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3-2.Motion in Plane
medium
An electric fan has blades of length $30\, cm$ as measured from the axis of rotation. If the fan is rotating at $1200 \,r.p.m.$ The acceleration of a point on the tip of the blade is approximately ........ $m/s^2$.
A
$1600 $
B
$4740$
C
$2370$
D
$5055 $
Solution
$\omega=2 \pi \times \frac{1200}{60} \mathrm{rad} / \mathrm{s}$
$=40 \pi \mathrm{rad} / \mathrm{s}$
$\mathrm{acc}=\omega^{2} \mathrm{r}=(40 \pi)^{2} \times \frac{30}{100}$
$=40 \times 40 \times 10 \times \frac{30}{100}=4800 \mathrm{m} / \mathrm{s}^{2}$
as we have taken $\pi^{2}=10$
so answer is slightly less than it hence
$4740 \mathrm{m} / \mathrm{s}^{2}$ is correct option.
Standard 11
Physics
