AS & A Level Physics (9702)
Motion in a circle Banker
16 questions · about 2% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Centripetal acceleration & force · Core · Paper 4 style · 5 marksA car of mass 1200 kg travels around a flat roundabout in Islamabad at a constant speed of 12 m s⁻¹ along a circular path of radius 25 m.
- (a) Calculate the centripetal acceleration of the car. [2]
- (b) Calculate the resultant force on the car and state its direction. [2]
- (c) State what provides this force. [1]
Show answer
- (a) 5.76 m s⁻²
a = v²/r (M1) = 144/25 = 5.76 m s⁻² (A1).
- (b) 6910 N towards the centre of the roundabout
F = ma = 1200 × 5.76 = 6910 N (A1); towards the centre (B1).
- (c) Friction between the tyres and the road
B1: (sideways) friction of road on tyres.
A ball on a string moves at constant speed in a horizontal circle on a smooth horizontal table, the other end of the string being fixed to a peg at the centre.
- (a) Explain why the ball is accelerating even though its speed is constant. [2]
- (b) State the direction of the resultant force on the ball and explain why this force does no work on the ball. [2]
- (c) The string breaks. Describe the subsequent path of the ball, as seen from above. [1]
Show answer
- (a) Its direction of motion (velocity) changes continuously; acceleration is rate of change of velocity.
B1: velocity is a vector / direction changes; B1: change in velocity means acceleration.
- (b) Towards the centre of the circle; it is always perpendicular to the velocity, so there is no displacement in the direction of the force.
B1: towards centre; B1: force perpendicular to motion → no work done, so kinetic energy (speed) constant.
- (c) A straight line along the tangent to the circle at the point of release.
B1: tangential straight line (no resultant horizontal force).
A ball of mass 0.25 kg is whirled in a vertical circle on a light string of length 0.80 m. At the top of the circle its speed is 4.0 m s⁻¹. g = 9.81 m s⁻².
- (a) Calculate the tension in the string when the ball is at the top of the circle. [3]
- (b) Calculate the minimum speed at the top for the string to remain taut. [2]
Show answer
- (a) 2.55 N
At top, T + mg = mv²/r (M1); T = 0.25 × 16/0.80 − 0.25 × 9.81 (M1) = 2.55 N (A1).
- (b) 2.80 m s⁻¹
T = 0 → mg = mv²/r (M1); v = √(gr) = 2.80 m s⁻¹ (A1).
The drum of a washing machine has radius 0.25 m and spins at 1200 revolutions per minute during the spin cycle.
- (a) Calculate the angular speed of the drum in rad s⁻¹. [1]
- (b) Calculate the centripetal force needed to keep a 0.10 kg piece of wet cloth moving with the drum wall. [2]
- (c) Explain why water leaves the clothes through the holes in the drum wall. [2]
Show answer
- (a) 126 rad s⁻¹
ω = 1200 × 2π/60 = 126 rad s⁻¹ (A1).
- (b) 395 N
F = mω²r (M1) = 0.10 × 126² × 0.25 = 395 N (A1).
- (c) The drum wall pushes the clothes towards the centre, but where there is a hole there is no force on the water, so it continues in a straight line (tangentially) out through the hole.
B1: water needs a centripetal force to stay in the circle; B1: at the holes no force acts on it, so it moves off along a tangent.
An athlete swings a hammer of mass 7.26 kg on a wire so that it moves in a circle of radius 1.9 m. Just before release the speed of the hammer is 25 m s⁻¹. g = 9.81 m s⁻².
- (a) Calculate the tension in the wire just before release. Assume the hammer moves in a horizontal circle and ignore its weight. [2]
- (b) Calculate the centripetal acceleration of the hammer and compare it with g. [2]
Show answer
- (a) 2390 N
T = mv²/r (M1) = 7.26 × 625/1.9 = 2390 N (A1).
- (b) 329 m s⁻², about 34 times g
a = v²/r = 329 m s⁻² (A1); ratio a/g ≈ 34 (A1).
A section of a racing track is a horizontal circular arc of radius 80 m, banked at 15° to the horizontal. A car of mass m travels round it. g = 9.81 m s⁻².
- (a) By considering the vertical and horizontal components of the normal contact force N, show that the speed at which no friction is needed is given by v² = rg tan θ. [3]
- (b) Calculate this speed. [1]
- (c) Explain what additional force is needed if the car travels faster than this speed, and state its direction. [2]
Show answer
- (a) N cos θ = mg and N sin θ = mv²/r; dividing gives tan θ = v²/(rg).
