Question

A wet open umbrella is held vertical and it whirled about the handle at a uniform rate of $21$ revolutions in $44\, s$. If the rim of the umbrella is a circle of $1\, m$ in diameter and the height of the rim above the flour is $4.9\, m$, the locus of the drop is a circle of radius:

• A. $\sqrt{2.5}\,m$
• B. $1\,m$
• C. $3\,\,m$
• D. $1.5\,\,m$

Answer 1

Answer: (A)

$\sqrt{2.5}\,m$

Answer 2

From equation of motion
$h=u t+\frac{1}{2} g t^{2}$

where $u$ is initial velocity and $t$ is time.
Since, $u=0$
$\therefore t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 4.9}{9.8}}=1 s$
The horizontal range of the drop $=x$,
then $x=\left(\frac{v_{t}}{0}\right) t$
Also, $\omega=\frac{\Delta \theta}{\Delta t}=\frac{21 \times 2 \pi}{44}=3 \,rad / s$
Tangential speed $v_{t}=r \omega$
$=0.5 \times 3 \times 1.5 \,m / s$
$\therefore x=1.5 \times 1=1.5 m$
Locus of drop $=\sqrt{x^{2}+r^{2}}$
$=\sqrt{(1.5)^{2}+(0.5)^{2}}$
$=\sqrt{2.5} \,m$