Questions on angular speed and horizontal circular motion

Question 1

Explain why a particle moving at constant speed in a circle is accelerating, and state the direction of that acceleration.

Accepted 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.

Question 2

A particle moves in a circle of radius 1.5 m with angular speed 4 rad/s. Find its linear speed and its acceleration.

Accepted 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.

Question 3

A wheel of radius 0.35 m turns at a steady 300 revolutions per minute. Find its angular speed in rad/s and the speed of a point on its rim.

Accepted Answer

One revolution is 2π radians, so ω = 300 × 2π/60 = 10π = 31.4 rad/s. Then v = rω = 0.35 × 31.4 = 11.0 m/s. M1 for converting revolutions per minute to radians per second, A1 for 31.4 rad/s, A1 for 11.0 m/s. The conversion is the whole of the difficulty: 300 rev/min is five revolutions a second, each of 2π radians, and a point on the rim covers 2π × 0.35 = 2.2 m in every 0.2 s.

Question 4

For each of the following, name the force, or the component of a force, that provides the acceleration towards the centre: a car rounding a flat horizontal bend; a satellite in a circular orbit; the bob of a conical pendulum.

Accepted Answer

For the car, the sideways friction between the tyres and the road, since the weight and the normal reaction are both vertical and cannot act towards the centre. For the satellite, the gravitational attraction of the planet, which is the only force acting. For the bob, the horizontal component of the tension in the string, the vertical component being what balances the weight. B1 B1 B1, one for each. In every case the force towards the centre is not a new force added to the diagram: it is the resultant of the forces already there, and writing it in as an extra force is the error to avoid.

Question 5

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.

Accepted Answer

Resolve vertically to obtain T cos 40° = 0.4g. The bob has no vertical acceleration, so T = 3.92/0.766 = 5.12 N.
The radius of the horizontal circle is 1.2 sin 40° = 0.771 m, not the string length.
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.

Question 6

A bend of radius 120 m is banked at 15°. Find the speed at which no sideways friction is needed.

Accepted 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.

Question 7

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.

Accepted 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.
In this ideal model the mass cancels because both relevant forces are proportional to mass.

Question 8

A particle moves so that its position vector at time t seconds, relative to the centre of its path, is r = (0.8 cos 5t)i + (0.8 sin 5t)j metres. Find its velocity vector, show that the velocity is perpendicular to r, and show that the acceleration is directed towards the centre with magnitude rω².

Accepted Answer

Differentiating component by component, v = (−4 sin 5t)i + (4 cos 5t)j, so the speed is 4 m/s because sin² + cos² = 1. The scalar product r · v = 0.8(−4) cos 5t sin 5t + 0.8(4) sin 5t cos 5t = 0, so the velocity is perpendicular to the position vector and points along the tangent.
Differentiating again, a = (−20 cos 5t)i + (−20 sin 5t)j = −25r. A negative multiple of r points from the particle straight back at the centre, and the magnitude is 25 × 0.8 = 20 m/s², which is rω² with r = 0.8 and ω = 5.
M1 for differentiating the position vector, A1 for the velocity, B1 for the scalar product being zero, M1 for differentiating again, A1 for a = −25r with the magnitude 20 m/s².

Question 9

A coin rests on a rough horizontal turntable, 0.2 m from the axis, and the turntable turns steadily at 45 revolutions per minute. The coefficient of friction between the coin and the surface is 0.5. Show that the coin does not slip, and find the greatest distance from the axis at which it could rest without slipping, taking g = 9.8 m/s².

Accepted Answer

Convert before anything else: ω = 45 × 2π/60 = 4.71 rad/s, so ω² = 22.2. The only horizontal force on the coin is friction, so friction must supply the whole radial acceleration, and the force needed is mrω² = 0.2 × 22.2m = 4.44m N.
The most friction available is μR = μmg = 0.5 × 9.8m = 4.9m N. Since 4.44m < 4.9m the friction required is within what the surface can provide, so the coin does not slip.
At the greatest distance the friction is limiting, μmg = mrω², so r = μg/ω² = 4.9/22.2 = 0.221 m.
B1 for converting to 4.71 rad/s, M1 for friction supplying mrω², M1 for comparing with limiting friction μmg, A1 for 4.44 < 4.9 with the conclusion stated, A1 for 0.221 m. The mass cancels throughout, so the answer is the same for any coin.

Question 10

A particle of mass 0.2 kg moves in a horizontal circle of radius 0.4 m on a smooth horizontal table. It is attached to a light inextensible string which passes through a small smooth hole in the table at the centre of the circle and carries a particle of mass 0.5 kg hanging at rest below the table. Find the angular speed and the speed of the moving particle, taking g = 9.8 m/s².

Accepted Answer

The hanging particle is at rest, so the tension is its weight: T = 0.5 × 9.8 = 4.9 N. The string is the only horizontal force on the moving particle, so the tension provides the whole of the acceleration towards the centre: T = mrω², giving 4.9 = 0.2 × 0.4 × ω², ω² = 61.25 and ω = 7.83 rad/s. Then v = rω = 0.4 × 7.83 = 3.13 m/s. B1 for T = 4.9 N, M1 for T = mrω², A1 for 7.83 rad/s, A1 for 3.13 m/s. The smooth hole means the tension is the same on both sides.

Question 11

A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 30 N. The other end is fixed to a point O on a smooth horizontal table, and the particle moves in a horizontal circle about O with angular speed 6 rad/s. Find the extension of the string and the radius of the circle.