B1: N cos θ = mg (no vertical acceleration); B1: N sin θ = mv²/r (provides centripetal force); B1: divide to obtain v² = rg tan θ.
- (b) 14.5 m s⁻¹
v = √(80 × 9.81 × tan 15°) = 14.5 m s⁻¹ (A1).
- (c) Friction acting down the slope (towards the centre side), to provide the extra centripetal force.
B1: N sin θ alone is too small to give mv²/r; B1: friction needed, acting down the banking (parallel to surface).
A large Ferris wheel at a funfair has radius 20 m and turns at a constant rate, making one revolution every 60 s. A passenger of mass 60 kg sits in a seat that stays upright. g = 9.81 m s⁻².
- (a) Calculate the centripetal acceleration of the passenger. [2]
- (b) Calculate the contact force of the seat on the passenger at the top and at the bottom of the wheel. [3]
- (c) Explain why the passenger feels lighter at the top. [1]
Show answer
- (a) 0.219 m s⁻²
ω = 2π/60 = 0.105 rad s⁻¹ (M1); a = ω²r = 0.219 m s⁻² (A1).
- (b) top 575 N; bottom 602 N
Top: mg − N = ma → N = 60(9.81 − 0.219) = 575 N (M1 A1); bottom: N − mg = ma → N = 602 N (A1).
- (c) The seat pushes up with a force less than her weight because part of her weight provides the centripetal force.
B1: contact force (felt as weight) is less than mg at the top.
In a 'well of death' show at a fair in Pakistan, a motorcyclist rides around the inside of a vertical cylindrical wooden wall of radius 5.0 m, moving in a horizontal circle. The total mass of rider and motorcycle is 250 kg and the speed is 15 m s⁻¹. g = 9.81 m s⁻².
- (a) Calculate the normal contact force of the wall on the motorcycle and rider. [2]
- (b) State the size of the frictional force on the tyres and explain why it has this value. [2]
- (c) The maximum frictional force available is 0.60 times the normal contact force. Calculate the minimum speed at which the rider can go round without sliding down. [2]
Show answer
- (a) 1.13 × 10⁴ N
N = mv²/r (M1) = 250 × 225/5.0 = 11 250 ≈ 1.13 × 10⁴ N (A1; accept 1.1 × 10⁴ N).
- (b) 2450 N
B1: there is no vertical acceleration, so friction (upwards) balances weight; B1: 250 × 9.81 = 2450 N.
- (c) 9.04 m s⁻¹
0.60 mv²/r = mg (M1); v = √(gr/0.60) = 9.04 m s⁻¹ (A1).
A passenger aircraft of mass 5.0 × 10⁴ kg flies at a constant speed of 120 m s⁻¹ and turns in a horizontal circle of radius 2.0 km by banking its wings. The lift force acts perpendicular to the wings. g = 9.81 m s⁻².
- (a) Explain how the lift force allows the aircraft to turn in a horizontal circle. [2]
- (b) Calculate the angle of bank. [2]
- (c) Calculate the lift force. [2]
Show answer
- (a) When banked, the lift (perpendicular to the wings) is tilted; its vertical component balances the weight and its horizontal component provides the centripetal force.
B1: vertical component of lift = weight; B1: horizontal component of lift towards centre = mv²/r.
- (b) 36.3°
tan θ = v²/(rg) = 14400/(2000 × 9.81) (M1); θ = 36.3° (A1).
- (c) 6.08 × 10⁵ N
L cos θ = mg (M1); L = 5.0 × 10⁴ × 9.81/cos 36.3° = 6.08 × 10⁵ N (A1).
A small coin is placed on a horizontal turntable 8.0 cm from its centre. The speed of rotation is increased slowly; the coin starts to slide when the turntable reaches 45 revolutions per minute.
- (a) Calculate the centripetal acceleration of the coin just before it slides. [2]
- (b) Hence find the maximum frictional force on the coin as a fraction of its weight. [1]
- (c) The turntable is now set to 33 revolutions per minute. Calculate the greatest distance from the centre at which the coin can be placed without sliding. [2]
Show answer
- (a) 1.78 m s⁻²
ω = 45 × 2π/60 = 4.71 rad s⁻¹ (M1); a = ω²r = 1.78 m s⁻² (A1).
- (b) 0.181
F/mg = a/g = 0.181 (A1).
- (c) 14.9 cm
Same maximum acceleration: r = a/ω² (M1) with ω = 3.46 rad s⁻¹ → r = 0.149 m (A1).
