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:


$T\cos \theta =mg$ ... (1)
$T\sin \theta =\frac{m{v}^2}{r}$ ... (2)
To find $v$:
Given, radius of horizontal circle $r=0.5\text{m}$
Time period $\left(T\right)=1.57\text{s}$
Speed $=\frac{\text{distance}}{\text{time}}=\frac{2\pi r}{T}$
$=\frac{2\pi \times 0.5}{1.57}$$=2\text{m/s}$

$\therefore$ Squaring & adding equation (1) & (2)
$(T\cos \theta {)}^2+(T\sin \theta {)}^2=(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2$
$\Rightarrow T^2=(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2$
$\Rightarrow T=\sqrt{(mg{)}^2+{\left(\frac{m{v}^2}{r}\right)}^2}$
$=\sqrt{(10\times 10{)}^2+{\left(\frac{10\times (2{)}^2}{0.5}\right)}^2}$
$=\sqrt{(100{)}^2+(80{)}^2}=\sqrt{16400}=128\text{N}$