The problems are on the circular motion worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Circular motion at constant speed involves a constant angular velocity ω = 2π/T and a centripetal acceleration a = v²/r = ω²r directed towards the centre. These worked answers correct the 20 problems of the Physics 11-12 worksheet on circular motion, from key relationships to laboratory centrifuge: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. If your answer is different, check that you converted rpm to rad/s, used radians for angular quantities and chose the correct direction for the net force before you substitute.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.6
Problem 1 Key relationships
1. ω = 2π ÷ T.2. v = r ω.
3. a = v² ÷ r = ω² r.
4. F = m v² ÷ r.
ω = 2π/T, v = rω, a = v²/r = ω²r, F = mv²/r.
Problem 2 Angles and rates
1. 180° = π rad = 3.14 rad.2. 120 × 2π ÷ 60 = 12.6 rad/s.
3. ω = 2π f = 2π × 0.50 = 3.14 rad/s.
4. 120 rpm is 2.0 revolutions per second, so T = 1 ÷ 2.0 = 0.50 s.
π rad; 12.6 rad/s; 3.14 rad/s; 0.50 s.
Problem 3 Fan blade tip
1. ω = 1500 × 2π ÷ 60 = 157 rad/s.2. v = r ω = 0.18 × 157.1 = 28.3 m/s.
ω = 157 rad/s; v = 28 m/s.
Problem 4 Rotation of the Earth
1. ω = 2π ÷ 86 400 = 7.27 × 10⁻⁵ rad/s.2. v = r ω = 6.37 × 10⁶ × 7.27 × 10⁻⁵ = 463 m/s.
ω = 7.27 × 10⁻⁵ rad/s; v = 463 m/s.
Problem 5 Toy on a string
1. a = v² ÷ r = 9.0 ÷ 0.60 = 15 m/s².2. T = m a = 0.25 × 15 = 3.75 N.
a = 15 m/s²; T = 3.8 N.
Problem 6 What supplies the force
1. Friction between the tyres and the road.2. Gravitational attraction of the Earth.
3. Tension in the string.
4. Normal reaction of the wall.
friction; gravity; tension; normal reaction.
Problem 7 Roundabout on a bicycle
1. F = m v² ÷ r = 70 × 36 ÷ 15 = 168 N.2. μ m g ≥ m v² ÷ r, so μ = v² ÷ (r g) = 36 ÷ (15 × 9.81) = 0.245.
3. v = √(μ g r) = √(0.20 × 9.81 × 15) = √29.43 = 5.4 m/s.
168 N; μ ≥ 0.24; 5.4 m/s.
Problem 8 Maximum speed on a bend
1. v = √(μ g r) = √(0.65 × 9.81 × 45) = √287 = 17 m/s.2. v = √(0.25 × 9.81 × 45) = √110 = 10.5 m/s.
3. The friction limit μ m g and the force needed m v² ÷ r both contain m, so it cancels.
17 m/s; 10.5 m/s.
Problem 9 Whirling a bung
1. F ÷ v² = 1.6 ÷ 4 = 0.40, 3.6 ÷ 9 = 0.40, 6.4 ÷ 16 = 0.40, 10.0 ÷ 25 = 0.40, 14.4 ÷ 36 = 0.40 N s²/m².2. F is proportional to v², so the graph is a straight line through the origin with gradient m ÷ r = 0.40 kg/m.
3. m = gradient × r = 0.40 × 0.50 = 0.20 kg.
gradient = 0.40 kg/m; m = 0.20 kg.
Problem 10 String breaks
1. Path P, along the tangent in the direction of motion.2. When the string breaks, there is no resultant force, so the ball continues with its velocity at that instant: a straight line at constant speed.
3. The force was directed towards the centre of the circle (along the string).
path P.
Problem 11 Rotor ride
1. N is the only force towards the centre, so N = m v² ÷ r.2. Friction balances weight: μ N = m g, so μ m v² ÷ r = m g and v = √(g r ÷ μ) = √(9.81 × 3.0 ÷ 0.40) = 8.6 m/s.
