is running at a speed of 100 rpm is (in N)

• A. 248
• B. 496
• C. 124
• D. 336

Answer 1

The correct option is A 248


$m_A=9\text{kg,}{m}_B=12\text{kg,}$

$r_A=37\text{cm,}{m}_A=44\text{cm}$

$N=100\text{rpm}$

$\omega =\frac{2\pi N}{60}=10.47\text{rad/sec}$

As the rotor is statically balanced

$200\times r=\sqrt{(9\times 37{)}^{+}(12\times 44{)}^2}$

$\Rightarrow r=3.12\text{cm}$

$\tan \theta =\frac{m_B{r}_B}{m_A{r}_A}=\frac{12\times 44}{9\times 37}$

$\Rightarrow \theta ={57.76}^{\circ }$

$\text{The angular position of balance mass is to be placed}=\theta +{180}^{\circ }=180+57.76={237.76}^{\circ }$

$\sum M_A=0$

$\sum \left(mrl\cos \theta \right){\omega }^2=(9\times \frac{37}{100}\times \frac{5}{100}\cos 0+12\times \frac{44}{100}\times \frac{90}{100}\times \cos 90+200\times \frac{3.12}{100}\times \frac{55}{100}\times \cos 237.76)\times (10.47{)}^2=-182.4484\text{N}$

$\sum \left(mrl\sin \theta \right){\omega }^2=(9\times \frac{37}{100}\times \frac{5}{100}\sin 0+12\times \frac{44}{100}\times \frac{90}{100}\times \sin 90+200\times \frac{3.12}{100}\times \frac{55}{100}\times \sin 237.76)\times (10.47{)}^2=202.7046\text{N}$

$R_B\times 1.1=\sqrt{(-182.4484{)}^2+(202.7046{)}^2}$

$R_B=247.927\text{N}$

$\text{The forces in bearing is}247.92\text{N}$