A mass of $1 \; kg$ is attached to the lower end of a string $1 \; m$ long whose upper end is fixed. The mass is made to rotate in a horizontal circle of radius $60 \; m$. If the circular speed of the mass is constant. Find the tension in the string and the period of motion.

Given, (need figure)

Mass of the object $(m) = 1\;kg$

Length of string $(l) = 1\;m$

Radius of circle $(r) = 60\;cm = 0.6\;m$

Tension on the string $(T) = \;?$

Period of motion $(t) = \;?$

From Figure,

$T \;Cos \theta = mg$ .......... (i)

$T \;Sin \theta = \; $$\frac{mv^{2}}{r}$ .......... (ii)

We have,

$tan \theta  = \;$$\frac{v^2}{r\;g}$ and            $Sin \theta = \;$ $ \frac{r}{l} = \frac{0.6}{1}$ $ = 0.6 $

⇒ $\theta = Sin^{-1}(0.6) = 36.87$

Now, From the relation,

$T =\;$ $ \frac{m\;*\;g}{Cos \theta} = \frac{1 \;*\;10}{Cos (36.87)}$ $ = 12.5\;N$

⇒ Tension on the string $(T) = 12.5\;N$

Again from relation,

$t = \;$$2 \pi \sqrt{\frac{l\;Cos \theta}{g}}$

$t = \; $ $2 \pi \sqrt{\frac{1 \;* \;Cos (36.87)}{10}}$ = $1.78 \;Sec$

⇒ Period of motion $(t) = 1.78 \;Sec$

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