Maths › Further Mechanics 2 › Angular speed and horizontal circular motion
Angular speed and horizontal circular motion
Something moving in a circle at steady speed is still accelerating, because its direction keeps changing. The acceleration points at the centre, and something has to supply it.
Builds on Forces and Newton's laws and Radians, arcs and small angles.
IN THIS TOPIC
- Convert between linear speed, angular speed and period.
- Identify the force providing the radial acceleration in a given situation.
- Solve conical pendulum and banked track problems.
WHAT YOU PROBABLY THINK
A particle moving in a circle at constant speed has no acceleration, since its speed is not changing.
Turning is accelerating
Angular speed ω is the rate at which the angle at the centre increases, in radians per second, and the linear speed is v = rω. Even at constant speed the velocity is changing, because its direction is, so there is an acceleration. It points towards the centre and has magnitude:
The opening claim confuses speed with velocity, which is the same slip as thinking a rebounding ball needs no impulse. Nothing moves in a circle by itself: some real force, tension, friction, a normal reaction or gravity, must be pointing at the centre to supply that acceleration, and naming it is the first line of every solution.
WORKED EXAMPLE
A conical pendulum
A bob of mass 0.2 kg on a string of length 0.5 m moves in a horizontal circle with the string at 30° to the vertical. Find the tension and the angular speed, taking g = 9.8 m/s².
Vertically: T cos30° = 0.2(9.8), so T = 1.96/0.866 = 2.26 N.
The radius is 0.5 sin30° = 0.25 m. Horizontally: T sin30° = 0.2(0.25)ω².
So 1.132 = 0.05ω², giving ω² = 22.63 and ω = 4.76 rad/s. The period is 2π/ω = 1.32 s.
Banked tracks
On a banked track the normal reaction is no longer vertical, so its horizontal component can supply the radial acceleration without any friction at all. Resolving vertically and horizontally and dividing one equation by the other removes both the mass and the reaction:
That is the design speed for the bend. Below it the vehicle tends to slide down the bank and friction acts up; above it the tendency reverses. Questions that mention friction are asking for the fastest or slowest safe speed, which needs the friction term added with the right sign.
YOUR TURN
Designing a bend
A bend of radius 80 m is to be banked so that a car travelling at 20 m/s needs no sideways friction. Find the banking angle, taking g = 9.8 m/s².
Show the working
Resolving vertically: R cos θ = mg. Horizontally: R sin θ = mv²/r.
Dividing: tan θ = v²/(rg) = 400/(80 × 9.8) = 0.510.
So θ = 27.0°.
The mass has cancelled, so the same bank suits a lorry and a motorcycle: only the speed matters.
THE EXAM BIT
- Name the force providing the radial acceleration before writing any equation.
- Resolve vertically and radially, not along and perpendicular to a string.
- Use rω² when ω is given and v²/r when the speed is; converting first wastes a line.
- For a banked track, divide the two equations to remove the reaction and the mass.
CHECK YOURSELF
A particle moves in a circle of radius 2 m at 3 rad/s. Find its speed and its acceleration.
Show a hint
One multiplication each.
Show the answer
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Circular motion at constant speed still accelerates, towards the centre, with magnitude rω² or v²/r.
Some real force must supply it; on a banked track the reaction alone can, at the design speed given by tan θ = v²/rg.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Convert between linear speed, angular speed and period.
- Identify the force providing the radial acceleration in a given situation.
- Solve conical pendulum and banked track problems.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.