Physiworld · Dynamics revision notes · www.physiworld.com/notes/dynamics/
All revision notes Revision notes · DynamicsDynamics revision notes
Motion, forces, momentum, energy, springs, moments and circular motion, from distance–time graphs to angular momentum. Each section follows one lesson: what you need to know, key terms, the equations with units, a worked example, common mistakes and an exam tip.
1 Describing Motion
What you need to know
Distance–Time Graphs
- Speed is the distance travelled each second. Average speed = total distance ÷ total time, stops included.
- On a distance–time graph the gradient is the speed. A straight sloping line is a constant speed; a horizontal line means the object is stationary; a line that gets steeper means it is speeding up.
- For a curved line, the speed at one instant is the gradient of the tangent drawn at that point.
Velocity–Time Graphs
- Velocity is speed in a stated direction, so it is a vector. Acceleration is the change in velocity each second, in m/s²; a negative acceleration is a deceleration.
- On a velocity–time graph the gradient is the acceleration: a horizontal line is a constant velocity, a straight sloping line a constant acceleration.
- The area under a velocity–time graph is the distance travelled. Split it into rectangles and triangles.
Equations of Motion
- For a constant acceleration, v² − u² = 2as links the start speed u, the final speed v, the acceleration a and the distance s without the time.
- Average velocity for constant acceleration is (u + v) ÷ 2, so distance = (u + v) ÷ 2 × t: the same as the area under the v–t line.
- Anything falling freely near the Earth with little air resistance accelerates at g ≈ 9.8 m/s².
Reaction Time
- Reaction time is the delay between seeing a hazard and acting; it is typically 0.2 s to 0.9 s.
- Measure it with a ruler drop: the further the ruler falls before you catch it, the longer your reaction time.
- Tiredness, alcohol, some drugs and distractions such as a phone make the reaction time longer.
Key terms
- Scalar
- a quantity with size only, such as distance or speed
- Vector
- a quantity with size and direction, such as displacement, velocity or force
- Acceleration
- the rate of change of velocity, in m/s²
- Gradient
- the rise divided by the run of a line on a graph
Equations
| Equation | What it says, symbols and units |
|---|---|
v = st | Speed is the distance travelled each second.
|
a = ΔvΔt | Acceleration is the change in velocity each second; a deceleration is a negative acceleration.
|
distance = area under a speed–time graph | For constant speed or constant acceleration, the area under the line is the distance travelled. |
v2 − u2 = 2as | Links the start and end speeds to the distance, for constant acceleration, without needing the time.
|
Diagram to know
A velocity–time graph with three parts: a straight line rising from the origin (constant acceleration), a horizontal line (constant velocity) and a straight line falling to zero (deceleration). Label the gradient as acceleration and shade the area as distance.
Worked example
A car speeds up at a constant rate from 4.0 m/s to 16.0 m/s in 6.0 s. Calculate (a) its acceleration and (b) the distance it travels in that time.
- (a) a = (v − u) ÷ t = (16.0 − 4.0) ÷ 6.0 = 2.0 m/s².
- (b) Distance = area under the v–t line = average velocity × time = (4.0 + 16.0) ÷ 2 × 6.0 = 60 m.
- Check with v² − u² = 2as: 16.0² − 4.0² = 240, and 2 × 2.0 × 60 = 240. The two agree.
Answer: (a) 2.0 m/s² (b) 60 m
Common mistakes
- Reading a velocity–time graph as a distance–time graph: a horizontal line means stationary on one and constant velocity on the other.
- Using km/h in an equation that needs m/s. Divide km/h by 3.6 to get m/s.
- Forgetting that the area under a v–t graph needs the time axis in seconds, not minutes.
Exam tip
When a question says the object starts from rest, write u = 0 m/s straight away; when it comes to rest, v = 0 m/s. List u, v, a, s and t before choosing an equation.
AP Physics and NGSS: AP Physics 1 uses all the constant-acceleration equations, including v = u + at and s = ut + ½at² (written with x, v₀ and v), and works with signed displacement and velocity throughout.
2 Forces & Motion
What you need to know
Newton's First Law
- Newton's first law: an object stays at rest, or keeps moving at a constant velocity, unless a resultant force acts on it.
