• # question_answer A shaft has an attached disc at the centre of its length. The disc has its centre of gravity located at a distance of 2 mm from the axis of the shaft. When the shaft is allowed to vibrate in its natural bow-shaded mode, it has a frequency of vibration of 10 radians second. When the shaft is rotated a 300 revolutions per minute, it will whirl with a radius of: A) 2 mm               B) 225 mmC) 2.50 mm                      D) 3.00 mm

$\omega =\frac{2\pi \times 300}{600}=10\pi \,\text{rad/s}$ $\beta =\frac{\omega }{{{\omega }_{n}}}\,\frac{10\pi }{10}=\pi$ $r=\frac{e{{\beta }^{2}}}{{{\beta }^{2}}-1}=\frac{2\times {{\pi }^{2}}}{{{\pi }^{2}}-1}=225\,\,mm$