Question

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

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

Answer 1

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

FBD of the ball:

Let the force exerted by the rod be $F$ whose components along the vertical and horizontal respectively are $F_1$ and $F_2$ as shown in the FBD.


So, from FBD

$F_1=mg$ ....... (1)

$F_2=\frac{m{v}^2}{r}$ ....... (2)

To find $v$:

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

Time period $\left(T\right)=1.57\text{s}$

Speed $=\frac{\text{distance}}{\text{time}}=\frac{2\pi r}{T}$

$v=\frac{2\pi \times 0.5}{1.57}$$=2\text{m/s}$

So magnitude of the net force is given by

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

$\therefore$ Squaring & adding equation $\left(1\right)$ & $\left(2\right)$

$F^2=(F_1{)}^2+(F_2{)}^2$

$F^2=(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2$

$\Rightarrow F^2=(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2$

$\Rightarrow F=\sqrt{(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2}$

$=\sqrt{(10\times 10{)}^2+{\left(\frac{10\times (2{)}^2}{0.5}\right)}^2}$

$=\sqrt{(100{)}^2+(80{)}^2}=\sqrt{16400}=128\text{N}$