Questions: A 0.200 kg stone whirled in a horizontal circle on the end of a 0.800-m-long string. The string makes an angle of 15.0° with the horizontal. Determine the speed v of the stone. v= m / s

Solution Steps

The stone is moving in a horizontal circle, and the tension in the string provides the necessary centripetal force. The string makes an angle of $15.0^\circ$ with the horizontal, so we need to consider the components of the tension force.

Let $T$ be the tension in the string. The vertical component of the tension balances the weight of the stone, and the horizontal component provides the centripetal force.

• Vertical component: $T \cos(15^\circ) = mg$
• Horizontal component: $T \sin(15^\circ) = \frac{mv^2}{r}$

where:

• $m = 0.200 \, \text{kg}$ (mass of the stone)
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)
• $r = 0.800 \, \text{m} \cdot \cos(15^\circ)$ (horizontal radius of the circle)

From the vertical component equation: $$T \cos(15^\circ) = mg$$ $$T = \frac{mg}{\cos(15^\circ)}$$

Substitute $T$ from the previous step into the horizontal component equation: $$\frac{mg}{\cos(15^\circ)} \sin(15^\circ) = \frac{mv^2}{r}$$

Cancel $m$ from both sides and solve for $v$: $$\frac{g \sin(15^\circ)}{\cos(15^\circ)} = \frac{v^2}{r}$$ $$v^2 = r \cdot g \cdot \tan(15^\circ)$$ $$v = \sqrt{r \cdot g \cdot \tan(15^\circ)}$$

Substitute the values: $$v = \sqrt{(0.800 \cdot \cos(15^\circ)) \cdot 9.81 \cdot \tan(15^\circ)}$$

Final Answer