Circular motion and solving problems

Every problem in this chapter is solved the same way:

• Choose the body (or system) you are looking at.
• Draw its free-body diagram: every force on that body from outside, such as weight, normal force, tension, friction. Leave out the forces it exerts on other bodies.
• Apply F = ma to it, in each direction.

A string pulls away from every body it is tied to. A light string has the same tension all along, and if it stays taut the bodies it joins share the same acceleration.

Example: blocks of 2 kg, 3 kg and 5 kg in a row on a smooth table, joined by strings, are pulled by 20 N on the 5 kg block. Treat all three as one system: a = 20/10 = 2 m s⁻². The string pulling the 2 kg block needs T = 2 × 2 = 4 N; the string pulling the 2 kg and 3 kg blocks together needs T = 5 × 2 = 10 N.

Over a pulley, a heavier mass m₂ and a lighter m₁: the heavier one goes down. For m₂: m₂g – T = m₂a; for m₁: T – m₁g = m₁a. Adding gives a = (m₂ – m₁)g/(m₁ + m₂). With 3 kg and 5 kg, a = 2 × 9.8/8 ≈ 2.45 m s⁻².

A body moving in a circle needs a net force mv²/R towards the centre. This is not a new kind of force: it must come from a real one, such as tension, friction or a normal force.

• On a level road, friction provides it, so a car can take the turn only if v² ≤ μ_s Rg. On a curve of radius 50 m with μ_s = 0.5 (g = 10), v_max = √250 ≈ 15.8 m/s, about 57 km/h. This is how speed limits on bends are set.
• A cyclist leans inwards, so that part of the push from the road points to the centre: tan θ = v²/Rg. At 10 m/s on a 20 m bend, tan θ ≈ 0.51, a lean of about 27°.

On a banked road the outer edge is raised, so the normal force itself leans towards the centre. At the optimum speed v₀ = √(Rg tan θ) no friction is needed at all, and the tyres wear least. A 100 m curve banked at 10° (tan 10° ≈ 0.18) has v₀ ≈ 13 m/s. With friction helping, the top speed rises to v_max = √[Rg(μ_s + tan θ)/(1 – μ_s tan θ)].