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Since both are lengths, their ratio gives a dimensionless angular measure."},{"text":"The angle subtended at the centre of a circle by an arc length equal to the diameter.","correct":false,"explanation":"Confusing the radius with the diameter makes the arc twice as long, producing an angle of two radians. One radian requires the arc length to equal the radius."},{"text":"The angle subtended at the centre of a circle by an arc length equal to the circumference.","correct":false,"explanation":"An arc equal to the full circumference subtends one complete revolution, which is $2\\pi$ radians. One radian instead corresponds to an arc equal to the radius."},{"text":"The angle subtended at the circumference of a circle by an arc length equal to the radius.","correct":false,"explanation":"A radian is defined using the angle at the centre, not an angle at the circumference. The relevant central angle subtends an arc whose length equals the radius."}]},{"id":"86d48286-a3fa-4e23-8071-7375060563ba","cardType":"basic","points":1,"front":"How is an angle in degrees converted into radians?","back":"Multiply by

$$\\frac{\\pi}{180}$$

","mcqOptions":[{"text":"Multiply by

$$\\frac{\\pi}{180}$$

","correct":true,"explanation":"A complete revolution is both $360^\\circ$ and $2\\pi$ radians, so one degree equals $\\frac{\\pi}{180}$ radians. Therefore, degrees are multiplied by this factor."},{"text":"Multiply by

$$\\frac{180}{\\pi}$$

","correct":false,"explanation":"This is the inverse conversion factor and converts radians into degrees. To convert degrees into radians, multiply by $\\frac{\\pi}{180}$."},{"text":"Multiply by

$$\\frac{\\pi}{360}$$

","correct":false,"explanation":"This incorrectly equates $360^\\circ$ with $\\pi$ radians rather than $2\\pi$ radians, halving the result. The correct multiplier is $\\frac{\\pi}{180}$."},{"text":"Multiply by