- Balanced forces do not stop an object; they keep its velocity the same. The tendency to keep moving is called inertia, and more mass means more inertia.
Newton's Second Law
- Newton's second law: the acceleration is proportional to the resultant force and inversely proportional to the mass, F = ma.
- The acceleration is in the direction of the resultant force. One newton is the force that gives 1 kg an acceleration of 1 m/s².
Resultant Force
- The resultant force is the single force with the same effect as all the forces together. Along one line, add forces in the same direction and subtract opposite ones.
- If the resultant is zero the object is in equilibrium: at rest or at constant velocity. Show the forces on a free-body diagram, one arrow per force.
Newton's Third Law
- Newton's third law: when A pushes or pulls on B, B pushes or pulls on A with a force of the same size in the opposite direction.
- The two forces are the same type and act on different objects, so they never cancel each other. A book's weight and the table's push on it are not a third-law pair.
Key terms
- Force
- a push or a pull, measured in newtons (N); a vector
- Resultant force
- the overall force when all forces on an object are combined
- Inertia
- the tendency of an object to stay at rest or keep moving at the same velocity
- Equilibrium
- zero resultant force (and, for a rigid body, zero resultant moment)
Equations
| Equation | What it says, symbols and units |
|---|---|
F = ma | A resultant force accelerates a mass, in the direction of the force.
|
resultant = sum of forces along a line | Forces in the same direction add; forces in opposite directions subtract. |
W = mg | Weight is the gravitational force on a mass; on Earth g is about 9.8 N/kg.
|
Diagram to know
A free-body diagram of a car: driving force forward, air resistance and friction backward, weight down, the road's normal contact force up. Arrow lengths show the sizes; the arrows start on the object.
Worked example
A car of mass 1200 kg has a driving force of 3000 N. Air resistance and friction together are 600 N. Find the resultant force and the acceleration.
- Resultant force = 3000 − 600 = 2400 N, forwards.
- a = F ÷ m = 2400 ÷ 1200 = 2.0 m/s², forwards.
Answer: 2400 N forwards; 2.0 m/s² forwards
Common mistakes
- Putting the driving force, not the resultant force, into F = ma.
- Thinking a moving object needs a resultant force to keep moving at constant speed. It only needs one to change its velocity.
- Saying that third-law forces cancel. They act on different objects.
Exam tip
Draw the free-body diagram before any number. Mark which direction is positive, find the resultant, and only then use F = ma.
AP Physics and NGSS: AP Physics 1 writes the second law as the net force, ΣF = ma, applied along each axis separately, and treats systems of connected objects (an Atwood machine, a train of carts).
3 Momentum & Collisions
What you need to know
Momentum
- Momentum = mass × velocity, p = mv, in kg m/s. It is a vector: an object moving the other way has negative momentum.
Conservation of Momentum
- In a closed system (no external force) the total momentum before a collision equals the total momentum after it.
- When objects stick together, they move off with one shared velocity. Kinetic energy is usually lost in a collision (to heat and sound) even though momentum is conserved.
Impulse
- Impulse = force × time = change in momentum, FΔt = Δ(mv), in N s. Force is the rate of change of momentum, F = Δp ÷ Δt.
- For the same change in momentum, a longer impact time means a smaller force. Crumple zones, air bags, seat belts, helmets and soft mats all work this way.
Explosions
- In an explosion or a push-apart starting from rest, the total momentum is zero before, so it is zero after: the parts move off with equal and opposite momenta.
- A gun's recoil is this effect: the light bullet moves fast one way, the heavy gun slowly the other.
Key terms
- Momentum
- mass × velocity, a vector in kg m/s
- Closed system
- a group of objects with no resultant external force on them
- Impulse
- force × the time it acts, equal to the change in momentum, in N s
- Inelastic collision
- a collision in which kinetic energy is not conserved (momentum still is)
Equations
| Equation | What it says, symbols and units |
|---|---|
p = mv | Momentum is mass times velocity, and it has a direction.
|
total momentum before = total momentum after | In a collision or explosion with no outside force, the total momentum does not change. |
impulse = FΔt = Δ(mv) | A force acting for a time changes the momentum by that amount.
|
F = ΔpΔt | Resultant force is the rate of change of momentum, so a longer impact time means a smaller force.
|
Worked example
A 2.0 kg trolley moving at 3.0 m/s hits a stationary 1.0 kg trolley. They stick together. Find their velocity after the collision and show that kinetic energy is lost.