Accepted Answer

Let the extension be x, so the radius is r = 0.5 + x. By Hooke's law the tension is T = λx/l = 30x/0.5 = 60x. The tension is the only horizontal force, so T = mrω²: 60x = 0.3(0.5 + x)(36) = 10.8(0.5 + x) = 5.4 + 10.8x. Then 49.2x = 5.4 and x = 0.110 m, so the radius is 0.5 + 0.110 = 0.610 m. M1 for Hooke's law T = 60x, M1 for T = mrω² with r = 0.5 + x, A1 for 60x = 5.4 + 10.8x, A1 for the extension, A1 for the radius. The radius is the stretched length, not the natural length, and x appears on both sides of the equation for that reason; the tension comes to 60 × 0.110 = 6.59 N.

Question 12

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.

Accepted 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. 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: 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.

Question 13

A particle of mass 0.5 kg is joined to two points A and B on a fixed vertical axis, A being 0.5 m above B, by light inextensible strings of length 0.3 m to A and 0.4 m to B. The particle moves in a horizontal circle at 8 rad/s with both strings taut. Find the tension in each string, and the least angular speed at which the lower string remains taut. Take g = 9.8 m/s².

Accepted Answer

Geometry first: 0.3² + 0.4² = 0.5², so the strings meet at right angles at the particle, the radius is 0.3 × 0.4 ÷ 0.5 = 0.24 m, and the particle sits 0.18 m below A and 0.32 m above B. That gives cos α = 0.6 and sin α = 0.8 for the upper string, cos β = 0.8 and sin β = 0.6 for the lower.
With mg = 4.9 N and mrω² = 0.5 × 0.24 × 64 = 7.68 N, resolving vertically (the upper string holds the particle up, the lower pulls it down): 0.6T₁ − 0.8T₂ = 4.9. Resolving horizontally towards the centre, both pull inwards: 0.8T₁ + 0.6T₂ = 7.68.
Multiply the first by 0.6 and the second by 0.8 and add, and T₂ drops out: T₁ = 0.6(4.9) + 0.8(7.68) = 9.08 N. Then T₂ = 0.6(7.68) − 0.8(4.9) = 0.688 N.
The lower string stays taut while T₂ ≥ 0, which is the condition rω² ≥ g tan α. So ω² ≥ 9.8(0.8/0.6)/0.24 = 54.4 and ω ≥ 7.38 rad/s. The given 8 rad/s is above it, which is why T₂ came out positive but small.
B1 for the radius 0.24 m and both angle pairs from the geometry, M1 for resolving vertically, A1 for 0.6T₁ − 0.8T₂ = 4.9, M1 for resolving horizontally, A1 for 0.8T₁ + 0.6T₂ = 7.68, A1 for both tensions, M1 for setting T₂ = 0 to reach rω² = g tan α, A1 for 7.38 rad/s.

Question 14

A particle of mass m moves in a horizontal circle with angular speed ω on the smooth inside surface of a fixed hemispherical bowl of radius R, the centre of the circle being a distance h below the centre of the bowl. Show that ω² = g/h. Given that R = 0.5 m and ω = 7 rad/s, find the radius of the circle and the magnitude of the normal reaction in terms of mg. Take g = 9.8 m/s².

Accepted Answer

The forces on the particle are its weight and the normal reaction N, which acts along the radius of the bowl towards its centre. Let that radius make angle θ with the vertical, so the circle has radius R sin θ and h = R cos θ.
Vertically: N cos θ = mg. Horizontally, towards the centre of the circle: N sin θ = m(R sin θ)ω². Dividing the second by the first, tan θ = R sin θ ω²/g, so ω² = g/(R cos θ) = g/h.
With ω = 7: h = 9.8/49 = 0.2 m below the centre, so the circle has radius √(0.5² − 0.2²) = 0.458 m. Then cos θ = h/R = 0.4 and N = mg/cos θ = 2.5mg.
M1 for resolving vertically, M1 for resolving towards the centre with radius R sin θ, A1 for ω² = g/h, A1 for h = 0.2 m, A1 for the radius 0.458 m, A1 for N = 2.5mg. The normal reaction does two jobs at once, holding the particle up and turning it, which is why it exceeds the weight; a faster particle rides higher, since h = g/ω² shrinks, and it can never reach the rim, where h would be zero.

Question 15

A conical pendulum consists of a particle on a light inextensible string of length l, moving in a horizontal circle with the string at angle θ to the vertical and angular speed ω. Show that ω² = g/(l cos θ). Deduce that the motion is possible only if ω exceeds a certain value, which you should find, and explain why the string can never be horizontal, however fast the particle moves.

Accepted Answer

Vertically, with no vertical acceleration: T cos θ = mg. Horizontally, towards the centre of a circle of radius l sin θ: T sin θ = m(l sin θ)ω². The sin θ cancels, so T = mlω², and substituting into the first equation gives mlω² cos θ = mg, that is ω² = g/(l cos θ).
Since cos θ ≤ 1, ω² ≥ g/l, with equality only when θ = 0 and the string hangs vertically with no circle at all. So a circle of any size needs ω > √(g/l).
As θ approaches 90°, cos θ approaches 0 and ω² = g/(l cos θ) grows without bound, as does T = mg/cos θ. A horizontal string would have no vertical component of tension to balance the weight, so no finite angular speed can produce it.
M1 for the two resolved equations, A1 for ω² = g/(l cos θ), M1 for cos θ ≤ 1, A1 for ω > √(g/l), B1 for the argument about the vertical component of the tension. The lower bound √(g/l) is also the angular frequency of small swings of the same pendulum: a slow spin is exactly what a small swing looks like from above.