$$\\frac{180}{2\\pi}$$

","correct":false,"explanation":"This factor comes from reversing and then halving the required conversion, so it does not convert degrees to radians. The correct multiplier is $\\frac{\\pi}{180}$."}]},{"id":"bcfa42ef-23fd-4bf1-b9b1-aa6749ec932c","cardType":"basic","points":1,"front":"Define angular speed ($\\omega$).","back":"The rate of change of angle, measured in $\\text{rad s}^{-1}$.","mcqOptions":[{"text":"The rate of change of angle, measured in $\\text{rad s}^{-1}$.","correct":true,"explanation":"Angular speed describes how quickly angular displacement changes with time. Dividing an angle in radians by a time in seconds gives the unit $\\text{rad s}^{-1}$."},{"text":"The rate of change of angular speed, measured in $\\text{rad s}^{-2}$.","correct":false,"explanation":"This defines angular acceleration, which measures how angular speed changes with time. Angular speed itself is the rate of change of angle, measured in $\\text{rad s}^{-1}$."},{"text":"The rate of change of linear displacement, measured in $\\text{m s}^{-1}$.","correct":false,"explanation":"This describes linear velocity rather than angular speed. Angular speed measures the rate of change of angle and therefore has units of $\\text{rad s}^{-1}$."},{"text":"The angle turned during one revolution, measured in $\\text{rad}$.","correct":false,"explanation":"The angle in one revolution is always $2\\pi$ radians and contains no time dependence. Angular speed is the rate of change of angle, measured in $\\text{rad s}^{-1}$."}]},{"id":"48392b43-cd50-4e1b-9851-b2749415ac53","cardType":"cloze","points":null,"front":"The relationship between angular speed ($\\omega$) and time period ($T$) is $\\omega = $ $$\\text{[...]}$$.","back":"The relationship between angular speed ($\\omega$) and time period ($T$) is $\\omega = $ $$\\boxed{\\frac{2\\pi}{T}}$$.","mcqOptions":[{"text":"$$$\\frac{2\\pi}{T}$$","correct":true,"explanation":"One complete cycle covers an angle of $2\\pi$ radians in time $T$. Dividing angular displacement by time therefore gives $\\omega=\\frac{2\\pi}{T}$."},{"text":"$$$\\frac{T}{2\\pi}$$","correct":false,"explanation":"This reverses the required ratio and has units of seconds rather than inverse seconds. Angular speed is angle divided by time, so $\\omega=\\frac{2\\pi}{T}$."},{"text":"$$$2\\pi T$$","correct":false,"explanation":"Multiplying by the period makes angular speed increase for slower rotations and gives the wrong units. Since one cycle takes time $T$, $\\omega=\\frac{2\\pi}{T}$."},{"text":"$$$\\frac{\\pi}{T}$$","correct":false,"explanation":"This uses the angle for half a revolution while $T$ is the period of a complete revolution. A full cycle covers $2\\pi$ radians, giving $\\omega=\\frac{2\\pi}{T}$."}]},{"id":"a62fdb78-c697-4675-9915-e1a02175fb06","cardType":"cloze","points":null,"front":"The relationship between angular speed ($\\omega$) and frequency ($f$) is $\\omega = $ $$\\text{[...]}$$.","back":"The relationship between angular speed ($\\omega$) and frequency ($f$) is $\\omega = $ $$\\boxed{2\\pi f}$$.","mcqOptions":[{"text":"$$$2\\pi f$$","correct":true,"explanation":"Frequency gives the number of complete cycles per second, and each cycle covers $2\\pi$ radians. Therefore, multiplying them gives $\\omega=2\\pi f$."},{"text":"$$$\\frac{f}{2\\pi}$$","correct":false,"explanation":"This divides by the angular displacement per cycle instead of multiplying by it. Each cycle contributes $2\\pi$ radians, so the angular speed is $\\omega=2\\pi f$."},{"text":"$$$\\frac{2\\pi}{f}$$","correct":false,"explanation":"This treats frequency as though it were the period and produces units of time. Since $f$ is cycles per second, angular speed is $\\omega=2\\pi f$."},{"text":"$$$\\pi f$$","correct":false,"explanation":"This uses the angle for half a revolution for each complete cycle, making the result half the correct value. One cycle spans $2\\pi$ radians, so $\\omega=2\\pi f$."}]},{"id":"2b24b94c-d30d-416e-8e52-b4c63ce21ec6","cardType":"basic","points":1,"front":"State the equation linking linear speed ($v$), radius ($r$), and angular speed ($\\omega$).","back":"$$$v = r\\omega$$","mcqOptions":[{"text":"$$$v = r\\omega$$","correct":true,"explanation":"An angular displacement $\\theta$ corresponds to arc length $s=r\\theta$. Differentiating with respect to time gives the linear speed relationship $v=r\\omega$."},{"text":"$$$v = \\frac{\\omega}{r}$$","correct":false,"explanation":"Dividing angular speed by radius gives incompatible units and incorrectly predicts lower linear speed at larger radii. The arc-length relationship gives $v=r\\omega$."},{"text":"$$$v = \\frac{r}{\\omega}$$","correct":false,"explanation":"This division has units that do not represent speed and incorrectly makes linear speed decrease as angular speed increases. The correct relationship is $v=r\\omega$."},{"text":"$$$v = r\\omega^2$$","correct":false,"explanation":"Squaring angular speed belongs in the centripetal acceleration equation $a=r\\omega^2$, not the linear speed equation. Linear speed is given by $v=r\\omega$."}]},{"id":"cfc92362-c1aa-4fa8-b444-92d23e6dae9e","cardType":"basic","points":2,"front":"Two objects are on a spinning disc at different distances from the centre. Compare their angular and linear speeds.","back":"

They share the same angular speed ($\\omega$).
The object at the larger radius has a higher linear speed ($v$).

","mcqOptions":[{"text":"

They share the same angular speed ($\\omega$).
The object at the larger radius has a higher linear speed ($v$).