- Momentum before = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg m/s.
- Mass after = 2.0 + 1.0 = 3.0 kg, so v = 6.0 ÷ 3.0 = 2.0 m/s in the original direction.
- Kinetic energy before = ½ × 2.0 × 3.0² = 9.0 J; after = ½ × 3.0 × 2.0² = 6.0 J. So 3.0 J is transferred to heat and sound.
Answer: 2.0 m/s in the original direction; kinetic energy falls from 9.0 J to 6.0 J
Common mistakes
- Ignoring direction. A ball hitting a wall at 5 m/s and bouncing back at 5 m/s has changed its velocity by 10 m/s, not zero.
- Using conservation of momentum when an external force (friction, a push) acts during the event.
- Assuming kinetic energy is conserved because momentum is.
Exam tip
Write "taking right as positive" at the start of every momentum answer, then give any velocity to the left a minus sign. Examiners give a mark for the direction of the final answer.
AP Physics and NGSS: AP Physics 1 adds elastic collisions (kinetic energy conserved too), collisions in two dimensions, the centre-of-mass velocity of a system, and impulse as the area under a force–time graph.
4 Work, Energy & Power
What you need to know
Kinetic Energy
- Kinetic energy Ek = ½mv². Doubling the speed makes the kinetic energy four times as big.
Energy Conservation
- Energy is never created or destroyed: it is transferred between stores (kinetic, gravitational potential, elastic, thermal, chemical, nuclear).
- For a falling object with no air resistance, the gravitational potential energy lost equals the kinetic energy gained: mgΔh = ½mv², so the mass cancels.
- In real machines some energy is dissipated, usually as heat, so the useful output is less than the input.
Work Done
- Work done = force × distance moved in the direction of the force, W = Fd, in joules. Work done is the energy transferred.
- Work done against friction heats the surfaces.
Power
- Power is the rate of doing work or transferring energy, P = W ÷ t = ΔE ÷ t, in watts (1 W = 1 J/s).
- Efficiency = useful output ÷ total input, for energy or for power, as a decimal or a percentage. It can never be more than 100%.
Key terms
- Energy store
- a way energy is held: kinetic, gravitational potential, elastic, thermal, chemical, nuclear
- Work done
- energy transferred by a force moving through a distance, in J
- Power
- energy transferred per second, in W
- Dissipated
- spread out to the surroundings, usually as heat, and no longer useful
Equations
| Equation | What it says, symbols and units |
|---|---|
Ek = ½mv2 | Energy of a moving object; double the speed and the kinetic energy is four times as big.
|
ΔEp = mgΔh | Energy gained by a mass lifted through a height, near the Earth's surface.
|
W = Fd = ΔE | The work a force does is the energy it transfers.
|
P = Wt = ΔEt | Power is how fast work is done or energy is transferred; 1 W is 1 J each second.
|
efficiency = useful energy outputtotal energy input (× 100%) | The share of the input that ends up useful; the same with useful power output over total power input. Never more than 100%.
|
Diagram to know
A Sankey diagram for a motor: one wide input arrow splitting into a useful-output arrow and a wasted-heat arrow, with the widths in proportion to the energies.
Worked example
A 0.50 kg ball is dropped from a height of 1.8 m. Ignoring air resistance, find the energy transferred to its kinetic store and its speed just before it lands.
- GPE lost = mgΔh = 0.50 × 9.8 × 1.8 = 8.82 J, so the kinetic energy gained is 8.82 J.
- ½ × 0.50 × v² = 8.82, so v² = 2 × 8.82 ÷ 0.50 = 35.28.
- v = √35.28 = 5.9 m/s.
Answer: 8.82 J (8.8 J to 2 s.f.); 5.9 m/s
Common mistakes
- Forgetting to square the speed in ½mv², or squaring the ½m as well.
- Using a mass in grams or a time in minutes. Convert to kg and s first.
- Using the whole distance in W = Fd when the force acts at an angle or only for part of the way.
