Laws of Motion · Lesson 11 of 14
Forces in Circular Motion
“Identify which real interaction supplies the radial force and apply force limits to strings and level-road turns.”
• Use the radial equation without adding a separate centripetal interaction. • Find tension and maximum speed for an ideal supported horizontal circle. • Explain release from a circular path. • Derive a level-road friction limit and distinguish real forces from apparent outward effects.
Centripetal describes a role, not a new interaction
The previous chapter established that a particle moving in a circle of radius R at speed v has inward radial acceleration v²/R. The second law therefore requires an inward net radial force mv²/R. This is called the centripetal force, but it must be supplied by identifiable interactions: tension, friction, gravity, normal reaction or their components. Do not add an extra “centripetal force” arrow alongside the forces already providing it.
A planet receives its inward gravitational force from the Sun. A particle moving on a smooth horizontal table and tied to a fixed peg receives its inward force from the string. Its weight and table normal balance vertically, while tension supplies the horizontal radial resultant. In that supported setup, the net force is T, with T = mv²/R; it is not T + mv²/R or T − mv²/R.
A string has a limiting tension
Problem
A 0.25 kg stone moves in an ideal supported horizontal circle of radius 1.5 m at 40 revolutions per minute. Find tension and the speed limit for a 200 N string.
- 1.Frequency is 40/60 = 2/3 Hz, so linear speed is 2π × 1.5 × 2/3 = 2π = 6.283 m/s.
- 2.T = 0.25 × (2π)²/1.5 = 6.580 N, about 6.6 N.
- 3.The limiting linear speed is √(200 × 1.5/0.25) = √1200 = 34.64 m/s, about 35 m/s.
- 4.These match the chapter’s intended radial-only calculation. If a stone is literally suspended with only gravity and a slanted string acting, the vertical tension component must also balance mg, so total tension exceeds the radial component. State the support model instead of silently ignoring gravity.
What happens if the string breaks?
Immediately after the string breaks, the object retains its instantaneous tangent velocity. It does not receive a radial outward launch merely because the inward force disappears. On a smooth horizontal table, with vertical weight and normal still balanced, it moves in a straight tangent line. If unsupported in the air, it initially departs along the tangent and then follows a gravity-curved projectile path. The initial direction and the entire later trajectory are different claims.
A level-road turn uses lateral static friction
For a car turning on a level road, vertical forces balance: N = mg when there is no vertical acceleration. Static tyre-road friction supplies the horizontal inward force while the tyres roll without sideways slip. At lower speeds the required friction is mv²/R; at the limit it reaches μ_sN. Beyond that limit, the assumed circular path cannot be maintained without slipping in this model.
Problem
A cyclist takes a 3 m radius level turn at 18 km/h. Let μ_s = 0.10 and g = 9.8 m/s². Is there enough static friction?
- 1.Convert 18 km/h to 5 m/s. Required v² is 25 m²/s².
- 2.The available limit μ_sRg is 0.10 × 3 × 9.8 = 2.94 m²/s², far below 25.
- 3.The maximum speed in this model is √2.94 = 1.715 m/s. At 5 m/s the static-friction condition fails, so the cyclist cannot maintain the turn without lateral slip.
- 4.This is a simplified lateral-traction test. Balancing the bicycle’s lean involves extended-body rotation, treated separately.
Why an outward push can be felt
In the ground inertial frame, a turning passenger needs an inward force from the seat, door or restraint. The body’s tendency to continue tangent to the path can feel like an outward push relative to the turning vehicle. In a rotating-frame analysis an outward centrifugal inertial term may be introduced, but it is not an additional material interaction in the ground-frame force diagram. Keep the frame consistent.
When speed changes around the circle
A tangential net force changes speed, while the radial component supplies v²/R. Therefore total force is purely inward only in uniform circular motion. If a car accelerates or brakes during a turn, tyre friction may have to supply both longitudinal and lateral components; the simple bound using only mv²/R no longer accounts for the entire friction demand.
Quiz
A particle circles a peg on a smooth horizontal table, with string tension T providing the only horizontal force. What is its net horizontal force?
Key Takeaways
• Look for real agents supplying the required inward force. • The radial equation is net inward force = mv²/R. • A string-tension calculation needs a stated vertical support model. • When radial support vanishes, initial velocity is tangent. • Level-road turns require sufficient static friction, and apparent outward effects depend on the frame.