","correct":true,"explanation":"Every point on a rigid spinning disc completes each revolution in the same time, so angular speed is shared. Since $v=r\\omega$, linear speed increases with radius."},{"text":"

They share the same angular speed ($\\omega$).
The object at the smaller radius has a higher linear speed ($v$).

","correct":false,"explanation":"The angular-speed statement is correct, but the linear-speed comparison is reversed. Because $v=r\\omega$, the object at the larger radius has the higher linear speed."},{"text":"

They share the same linear speed ($v$).
The object at the larger radius has a higher angular speed ($\\omega$).

","correct":false,"explanation":"This swaps the quantities that are shared and radius-dependent. A rigid disc gives both objects the same angular speed, while $v=r\\omega$ makes linear speed higher at the larger radius."},{"text":"

They share the same linear speed ($v$).
The object at the smaller radius has a higher angular speed ($\\omega$).

","correct":false,"explanation":"Points on a rigid disc do not have different angular speeds; they complete revolutions together. Their linear speeds differ because $v=r\\omega$, with the larger-radius object moving faster."}]},{"id":"6160bf72-030b-4bd5-84e6-20889e4d7ce4","cardType":"basic","points":3,"front":"Explain why an object moving at a constant speed in a circle is accelerating.","back":"

Velocity is a vector.
The direction is constantly changing.
A change in velocity over time constitutes acceleration.

","mcqOptions":[{"text":"

Velocity is a vector.
The direction is constantly changing.
A change in velocity over time constitutes acceleration.

","correct":true,"explanation":"Although the magnitude of velocity remains constant, its direction continually changes around the circle. Since acceleration is the rate of change of velocity, the object is accelerating."},{"text":"

Speed is a vector.
The magnitude is constantly changing.
A change in speed over time constitutes acceleration.

","correct":false,"explanation":"Speed is a scalar, and its magnitude is stated to be constant, so this reasoning contradicts the situation. Acceleration occurs because the direction of the velocity vector changes."},{"text":"

Velocity is a vector.
The direction remains constant.
A change in position over time constitutes acceleration.

","correct":false,"explanation":"In circular motion the velocity direction does not remain constant, and change in position per unit time defines velocity rather than acceleration. Changing velocity direction produces the acceleration."},{"text":"

Velocity is a scalar.
The direction has no effect on velocity.
Only a change in speed over time constitutes acceleration.