Exam tip
In an energy question, write one line naming the stores at the start and at the end ("GPE at the top becomes KE at the bottom") before the arithmetic. That line is often worth a mark on its own.
AP Physics and NGSS: AP Physics 1 defines work with the angle, W = Fd cos θ, uses the work–energy theorem (net work = change in kinetic energy) and finds work as the area under a force–displacement graph.
5 Newton's Laws in Practice
What you need to know
Resultant Force
- When the resultant force is zero, a moving object keeps a constant velocity. A falling skydiver reaches terminal velocity when air resistance has grown to equal the weight.
- Opening the parachute increases air resistance, giving a resultant force upwards: the skydiver decelerates to a new, lower terminal velocity.
Force, Mass & Acceleration
- With the same resultant force, a bigger mass has a smaller acceleration: a = F ÷ m.
- In the practical, a trolley is pulled by a falling mass; light gates or a ticker timer measure its acceleration while the force or the mass is changed, one at a time.
Action & Reaction
- Third-law pairs make things move: a rocket pushes exhaust gas backwards and the gas pushes the rocket forwards; your foot pushes the ground back and the ground pushes you forward.
Braking Distance
- Stopping distance = thinking distance + braking distance. Thinking distance = speed × reaction time.
- The brakes do work to remove the kinetic energy: F × d = ½mv². For the same braking force, the braking distance is proportional to speed², so doubling the speed makes it four times as long.
- Braking distance grows with speed, mass, wet or icy roads, worn tyres and worn brakes. Large decelerations are dangerous for the people inside and can overheat the brakes.
Key terms
- Terminal velocity
- the steady speed reached when the resistive force equals the driving force or weight
- Thinking distance
- the distance travelled during the driver's reaction time
- Braking distance
- the distance travelled while the brakes act, until the vehicle stops
- Stopping distance
- thinking distance plus braking distance
Equations
| Equation | What it says, symbols and units |
|---|---|
F = ma | A resultant force accelerates a mass, in the direction of the force.
|
W = Fd = ΔE | The work a force does is the energy it transfers.
For braking: braking force × braking distance = the car's kinetic energy, ½mv². |
Ek = ½mv2 | Energy of a moving object; double the speed and the kinetic energy is four times as big.
|
Worked example
A 900 kg car travels at 20 m/s. The driver's reaction time is 0.7 s and the brakes give a constant braking force of 6000 N. Find the stopping distance.
- Thinking distance = 20 × 0.7 = 14 m.
- Kinetic energy = ½ × 900 × 20² = 180 000 J.
- Braking distance = kinetic energy ÷ braking force = 180 000 ÷ 6000 = 30 m.
- Stopping distance = 14 + 30 = 44 m.
Answer: 44 m
Common mistakes
- Saying alcohol or tiredness increases the braking distance. They increase the reaction time, so the thinking distance.
- Doubling the braking distance when the speed doubles. It goes up four times.
- Saying the skydiver goes upwards when the parachute opens. The velocity is still downwards; the acceleration is upwards.
Exam tip
For a "suggest factors" question on stopping distance, sort your answer under two headings, thinking distance (the driver) and braking distance (the car and the road). Examiners mark a factor under the wrong heading as wrong.
AP Physics and NGSS: AP Physics 1 solves braking with the constant-acceleration equations and friction (f = μN), and treats a skydiver's drag only qualitatively.
6 Forces as Vectors
What you need to know
Forces at an Angle
- To add two vectors at right angles, use Pythagoras for the size of the resultant and trigonometry (tan θ) for its direction. A scale drawing works for any angle.
- Distance, speed, mass and energy are scalars; displacement, velocity, acceleration, force and momentum are vectors.
Resolving Forces
- A force F at an angle θ to a direction has a component F cos θ along it and F sin θ at right angles to it.
- On a slope at angle θ, the weight splits into mg sin θ down the slope and mg cos θ into the slope.
Friction
- Friction acts along the surface, opposite to the motion or to the way the object would slide. It heats the surfaces.
- Static friction grows to match the push, up to a maximum; once the object slides, kinetic friction is roughly constant.