A laboratory centrifuge in a hospital spins blood samples. A sample is 9.0 cm from the axis and must have a centripetal acceleration of 3000g, where g = 9.81 m s⁻².
- (a) Calculate the angular speed needed. [2]
- (b) Express this as revolutions per minute. [1]
- (c) Calculate the resultant force needed on 2.0 g of material in the tube at this radius. [1]
Show answer
- (a) 572 rad s⁻¹
ω²r = 3000 × 9.81 (M1); ω = √(29430/0.090) = 572 rad s⁻¹ (A1).
- (b) 5460 rev min⁻¹
572 × 60/(2π) = 5460 (A1).
- (c) 58.9 N
F = ma = 2.0 × 10⁻³ × 29430 = 58.9 N (A1).
A small metal bob of mass 0.15 kg hangs from a light string 1.2 m long. It moves in a horizontal circle at constant speed so that the string makes an angle of 30° with the vertical (a conical pendulum). g = 9.81 m s⁻².
- (a) Calculate the tension in the string. [2]
- (b) Calculate the speed of the bob. [3]
- (c) Calculate the time for one revolution. [1]
Show answer
- (a) 1.70 N
Vertical: T cos 30° = mg (M1); T = 0.15 × 9.81/cos 30° = 1.70 N (A1).
- (b) 1.84 m s⁻¹
r = 1.2 sin 30° = 0.60 m (B1); T sin 30° = mv²/r (M1); v = √(gr tan 30°) = 1.84 m s⁻¹ (A1).
- (c) 2.05 s
T = 2πr/v = 2.05 s (A1).
A car of mass 900 kg drives over a humpbacked bridge whose top is a vertical circular arc of radius 30 m. g = 9.81 m s⁻².
- (a) Calculate the normal contact force on the car at the top of the bridge when its speed is 12 m s⁻¹. [3]
- (b) Determine the maximum speed at which the car can cross the top of the bridge without losing contact with the road. [2]
- (c) Explain why the driver finds it harder to steer and brake near the top of the bridge at high speed. [2]
Show answer
- (a) 4510 N
mg − N = mv²/r (M1); N = 900 × 9.81 − 900 × 144/30 (M1) = 4510 N (A1).
- (b) 17.2 m s⁻¹
N = 0 → v² = gr (M1); v = √(9.81 × 30) = 17.2 m s⁻¹ (A1).
- (c) The normal contact force is reduced, so the maximum friction between tyres and road is reduced.
B1: normal force smaller at higher speed; B1: friction (grip) depends on the normal force, so less force available for steering/braking.
A ball of mass 0.20 kg on a light inextensible string of length 0.70 m is swung in a vertical circle. At the lowest point its speed is 6.0 m s⁻¹. Air resistance is negligible; g = 9.81 m s⁻².
- (a) Use conservation of energy to calculate the speed of the ball at the top of the circle. [2]
- (b) Calculate the tension in the string at the top and at the bottom of the circle. [4]
- (c) State where in the circle the string is most likely to break, with a reason. [1]
Show answer
- (a) 2.92 m s⁻¹
½mv_b² = ½mv_t² + mg(2r) (M1); v_t² = 36 − 4 × 9.81 × 0.70 = 8.53 → v_t = 2.92 m s⁻¹ (A1).
- (b) top 0.476 N; bottom 12.2 N
Top: T + mg = mv²/r (M1) → T = 0.20 × 8.53/0.70 − 1.962 = 0.476 N (A1); bottom: T − mg = mv²/r (M1) → T = 0.20 × 36/0.70 + 1.962 = 12.2 N (A1).
- (c) At the bottom, where the tension is greatest.
B1: at the bottom — weight acts away from the centre so tension must supply mg + mv²/r and speed is greatest.
A student investigates circular motion with a rubber bung of mass m tied to a string. The string passes through a vertical glass tube held in the hand and a mass of 0.100 kg hangs from its lower end, so the tension in the string is F = 0.981 N. The bung is whirled in a horizontal circle of radius r and the period T is measured:r = 0.300 m, T = 0.49 sr = 0.400 m, T = 0.58 sr = 0.500 m, T = 0.62 sr = 0.600 m, T = 0.69 sr = 0.700 m, T = 0.76 sr = 0.800 m, T = 0.79 s
- (a) Show that T² = (4π²m/F) r, where F is the tension in the string. [2]
- (b) Calculate T² for each reading. [1]
- (c) The best-fit straight line through the origin of a graph of T² against r has gradient 0.798 s² m⁻¹. Determine the mass of the bung. [2]
- (d) Suggest two reasons why the tension in the string may not equal the weight of the hanging mass, and explain one way to reduce the uncertainty in T. [3]
Show answer
- (a) F = mω²r and ω = 2π/T, so F = 4π²mr/T², giving T² = (4π²m/F)r.
