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-shaped mode, it has a frequency of vibration of 10 rad/s. When the shaft is rotated at 300 r.p.m., it will whirl with a radius of
A. 2 mm
B. 2.22 mm
C. 2.50 mm
D. 3.0 mm
Answer: Option B
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y=e/(((wn/w)^2)-1).
Given,
e=2mm,
Wn=10rad/sec.
W = 300rpm = 300 * ((2*pi)/60) rad/sec =10pi = 31.4rad/sec.
Substituting the values of e, Wn and W in y formula we get;
y=-2.247mm.
Since the length cannot be in negative we select the option choice 2. ie 2.22mm.
pls give solution...