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Angular speed and horizontal circular motion questions
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.
6 original questions · 25 marks · the angular speed and horizontal circular motion notes · Further Mechanics 2
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Explain why a particle moving at constant speed in a circle is accelerating, and state the direction of that acceleration.
Worked answer
Velocity is a vector, and although its magnitude is fixed its direction is continually changing, so the velocity is changing and there is an acceleration. B1 for the changing direction of the velocity, B1 for the direction of the acceleration. It points towards the centre of the circle, with magnitude rω² or v²/r.A particle moves in a circle of radius 1.5 m with angular speed 4 rad/s. Find its linear speed and its acceleration.
Worked answer
v = rω = 1.5 × 4 = 6 m/s. Acceleration = rω² = 1.5 × 16 = 24 m/s², directed towards the centre. B1 for the speed, M1 A1 for the acceleration. Using v²/r gives 36/1.5 = 24 as well.A bob of mass 0.4 kg is attached to one end of a light inextensible string of length 1.2 m, and moves in a horizontal circle with the string at 40° to the vertical. Find the tension in the string and the period of the motion, taking g = 9.8 m/s².
Worked answer
Resolving vertically, the bob has no vertical acceleration, so T cos 40° = 0.4(9.8) and T = 3.92/0.766 = 5.12 N. Two of the marks are for that equation; writing T = mg with no resolving loses both.
The radius is not the string length. It is 1.2 sin 40° = 0.771 m, and using 1.2 instead is the commonest error in the question.
Resolving horizontally towards the centre, T sin 40° = 0.4rω². Substituting T and r, the 0.4 and the sin 40° both cancel, leaving ω² = g/(l cos 40°) = 9.8/0.919 = 10.66, so ω = 3.27 rad/s.
The period is 2π/ω = 1.92 s.
M1 A1 for resolving vertically and the tension, B1 for the radius 1.2 sin 40°, M1 A1 for the horizontal equation and ω, A1 for the period.A bend of radius 120 m is banked at 15°. Find the speed at which no sideways friction is needed.
Worked answer
Vertically R cos θ = mg and horizontally R sin θ = mv²/r. Dividing removes both R and m: tan θ = v²/(rg), so v² = 120 × 9.8 × tan15° = 315.1 and v = 17.8 m/s. M1 for both resolved equations, M1 for dividing to reach tan θ = v²/(rg), A1 for v² = 315.1, A1 for 17.8 m/s. Below that speed friction must act up the slope, and above it down.A car travels round a flat bend of radius 50 m on a road where the coefficient of friction is 0.6. Find the greatest speed at which it can do so without slipping, taking g = 9.8 m/s².
Worked answer
On a flat bend the only horizontal force is friction, so it supplies the whole of the radial acceleration: F = mv²/r. At the greatest speed the friction is limiting, F = μR, and resolving vertically gives R = mg, so F = 0.6mg.
Setting the two expressions equal, v² = μgr = 0.6 × 9.8 × 50 = 294, so v = 17.1 m/s.
M1 for F = mv²/r, B1 for stating that the friction is limiting, M1 for R = mg and F = μR, A1 for 17.1 m/s.
The mass cancels, so a loaded lorry and an empty car slip at the same speed. State that the friction is at its limiting value; the assumption is worth a mark and is often left unsaid.A bend of radius 60 m is banked at 20° to the horizontal, and the coefficient of friction between the tyres and the road surface is 0.3. Find the greatest speed at which a car can go round the bend without slipping, taking g = 9.8 m/s².
Worked answer
At the greatest speed the car is on the point of sliding up the slope, so the friction acts down the slope and is limiting, F = 0.3R. Getting that direction the wrong way round gives the least speed instead, and it is the decision the whole question turns on. Draw the forces before writing anything.
Resolving vertically, with no vertical acceleration: R cos 20° − 0.3R sin 20° = mg.
Resolving horizontally towards the centre: R sin 20° + 0.3R cos 20° = mv²/60.
Dividing the second equation by the first removes both R and m, which is the method mark: v²/(60g) = (sin 20° + 0.3 cos 20°)/(cos 20° − 0.3 sin 20°) = (tan 20° + 0.3)/(1 − 0.3 tan 20°).
With tan 20° = 0.364 that ratio is 0.664/0.891 = 0.745, so v² = 60(9.8)(0.745) = 438.3 and v = 20.9 m/s.
B1 for the friction acting down the slope at its limiting value, M1 A1 for resolving vertically, M1 A1 for resolving horizontally, A1 for 20.9 m/s.
As a check, reversing the sign of the friction gives the least speed, 5.82 m/s, and the frictionless design speed of 14.6 m/s lies between the two, as it must.
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