3. ω = v ÷ r = 2.86 rad/s, so rpm = 2.86 × 60 ÷ 2π = 27 rpm.
v = 8.6 m/s; 27 rpm.
Problem 12 Hammer throw
1. F = m v² ÷ r = 7.26 × 625 ÷ 1.8 = 2521 N.2. Weight = 7.26 × 9.81 = 71.2 N, so the force is 2521 ÷ 71.2 = 35 times the weight.
F = 2.5 × 10³ N; 35 × weight.
Problem 13 Ball at the bottom
1. T − m g = m v² ÷ r, so T = 0.50 × 9.81 + 0.50 × 25 ÷ 1.2 = 4.905 + 10.42 = 15.3 N.2. At the limit T = 0, so v = √(g r) = √(9.81 × 1.2) = 3.4 m/s.
T = 15 N; vmin = 3.4 m/s.
Problem 14 Railway curve
1. N cos θ = m g and N sin θ = m v² ÷ r. Dividing gives tan θ = v² ÷ (r g).2. tan θ = 900 ÷ (900 × 9.81) = 0.102, so θ = 5.8°.
θ = 5.8°.
Problem 15 Centrifugal force myth
1. The passenger tends to move in a straight line. The car turns inward, so the door moves into the passenger and pushes inward on the passenger, who experiences this as being pushed outward.2. F = m v² ÷ r = 80 × 144 ÷ 30 = 384 N.
F = 384 N inward.
Problem 16 Banked bend with a car
1. v = √(r g tan θ) = √(100 × 9.81 × tan 15°) = √(981 × 0.2679) = √262.9 = 16.2 m/s.2. N = m g ÷ cos θ = 1200 × 9.81 ÷ 0.9659 = 12 190 N.
3. N sin θ = 12 190 × 0.2588 = 3155 N. m v² ÷ r = 1200 × 262.9 ÷ 100 = 3155 N. They are equal.
v = 16 m/s; N = 1.22 × 10⁴ N.
Problem 17 Conical pendulum
1. r = L sin θ = 0.80 × 0.50 = 0.40 m.2. Vertically T cos θ = m g, so T = 0.15 × 9.81 ÷ 0.866 = 1.70 N.
3. Horizontally T sin θ = m ω² r, so ω² = 1.70 × 0.50 ÷ (0.15 × 0.40) = 14.2 and ω = 3.76 rad/s. Tperiod = 2π ÷ 3.76 = 1.67 s.
T = 1.70 N; ω = 3.8 rad/s; period = 1.7 s.
Problem 18 Loop-the-loop
1. At the top N = 0, so v = √(g r) = √(9.81 × 8.0) = 8.86 m/s.2. The top is 2r = 16 m above the bottom: ½ vb² = ½ vt² + g (2 r), so vb² = 78.5 + 4 × 9.81 × 8.0 = 392 and vb = 19.8 m/s.
3. N − m g = m v² ÷ r, so N = 60 × 9.81 + 60 × 392 ÷ 8.0 = 589 + 2943 = 3532 N = 6.0 times the weight.
8.9 m/s; 19.8 m/s; N = 3.5 kN (6 mg).
Problem 19 Leaning motorcycle
1. tan θ = 225 ÷ (40 × 9.81) = 0.573, so θ = 29.8°.2. Friction must supply m v² ÷ r, with N = m g, so μ ≥ tan θ = 0.57.
3. tan θ = 400 ÷ 392.4 = 1.019, so θ = 45.5°. Raising the speed by a third needs a much larger lean and more friction (μ ≥ 1.02, which is above the value for most wet or dry roads).
θ = 30°; μ ≥ 0.57; 46°.
Problem 20 Laboratory centrifuge
1. ω = 3000 × 2π ÷ 60 = 314 rad/s.2. a = ω² r = 314.2² × 0.10 = 9870 m/s², which is 9870 ÷ 9.81 = 1.0 × 10³ g.
3. F = m a = 0.012 × 9870 = 118 N.
4. a is proportional to ω², so the acceleration is four times as large.
ω = 314 rad/s; a = 9.9 × 10³ m/s² (1.0 × 10³ g); F = 118 N.
