Explanation:
The ideal speed for a cyclist to navigate a curve on an inclined track without friction can be calculated using the centripetal force formula:
\[ F_{\text{centripetal}} = \frac{m \cdot v^2}{r} \]
Where:
- \( F_{\text{centripetal}} \) is the centripetal force,
- \( m \) is the mass of the cyclist,
- \( v \) is the velocity of the cyclist,
- \( r \) is the radius of the curve.
Additionally, the force due to gravity component acting down the incline is \( F_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \), where \( \theta \) is the slope angle.
For equilibrium on the inclined track, the centripetal force equals the force due to gravity component:
\[ F_{\text{centripetal}} = F_{\text{gravity}} \]
\[ \frac{m \cdot v^2}{r} = m \cdot g \cdot \sin(\theta) \]
The mass \( m \) cancels out:
\[ v^2 = r \cdot g \cdot \sin(\theta) \]
\[ v = \sqrt{r \cdot g \cdot \sin(\theta)} \]
Given:
- \( r = 35 \) m (radius of the curve),
- \( \theta = 33^\circ \) (slope angle),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
Let's calculate \( v \):
\[ v = \sqrt{35 \cdot 9.81 \cdot \sin(33^\circ)} \]
\[ v \approx \sqrt{35 \cdot 9.81 \cdot 0.545} \]
\[ v \approx \sqrt{188.97} \]
\[ v \approx 13.75 \, \text{m/s} \]
Therefore, the ideal speed at which the cyclist can travel that track is approximately \( 13.75 \, \text{m/s} \).
