Question

A $10\text{kg}$ ball attached at the end of a rigid massless rod of length $1\text{m}$ rotates at constant speed in a horizontal circle of radius $0.5\text{m}$ with a period of $1.58\text{s}$, as shown in the figure. The force exerted by the rod on the ball is
[Take $g=10{\text{m/s}}^2$]

• A. $158\text{N}$
• B. $128\text{N}$
• C. $110\text{N}$
• D. $98\text{N}$

Answer 1

The correct option is B $128\text{N}$

Drawing the forces acting on the ball, we get the following figure. Here, the force exerted by the rod is $F$ whose component along the vertical direction is $F_1$ and along the horizontal direction is $F_2$.


Given, radius of the horizontal circle $r=0.5\text{m}$

Time period $T=1.58\text{s}$

Angular velocity of the ball,

$\omega =\frac{2\pi }{T}$

$\Rightarrow \omega =\frac{2\times 3.14}{1.58}\approx 4\text{rad/s}$

From the free body diagram of the ball, we observe




$F_1=mg=100\text{N}$

$F_2=m{\omega }^{2}r=10\times 4^2\times 0.5=80\text{N}$

So, the net force exerted by the rod on the ball is

$F=\sqrt{F_1^2+F_2^2}$

$\Rightarrow F=\sqrt{{100}^2+{80}^2}$

$\Rightarrow F=128\text{N}$

$\begin{array}{c}\text{Why this question?}\\ \text{Concept - The net force exerted by the rod}\\ \text{will balance the gravitational force and}\\ \text{provide the required centripetal force.}\end{array}$