Question

A $10\text{kg}$ ball attached at the end of a rigid massless rod of length $(L=1\text{m})$ rotates at constant speed in a horizontal circle of radius $0.5\text{m}$ with a period of $1.57\text{s}$, as shown in the figure. The force exerted by the rod on the ball is (Take $g=10{\text{ms}}^{-2}$ & $\pi =3.14$)

• A. $20\sqrt{41}\text{N}$
• B. $10\sqrt{41}\text{N}$
• C. $20\sqrt{20}\text{N}$
• D. $10\sqrt{20}\text{N}$

Answer 1

The correct option is A $20\sqrt{41}\text{N}$

Normal reaction from rod on ball has $2$ components $N_x$ and $N_y$


$x$ - component of normal reaction will provide necessary centripetal force for the ball.
$\Rightarrow N_x=m{\omega }^{2}R$ ...(i)

For vertical equilibrium of ball:-
$N_y=mg=100\text{N}$ ...(ii)

$\because$ Time period; $T=1.57\text{s}$
$\Rightarrow T=\frac{2\pi }{\omega }=1.57\text{s}$
$\therefore \omega =\frac{2\pi }{1.57}=4\text{rad/s}$

Putting in Eq (i):
$N_x=m{\omega }^{2}R=\left(10\right)\left(4{)}^2\right(0.5)$
$=80\text{N}$

$\therefore$ Net force exerted by the rod is:
$N=\sqrt{N_x^2+N_y^2}=\sqrt{6400+10000}$
$=20\sqrt{41}\text{N}$