Section 3.4: Circular Motion and Normal Force
When a car or object moves over a curved path such as a hill or dip, the normal force varies depending on speed and curvature. At the top of a hill, the normal force decreases; at the bottom of a dip, it increases.
Top of a hill: \( N_{\text{top}} = mg - m \frac{v^2}{r} \)
Bottom of a dip: \( N_{\text{bottom}} = mg + m \frac{v^2}{r} \)
Key Concepts:
- Centripetal acceleration is toward the center of curvature.
- Weight acts downward; normal force adjusts to provide net centripetal force.
- At maximum speed over a hill, normal force can reach zero — the object is momentarily weightless.
Example 1 — Car Cresting a Hill
A car of mass 800 kg passes over the crest of a rounded hill of radius 20 m with speed 12 m/s. Determine the normal force and whether the car stays in contact.
Normal force: \( N = mg - m \frac{v^2}{r} = 800 \cdot 9.8 - 800 \cdot \frac{12^2}{20} = 7840 - 5760 = 2080 \text{ N} \)
Since \( N > 0 \), the car stays in contact with the road.
Practice Problems
- A 500 kg car goes over a hill of radius 15 m at 20 m/s. Find the normal force.
- Determine the speed at which a 1000 kg car would lose contact at the top of a 25 m radius hill.
- A roller coaster of mass 400 kg is at the bottom of a circular dip of radius 10 m, moving at 15 m/s. Find the normal force.
- A car moves over a hill of radius 30 m at 18 m/s. Compute normal force at the top and bottom.
- A motorbike of 250 kg goes over a bump of radius 8 m at 12 m/s. Find if it remains in contact.