M1: F = mrω² with ω = 2π/T; A1: rearrangement shown.
- (b) T² / s²: 0.240, 0.336, 0.384, 0.476, 0.578, 0.624
B1: all correct to 3 s.f.
- (c) 0.020 kg
m = gradient × F/(4π²) (M1) = 0.798 × 0.981/39.5 = 0.020 kg (A1).
- (d) Friction between string and tube; string not exactly horizontal; time 10 revolutions and repeat.
B1 each for any two: friction at the tube; string not horizontal (only a component of tension is horizontal / r measured to the top of the tube is not the true radius); mass of string. B1: time 10 or 20 revolutions and divide; repeat and average; video with frame-by-frame timing.
- (a) Define the radian. [1]
- (b) Convert 270° to radians. [1]
- (c) A point on a circle of radius 0.35 m moves through an angle of 2.4 rad. Calculate the length of arc travelled. [1]
Show answer
- (a) The angle subtended at the centre of a circle by an arc equal in length to the radius.
B1: arc length = radius.
- (b) 4.71 rad (3π/2)
270 × π/180 = 4.71 rad (A1).
- (c) 0.84 m
s = rθ = 0.35 × 2.4 = 0.84 m (A1).
A ceiling fan has blades of length 0.60 m measured from the axis and rotates at 180 revolutions per minute.
- (a) Calculate the angular speed of the blades. [2]
- (b) Calculate the speed of a blade tip. [1]
- (c) State how the angular speed and the linear speed of a point halfway along a blade compare with those of the tip. [2]
Show answer
- (a) 18.8 rad s⁻¹
f = 180/60 = 3.0 rev s⁻¹ (M1); ω = 2πf = 18.8 rad s⁻¹ (A1).
- (b) 11.3 m s⁻¹
v = ωr = 18.8 × 0.60 = 11.3 m s⁻¹ (A1).
- (c) Same angular speed; half the linear speed.
B1: ω same for all points on a rigid blade; B1: v = ωr so half.
A large clock on a clock tower has hands that move steadily.
- (a) Calculate the angular speed of the minute hand. [1]
- (b) Calculate the angular speed of the hour hand. [1]
- (c) The minute hand is 1.2 m long. Calculate the speed of its tip. [1]
- (d) Calculate the angle, in radians, that the hour hand turns through between 12:00 and 16:30. [1]
Show answer
- (a) 0.00175 rad s⁻¹
ω = 2π/3600 = 0.00175 rad s⁻¹ (A1).
- (b) 1.45 × 10⁻⁴ rad s⁻¹
ω = 2π/(12 × 3600) = 1.45 × 10⁻⁴ rad s⁻¹ (A1).
- (c) 0.00209 m s⁻¹
v = ωr = 0.00175 × 1.2 = 0.00209 m s⁻¹ (A1).
- (d) 2.36 rad
4.5 h of 12 h → (4.5/12) × 2π = 2.36 rad (A1).
A wind turbine at a wind farm near Jhimpir, Sindh, has blades 40 m long and turns at 15 revolutions per minute.
- (a) Calculate the angular speed of the rotor. [1]
- (b) Calculate the speed of a blade tip. [1]
- (c) Calculate the time for the rotor to turn through 1.0 rad. [1]
Show answer
- (a) 1.57 rad s⁻¹
ω = 15 × 2π/60 = 1.57 rad s⁻¹ (A1).
- (b) 62.8 m s⁻¹
v = ωr = 1.57 × 40 = 62.8 m s⁻¹ (A1).
- (c) 0.637 s
t = θ/ω = 1/1.57 = 0.637 s (A1).
The bob of a simple pendulum of length 0.80 m moves along an arc of length 0.12 m from its lowest point to one end of its swing.
- (a) Calculate the angle swung through, in radians. [1]
- (b) Convert this angle to degrees. [1]
- (c) Calculate the percentage difference between θ in radians and sin θ, and comment on the result. [2]
Show answer
- (a) 0.15 rad
θ = s/r = 0.12/0.80 = 0.15 rad (A1).
- (b) 8.59°
0.15 × 180/π = 8.59° (A1).
