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