Projectiles
- A projectile's horizontal and vertical motions are independent. With no air resistance the horizontal velocity stays constant and the vertical velocity changes by 9.8 m/s every second.
- The time in the air is set by the vertical motion alone.
Relative Motion
- Velocity is always measured relative to something. Add velocity vectors to find a velocity relative to the ground, such as a boat crossing a flowing river.
Key terms
- Resultant
- the single vector equal to two or more vectors added together
- Component
- the part of a vector that acts along a chosen direction
- Friction
- a force that opposes sliding between two surfaces
- Projectile
- an object moving under gravity alone after it is launched
Equations
| Equation | What it says, symbols and units |
|---|---|
R = √(F12 + F22) | The resultant of two forces (or velocities) at right angles, by Pythagoras; or find it with a scale drawing.
|
Fx = F cos θ, Fy = F sin θ | The two components of a force at angle θ to the x-direction.
|
f ≤ μN | Friction is at most the coefficient of friction times the normal force; equal to μN once sliding.
AP Physics 1 only. |
Diagram to know
A right-angled vector triangle: one force drawn along the bottom, the second drawn from its tip at right angles, and the resultant from the start of the first to the end of the second, with the angle θ marked at the start.
Worked example
Two forces act on a box: 12 N to the east and 5.0 N to the north. Find the size and direction of the resultant force.
- The forces are at right angles, so R = √(12² + 5.0²) = √169 = 13 N.
- tan θ = 5.0 ÷ 12, so θ = 23° north of east (to 2 s.f.).
Answer: 13 N at 23° north of east
Common mistakes
- Adding vectors at right angles as plain numbers (12 + 5 = 17 N).
- Giving the size of a resultant but no direction.
- Mixing up sin and cos: the component next to the angle uses cos.
Exam tip
If the question says "scale drawing", state your scale (for example 1 cm = 1 N), draw with a ruler and protractor, and measure. A calculated answer alone loses the drawing marks.
AP Physics and NGSS: Projectiles, resolving on slopes, friction coefficients and relative velocity are AP Physics 1 (and NGSS HS-PS2-1) content; IGCSE and GCSE stop at resultants and components.
7 Springs & Stretching
What you need to know
Stretching and Squashing
- Stretching, squashing or bending an object needs more than one force on it.
- A change of shape is elastic if the object returns to its original shape when the forces are removed, and plastic (inelastic) if it stays changed.
- Extension = stretched length − original length.
Hooke's Law
- Hooke's law: the extension of a spring is directly proportional to the force on it, up to the limit of proportionality: F = kx.
- The spring constant k is the gradient of the straight part of a force–extension graph; a stiffer spring has a bigger k. Past the limit of proportionality the graph curves.
Elastic Potential Energy
- A stretched spring stores elastic potential energy: E = ½kx² while it obeys Hooke's law.
- The work done stretching it is the area under the force–extension graph.
Key terms
- Extension
- how much longer an object is than its original length, in m
- Spring constant
- force per unit extension, in N/m
- Limit of proportionality
- the point beyond which force and extension are no longer proportional
- Elastic deformation
- a change of shape that is fully undone when the force is removed
Equations
| Equation | What it says, symbols and units |
|---|---|
F = kx | Up to the limit of proportionality, the extension of a spring is proportional to the force on it.
|
Ee = ½ke2 | Energy stored in a stretched spring, while it obeys Hooke's law.
AQA writes the extension as e; other courses use x. |
Diagram to know
A force–extension graph: a straight line through the origin up to the limit of proportionality, then curving towards the extension axis. Mark the limit of proportionality and label the gradient k.
Worked example
A spring is 12.0 cm long. With a 4.0 N load it is 17.0 cm long. Find the spring constant and the energy stored, assuming Hooke's law holds.
- Extension = 17.0 − 12.0 = 5.0 cm = 0.050 m.
- k = F ÷ x = 4.0 ÷ 0.050 = 80 N/m.
- E = ½kx² = ½ × 80 × 0.050² = 0.10 J.
Answer: 80 N/m; 0.10 J
Common mistakes
- Using the stretched length instead of the extension.
- Leaving the extension in cm when k is wanted in N/m.
- Using F = kx beyond the limit of proportionality.