- (c) 0.37%; for small angles sin θ ≈ θ (in radians)
sin 0.15 = 0.1494 (M1); difference = 0.37% of θ, so small-angle approximation is good (A1).
The Earth (radius 6.37 × 10⁶ m) rotates once on its axis in 24 hours. Treat it as a sphere.
- (a) Calculate the angular speed of the Earth about its axis. [1]
- (b) Calculate the speed of a point on the equator due to this rotation. [1]
- (c) Karachi is at latitude 24.9° N. Calculate the speed of a building in Karachi due to the Earth's rotation. [2]
- (d) State the angular speed of the building in Karachi compared with a point on the equator. [1]
Show answer
- (a) 7.27 × 10⁻⁵ rad s⁻¹
ω = 2π/(24 × 3600) = 7.27 × 10⁻⁵ rad s⁻¹ (A1).
- (b) 463 m s⁻¹
v = ωR = 463 m s⁻¹ (A1).
- (c) 420 m s⁻¹
Radius of its circle = R cos 24.9° (M1); v = ωR cos 24.9° = 420 m s⁻¹ (A1).
- (d) The same
B1: every point on the Earth turns through 2π in the same time.
A car travels at a constant 30 m s⁻¹ on the motorway. Its wheels have a diameter of 0.62 m and roll without slipping.
- (a) Calculate the angular speed of the wheel. [2]
- (b) Calculate the number of revolutions per minute. [1]
- (c) Calculate the time taken for one revolution. [1]
Show answer
- (a) 96.8 rad s⁻¹
r = 0.31 m (M1); ω = v/r = 30/0.31 = 96.8 rad s⁻¹ (A1).
- (b) 924
96.8 × 60/2π = 924 (A1).
- (c) 0.0649 s
T = 2π/ω = 0.0649 s (A1).
- (a) Define angular speed. [1]
- (b) Using the definition of the radian, show that the speed v of an object moving in a circle of radius r with angular speed ω is v = rω. [2]
- (c) A point on the rim of a potter's wheel of radius 0.18 m makes 2.5 revolutions per second. Calculate its speed. [2]
Show answer
- (a) The angle swept out per unit time by the radius (rate of change of angle).
B1: rate of change of angular displacement (angle in radians per unit time).
- (b) In time Δt the angle is Δθ = ωΔt; arc length s = rΔθ; v = s/Δt = rω.
M1: s = rθ (radian definition); A1: divide by time: v = rΔθ/Δt = rω.
- (c) 2.83 m s⁻¹
ω = 2π × 2.5 = 15.7 rad s⁻¹ (M1); v = 0.18 × 15.7 = 2.83 m s⁻¹ (A1).
A cyclist turns the front chain-wheel (radius 0.10 m) at 1.5 revolutions per second. The chain connects it to a rear sprocket of radius 0.040 m fixed to the rear wheel. The bicycle does not skid.
- (a) Calculate the speed of the chain. [1]
- (b) Calculate the angular speed of the rear sprocket. [2]
- (c) The rear wheel, radius 0.34 m, turns with the rear sprocket. Calculate the speed of the bicycle. [2]
Show answer
- (a) 0.942 m s⁻¹
v = 2πfr = 2π × 1.5 × 0.10 = 0.942 m s⁻¹ (A1).
- (b) 23.6 rad s⁻¹
Chain speed same at both sprockets (M1); ω = 0.942/0.040 = 23.6 rad s⁻¹ (A1).
- (c) 8.01 m s⁻¹
Same ω as sprocket (M1); v = 23.6 × 0.34 = 8.01 m s⁻¹ (A1).
A child sits 2.5 m from the centre of a playground roundabout that makes 6 complete revolutions in 50 s at a steady rate.
- (a) Calculate the angular speed of the roundabout. [2]
- (b) Calculate the speed of the child. [1]
- (c) Calculate the angle the child turns through in 3.0 s, in radians and in degrees. [2]
Show answer
- (a) 0.754 rad s⁻¹
Period = 50/6 = 8.33 s (M1); ω = 2π/T = 0.754 rad s⁻¹ (A1).
- (b) 1.88 m s⁻¹
v = ωr = 1.88 m s⁻¹ (A1).
- (c) 2.26 rad = 130°
θ = ωt = 2.26 rad (A1); × 180/π = 130° (A1).
A potter in Multan shapes clay on a wheel. Clay at the edge of the wheel, 0.18 m from the axis, moves at 2.4 m s⁻¹.
