Solved Examples

These examples walk through how the calculator converts your inputs into standard SI units—kilograms (kg), meters per second (m/s), and meters (m)—before computing the required inward force using the core physics formula:

F_c = (m × v²) ÷ r (or F_c = m × v² / r)

1. The Swinging Tetherball (Standard Metric)

A ball with a mass of 2 kg is swung at the end of a rope in a horizontal circle with a radius of 1.5 m at a speed of 6 m/s.

• Input: Mass = 2 kg, Velocity = 6 m/s, Radius = 1.5 m
• Step 1 (Unit Check): All values are already in SI units (kg, m/s, and m).
• Step 2 (Calculate):F_c = (2 × 6²) ÷ 1.5F_c = (2 × 36) ÷ 1.5 = 72 ÷ 1.5 = 48 N
• Calculator Output: 48.00 N (48 Newtons of inward tension force)

2. The Highway Off-Ramp (Metric Conversions)

A car weighing 1.5 tonnes takes a circular highway exit ramp with a curve radius of 50 m at a speed of 54 km/h.

• Input: Mass = 1.5 tonne, Velocity = 54 km/h, Radius = 50 m
• Step 1 (Convert Mass): Multiply tonnes by 1,000 → 1,500 kg
• Step 2 (Convert Speed): Multiply km/h by 0.277778 → 15 m/s
• Step 3 (Calculate):F_c = (1,500 × 15²) ÷ 50F_c = (1,500 × 225) ÷ 50 = 6,750 N (or 6.75 kN)
• Calculator Output: 6,750.00 N (This is the sideways friction required from the tires to prevent a skid!)

3. The Amusement Park Ride (Imperial to SI Conversions)

A 160 lb rider on a spinning carnival ride moves in a circle with a radius of 15 ft at a linear speed of 30 mph.

• Input: Mass = 160 lb, Velocity = 30 mph, Radius = 15 ft
• Step 1 (Convert Mass): Multiply pounds by 0.453592 → 72.57 kg
• Step 2 (Convert Speed): Multiply mph by 0.44704 → 13.41 m/s
• Step 3 (Convert Radius): Multiply feet by 0.3048 → 4.57 m
• Step 4 (Calculate):F_c = (72.57 × 13.41²) ÷ 4.57F_c = (72.57 × 179.83) ÷ 4.57 = 2,855.12 N (roughly 642 lbf pressing them against the wall!)
• Calculator Output: 2,855.12 N

Practice Problems (Unsolved)

Test your grasp of circular motion by running these real-world scenarios through the calculator or solving them by hand!

1. The Toy Drone: A small drone with a mass of 250 g flies in a tight circle of radius 4 m at a speed of 8 m/s. How much centripetal force must the propellers generate to maintain this loop?
2. The Olympic Hammer Throw: An athlete spins a 7.26 kg metal ball on a wire with a radius of 1.8 m at an impressive speed of 25 m/s right before release. What is the tension force in the wire?
3. The Race Car Curve: A 1,200 kg Formula car takes a high-speed corner with a radius of 80 m at 180 km/h. What is the centripetal force required to keep the car on the track?
4. The Velocity Doubler: A 5 kg object moves in a 2 m circle at 10 m/s, requiring 250 N of force. If the speed is doubled to 20 m/s without changing the radius, what happens to the required force?

Answer Key

• Problem 1: 4.00 N (Mass → 0.25 kg × 64 ÷ 4)
• Problem 2: 2,520.83 N (7.26 × 625 ÷ 1.8)
• Problem 3: 37,500.00 N or 37.5 kN (Speed → 50 m/s; 1,200 × 2,500 ÷ 80)
• Problem 4: It quadruples to 1,000 N! Because velocity is squared in the formula ($v^2$), doubling your speed increases the required holding force by a factor of four ($2^2 = 4$).

Common Mistakes & Pitfalls

When calculating centripetal force by hand or setting up physics models, keep an eye out for these frequent calculation errors:

1. Forgetting to Square the Velocity (v^2)

The most common mathematical slip is multiplying m times v instead of m times v^2. Because speed is squared, it has a much larger impact on holding force than mass or radius. This is why taking a highway curve even slightly too fast dramatically increases your risk of skidding off the road!

2. Confusing Centripetal vs. Centrifugal Force

In physics, Centripetal Force (Fc) is the real, measurable inward force pulling an object toward the center of a circle (like rope tension or tire friction). Centrifugal force is the apparent outward sensation you feel when inside a spinning vehicle. That outward sensation is actually just your body’s inertia trying to travel in a straight line while the vehicle pushes you inward!

3. Using Diameter Instead of Radius

In circular problems, measurements are often given across the entire width of a loop or track (the diameter). The formula strictly requires the radius (r), which is exactly half of the diameter (r = d \2). If you accidentally plug in the diameter, your calculated force will be cut in half!

4. Dividing by Zero (The Flat Line Trap)

In geometry, a circle with a radius of 0 meters cannot exist, and mathematically, dividing by zero causes an error. As the radius gets smaller and tighter, the required force skyrockets toward infinity. Always ensure your radius is greater than zero ($r > 0$).bined forces acting on the object.