Exam tip
In the spring practical, say how you measure the extension accurately: read the ruler at eye level against a fixed pointer, and add masses one at a time. These details are where the method marks are.
AP Physics and NGSS: AP Physics 1 uses the spring force as F = −kx (the minus sign showing it acts back towards equilibrium) and the energy as U = ½kx², and links springs to simple harmonic motion.
8 Turning Forces
What you need to know
Moments
- The moment of a force is its turning effect: moment = force × perpendicular distance from the pivot, in N m.
- A bigger force, or the same force further from the pivot, gives a bigger moment.
The Principle of Moments
- Principle of moments: for an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments.
- In equilibrium there is also no resultant force: the upward forces equal the downward forces.
Levers and Gears
- A lever lets a small force far from the pivot balance a large force close to it.
- Gears pass a moment on: a small gear driving a larger one makes it turn more slowly but with a bigger moment.
Centre of Gravity
- The centre of gravity (centre of mass) is the point where the whole weight seems to act.
- An object topples when the line of action of its weight falls outside its base. A low centre of gravity and a wide base make it more stable.
- To find the centre of gravity of a flat shape, hang it from two points in turn with a plumb line and mark where the lines cross.
Key terms
- Moment
- force × perpendicular distance from the pivot, in N m
- Pivot
- the point an object turns about
- Principle of moments
- in equilibrium, clockwise moments = anticlockwise moments
- Centre of gravity
- the point through which the weight of an object acts
Equations
| Equation | What it says, symbols and units |
|---|---|
moment = F × d | The turning effect of a force grows with the force and with its perpendicular distance from the pivot.
|
sum of clockwise moments = sum of anticlockwise moments | In equilibrium the turning effects about any pivot cancel, and there is no resultant force either. |
Diagram to know
A beam on a pivot with a force on each side. Draw each force as a vertical arrow and mark its perpendicular distance from the pivot.
Worked example
A child of weight 300 N sits 2.0 m from the pivot of a balanced seesaw. Where must a child of weight 400 N sit on the other side? Ignoring the plank's weight, what force does the pivot exert?
- Anticlockwise moment = 300 × 2.0 = 600 N m.
- Clockwise moment must equal it: 400 × d = 600, so d = 600 ÷ 400 = 1.5 m.
- No resultant force: the pivot pushes up with 300 + 400 = 700 N.
Answer: 1.5 m from the pivot; 700 N upwards
Common mistakes
- Using the distance along a slanted force instead of the perpendicular distance from the pivot.
- Forgetting the weight of the beam itself, which acts at its centre.
- Mixing cm and m in the same moment equation.
Exam tip
Take moments about the point where an unknown force acts: that force then has no moment and drops out of the equation.
AP Physics and NGSS: AP Physics 1 calls the moment torque, τ = rF sin θ, and uses rotational equilibrium (net torque zero) together with translational equilibrium.
9 Circular Motion
What you need to know
Speed Stays, Velocity Changes
- An object moving in a circle at constant speed is always changing direction, so its velocity changes: it is accelerating, towards the centre of the circle.
- That acceleration needs a resultant force towards the centre, the centripetal force.
What Sets the Force
- A bigger centripetal force is needed for a bigger mass, a higher speed or a smaller radius.
- The size is F = mv² ÷ r and the acceleration a = v² ÷ r; these equations are AP Physics 1 content.
What Provides the Force
- Centripetal force is not an extra force: it is provided by tension (a ball on a string), friction (a car on a bend) or gravity (a planet or satellite in orbit).
- If the force stops, the object carries on in a straight line along the tangent to the circle.
Key terms
- Centripetal force
- the resultant force towards the centre that keeps an object moving in a circle
- Centripetal acceleration
- the acceleration towards the centre of a circle, v²/r
- Tangent
- a straight line that touches the circle at one point; the direction of the velocity
- Period
- the time for one full circle, in s
Equations
| Equation | What it says, symbols and units |
|---|---|
v = 2πrT | Average orbital speed is the distance round the orbit divided by the time for one orbit.
|
a = v2r, F = mv2r | Centripetal acceleration and the resultant force needed, for speed v on a circle of radius r.
AP Physics 1. IGCSE and GCSE need only how the force depends on m, v and r. |
Diagram to know
A circle with the object on it: the velocity arrow along the tangent, and the force and acceleration arrows pointing to the centre.
