Question

$\text{A shaft, rotating at a uniform speed, carries two discs A and B of masses 5 kg and 4 kg, respectively. The CG of each disc is 2.5 mm from the axis of rotation, and the angle between the radii containing the centres of gravity is}{90}^{\circ }.\text{The shaft has bearings at C and D, between A and B, such that}AC=300mm,AD=900mm,\text{and}AB=1200mm.\text{It is desired to make the dynamic forces on the bearings equal and opposite, and to have a minimum value for a given speed by means of a mass in the plane E at a radius of 25 mm.}$

The distance of the plane E from the plane through A.

• A. 428 mm
• B. 468.5 mm
• C. 481.421 mm
• D. 528.421 mm

Answer 1

The correct option is B 468.5 mm


$m_A=5\text{kg,}{m}_B=4\text{kg}$

$r_A=2.5\text{mm,}{r}_B=2.5\text{mm,}{r}_E=25\text{mm}$

$\omega -\frac{2\pi \times 250}{60}=26.18\text{rad/sec}$


$(m_E{r}_E{)}^2=m_B{r}_B{)}^2+m_A{r}_A{)}^2$

$\Rightarrow m_E=m_E=\sqrt{\frac{(4\times 2.5{)}^2+(5\times 2.5{)}^2}{(25{)}^2}}=0.64\text{kg}$

$\tan \theta =\frac{m_B{r}_B}{m_A{r}_A}=\frac{4\times 2.5}{5\times 2.5}=\frac{4}{5}$

$\theta ={38.65}^{\circ }$

The angular position of mass 'E' wrt mass

${}^{\prime }{A}^{\prime }=180+\theta =180+36.65={218.65}^{\circ }$

$\sum A_A=0$

$m_E{l}_{AE}{r}_E{\omega }^2=m_B{r}_B{l}_{AB}{\omega }^2\cos 51.35$

$\Rightarrow 0.64\times l_{AE}\times 25=4\times 2.5\times 1200\cos 51.35$

$\Rightarrow l_{AE}=468.421\text{mm}$