# A mass of 0.2kg is whirled in a horizontal circle of radius 0.5m by a string inclined at 30° to the vertical. Calculate the tension in the string and the speed of the mass in the horizontal circle?

Jul 28, 2018

The tension is $= 2.26 N$ and the speed is $= 1.68 m {s}^{-} 1$

#### Explanation:

Let the tension in the string be $= T \left(N\right)$

Then,

Resolving in the vertical direction

$T \cos \theta = m g$

The acceleration due to gravity is $= 9.8 m {s}^{-} 2$

T*cos30º=0.2*9.8

$T = \frac{0.2 \cdot 9.8}{\cos 30} = 2.26 N$

Resolving in the horizontal direction

$T \sin \theta = m {v}^{2} / r$

Therefore,

${v}^{2} = \frac{r T \sin \theta}{m}$

$v = \sqrt{\frac{r T \sin \theta}{m}}$

$= \sqrt{\frac{0.5 \cdot 2.26 \cdot \sin 30}{0.2}}$

$= 1.68 m {s}^{-} 1$