Worked example
A 0.20 kg ball on a 0.50 m string is swung in a horizontal circle, making 4 turns in 2.0 s. Find its speed and the tension in the string (AP Physics 1).
- Period T = 2.0 ÷ 4 = 0.50 s.
- v = 2πr ÷ T = 2π × 0.50 ÷ 0.50 = 6.3 m/s.
- F = mv² ÷ r = 0.20 × 6.28² ÷ 0.50 = 16 N.
Answer: 6.3 m/s; 16 N
Common mistakes
- Drawing a "centrifugal" force outwards. The only resultant force is inwards.
- Saying there is no acceleration because the speed is constant.
- Saying the object flies outwards along the radius when the string breaks. It goes along the tangent.
Exam tip
When asked what provides the centripetal force, name the actual force: "the tension in the string", "friction between the tyres and the road", "the gravitational pull of the Earth". "Centripetal force" alone gets no mark.
AP Physics and NGSS: AP Physics 1 adds vertical circles (the force at the top and the bottom of a loop) and banked curves.
10 Rotation
What you need to know
Rotational Kinematics
- Rotation has its own versions of the motion quantities: angle θ in radians, angular velocity ω = Δθ ÷ Δt in rad/s and angular acceleration α in rad/s².
- A point at radius r on a turning object moves at v = rω. The constant-acceleration equations work for rotation with θ, ω and α in place of s, v and a.
Torque and Rotational Inertia
- Torque τ = rF sin θ is the turning effect of a force. A resultant torque gives an angular acceleration: τnet = Iα.
- Rotational inertia I depends on the mass and how far it is from the axis: the same mass further out is harder to spin up.
Rolling and Rotational Energy
- A wheel rolling without slipping has v = rω. Its kinetic energy is shared: ½mv² for moving along plus ½Iω² for turning.
- So a rolling object goes down a slope more slowly than one sliding without friction: some of the energy goes into spinning.
Angular Momentum
- Angular momentum L = Iω. With no external torque it is conserved: a skater who pulls their arms in lowers I and spins faster.
Key terms
- Angular velocity
- the angle turned per second, in rad/s
- Torque
- the turning effect of a force about an axis, in N m
- Rotational inertia
- how hard an object is to start or stop turning, in kg m²
- Angular momentum
- rotational inertia × angular velocity, in kg m²/s
Equations
| Equation | What it says, symbols and units |
|---|---|
v = rω | Speed of a point at radius r on an object turning at angular velocity ω.
|
τ = rF sin θ, τnet = Iα | Torque of a force, and Newton's second law for rotation.
|
L = Iω, K = ½Iω2 | Angular momentum and rotational kinetic energy.
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Worked example
A skater spins at 2.0 rad/s with rotational inertia 3.0 kg m². She pulls her arms in, reducing it to 1.2 kg m². Find her new angular velocity and her kinetic energy before and after.
- Angular momentum before = Iω = 3.0 × 2.0 = 6.0 kg m²/s, and it is conserved.
- New ω = 6.0 ÷ 1.2 = 5.0 rad/s.
- K before = ½ × 3.0 × 2.0² = 6.0 J; K after = ½ × 1.2 × 5.0² = 15 J. The extra energy is the work her muscles do pulling her arms in.
Answer: 5.0 rad/s; 6.0 J before, 15 J after
Common mistakes
- Using degrees in v = rω. The angle must be in radians.
- Assuming kinetic energy is conserved when angular momentum is.
- Treating rotational inertia as a fixed property of the mass; it changes when the mass moves relative to the axis.
Exam tip
AP questions often ask you to compare two objects (a hoop and a disc) without numbers. Say which has more mass far from the axis, so a larger I, and follow that through to the answer in words.
AP Physics and NGSS: This lesson is mostly AP Physics 1 (Units 5 and 6). IGCSE and GCSE meet only the moment of a force, which is the torque idea without the angle.
Written in Physiworld's own words. The course tags come from the same mapping as the syllabus maps; equations are on the IGCSE, GCSE, AP Physics 1 and NGSS formula sheets. Values use g = 9.8 N/kg.
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