- (a) Calculate the angular speed of the wheel. [1]
- (b) Calculate the number of revolutions per minute. [1]
- (c) Calculate the angle turned in 0.50 s and the number of revolutions this represents. [2]
Show answer
- (a) 13.3 rad s⁻¹
ω = v/r = 2.4/0.18 = 13.3 rad s⁻¹ (A1).
- (b) 127
13.3 × 60/2π = 127 (A1).
- (c) 6.67 rad; 1.06 revolutions
θ = ωt = 6.67 rad (A1); ÷ 2π = 1.06 rev (A1).
On an athletics track the inside lane (lane 1) running line is a semicircular bend of radius 36.5 m. Each lane outward adds 1.22 m to the radius, so lane 8 is 7 lanes out. A runner in lane 1 runs round the bend at a constant speed of 9.0 m s⁻¹.
- (a) Calculate the angular speed of the runner in lane 1 about the centre of the bend. [1]
- (b) A runner in lane 8 stays level with the lane-1 runner (on the same radius line) throughout the bend. Calculate the speed needed in lane 8. [2]
- (c) Each bend is a semicircle. Calculate how much further a runner in lane 8 travels round one bend than a runner in lane 1, and hence explain why starting positions are staggered. [2]
Show answer
- (a) 0.247 rad s⁻¹
ω = v/r = 9.0/36.5 = 0.247 rad s⁻¹ (A1).
- (b) 11.1 m s⁻¹
Same ω; r₈ = 36.5 + 7 × 1.22 = 45.0 m (M1); v = ωr = 11.1 m s⁻¹ (A1).
- (c) 26.8 m
Δs = πΔr = π × 8.54 = 26.8 m (M1 A1); stagger makes all lanes the same total distance.
A computer hard disc spins at 7200 revolutions per minute. The read head is over a track 3.0 cm from the axis.
- (a) Calculate the speed of the disc surface under the head. [2]
- (b) Each bit of data occupies 1.0 × 10⁻⁷ m along the track. Calculate the time for one bit to pass under the head. [1]
- (c) The head moves to a track at radius 0.015 m. State and explain the effect on the rate at which bits pass the head if bits are the same length. [2]
Show answer
- (a) 22.6 m s⁻¹
ω = 7200 × 2π/60 = 754 rad s⁻¹ (M1); v = ωr = 22.6 m s⁻¹ (A1).
- (b) 4.42 × 10⁻⁹ s
t = 1.0 × 10⁻⁷/22.6 = 4.42 × 10⁻⁹ s (A1).
- (c) The rate halves, because the linear speed v = ωr halves at the same angular speed.
B1: ω unchanged; B1: v ∝ r, so half the speed → half the bit rate (time per bit doubles).
An electric motor in a flour mill turns a small pulley of radius 2.5 cm at 1450 revolutions per minute. A belt that does not slip connects it to a larger pulley of radius 10.0 cm on the machine.
- (a) Calculate the speed of the belt. [2]
- (b) Calculate the rotation rate of the machine pulley in revolutions per minute. [2]
- (c) Calculate the centripetal acceleration of a small piece of belt as it passes round the motor pulley. [1]
Show answer
- (a) 3.80 m s⁻¹
ω = 1450 × 2π/60 = 152 rad s⁻¹ (M1); v = ωr = 3.80 m s⁻¹ (A1).
- (b) 363 rev min⁻¹ (362.5)
Belt speed same, so ω ∝ 1/r (M1): 1450 × 2.5/10.0 = 362.5 ≈ 363 rev min⁻¹ (A1; accept 362).
- (c) 576 m s⁻²
a = v²/r = 3.80²/0.025 = 576 m s⁻² (A1).
A fan with three identical, equally spaced blades is lit by a stroboscope flashing 25 times per second.
- (a) Explain why the fan appears stationary when it turns at certain speeds. [2]
- (b) Calculate the lowest angular speed of the fan for which it appears stationary. [2]
- (c) State the next higher rotation frequency, in revolutions per second, at which the fan appears stationary. [1]
Show answer
- (a) Between flashes each blade moves exactly to the position of a blade (the same or another) at the previous flash, so every flash shows the same pattern.
B1: between flashes the fan turns through a multiple of 120° (the angle between blades); B1: so the image looks identical at each flash.
- (b) 52.4 rad s⁻¹
Turns 1/3 rev in 1/25 s (M1) → f = 25/3 = 8.33 rev s⁻¹ → ω = 52.4 rad s⁻¹ (A1).
- (c) 16.7 rev s⁻¹
2/3 rev per flash → 16.7 rev s⁻¹ (A1).