","correct":false,"explanation":"This confuses velocity with speed: velocity is a vector, so direction matters. Even at constant speed, its changing direction means velocity changes and the object accelerates."}]},{"id":"09876b32-4485-48df-91fa-af84735be950","cardType":"basic","points":1,"front":"What is the direction of centripetal acceleration and centripetal force?","back":"Towards the centre of the circle.","mcqOptions":[{"text":"Towards the centre of the circle.","correct":true,"explanation":"In circular motion, the velocity continually changes direction because the acceleration and resultant force point radially inwards, towards the circle’s centre."},{"text":"Away from the centre of the circle.","correct":false,"explanation":"This reverses the required direction. Centripetal acceleration and force point towards the centre, while an outward effect is only apparent in a rotating reference frame."},{"text":"Along the tangent to the circle.","correct":false,"explanation":"The instantaneous velocity is tangential, but the centripetal acceleration and resultant force are perpendicular to it and directed towards the centre of the circle."},{"text":"In the direction of motion.","correct":false,"explanation":"The direction of motion is tangential rather than radial. Centripetal acceleration and force point towards the centre, continually turning the object’s velocity."}]},{"id":"cd580334-1276-4ab6-a7b5-b3d5dd9172af","cardType":"cloze","points":null,"front":"Centripetal acceleration ($a$) in terms of linear speed ($v$) is given by $a = $ $$\\text{[...]}$$.","back":"Centripetal acceleration ($a$) in terms of linear speed ($v$) is given by $a = $ $$\\boxed{\\frac{v^2}{r}}$$.","mcqOptions":[{"text":"$$$\\frac{v^2}{r}$$","correct":true,"explanation":"For uniform circular motion, the radial acceleration equals the square of the linear speed divided by the radius, so increasing speed has a squared effect."},{"text":"$$$\\frac{v}{r}$$","correct":false,"explanation":"This omits the required square of the linear speed and is not dimensionally an acceleration. The correct relationship is $a = \\frac{v^2}{r}$."},{"text":"$$$\\frac{v^2}{r^2}$$","correct":false,"explanation":"Squaring the radius introduces an extra division by length and gives incorrect dimensions. Centripetal acceleration is $a = \\frac{v^2}{r}$."},{"text":"$$$v^2r$$","correct":false,"explanation":"Multiplying by the radius reverses the correct radius dependence and gives incorrect dimensions. Centripetal acceleration is $a = \\frac{v^2}{r}$."}]},{"id":"a276ba0e-5442-4f73-8d85-d72b4b1a9e0c","cardType":"cloze","points":null,"front":"Centripetal acceleration ($a$) in terms of angular speed ($\\omega$) is given by $a = $ $$\\text{[...]}$$.","back":"Centripetal acceleration ($a$) in terms of angular speed ($\\omega$) is given by $a = $ $$\\boxed{\\omega^2 r}$$.","mcqOptions":[{"text":"$$$\\omega^2 r$$","correct":true,"explanation":"Using $v = \\omega r$ in $a = \\frac{v^2}{r}$ gives $a = \\omega^2r$, showing that acceleration depends on angular speed squared."},{"text":"$$$\\omega r$$","correct":false,"explanation":"This misses the square on angular speed and therefore has the dimensions of velocity, not acceleration. The correct expression is $a = \\omega^2r$."},{"text":"$$$\\frac{\\omega^2}{r}$$","correct":false,"explanation":"This incorrectly divides by radius and gives the wrong dimensions. Substituting $v = \\omega r$ into the linear-speed formula gives $a = \\omega^2r$."},{"text":"$$$\\omega^2 r^2$$","correct":false,"explanation":"This incorrectly squares the radius, producing dimensions inconsistent with acceleration. Only angular speed is squared, giving the correct relationship $a = \\omega^2r$."}]},{"id":"415ed2d8-4f1f-42b1-bb75-7f3ce04244f2","cardType":"basic","points":2,"front":"State the two equations for centripetal force ($F$).","back":"

$$F = \\frac{mv^2}{r}$$
$$F = m\\omega^2 r$$

","mcqOptions":[{"text":"

$$F = \\frac{mv^2}{r}$$
$$F = m\\omega^2 r$$

","correct":true,"explanation":"Multiplying either centripetal-acceleration formula by mass gives these force equations. They are equivalent because linear and angular speed are related by $v = \\omega r$."},{"text":"

$$F = \\frac{mv}{r}$$
$$F = m\\omega r$$

","correct":false,"explanation":"Both equations omit the required square on speed. Since centripetal force equals mass times acceleration, the correct forms contain $v^2$ and $\\omega^2$."},{"text":"

$$F = mv^2r$$
$$F = \\frac{m\\omega^2}{r}$$

","correct":false,"explanation":"These expressions reverse the correct dependence on radius. At fixed linear speed force decreases with radius, while at fixed angular speed it increases with radius."},{"text":"

$$F = \\frac{mr}{v^2}$$
$$F = \\frac{mr}{\\omega^2}$$

","correct":false,"explanation":"These equations invert the speed dependence, incorrectly predicting less force at greater speed. The correct equations are $F = \\frac{mv^2}{r}$ and $F = m\\omega^2r$."}]},{"id":"71d11a06-58c1-468a-aa55-f03a13f77c0f","cardType":"basic","points":1,"front":"What physical force provides the centripetal force for a planet orbiting a star?","back":"Gravity (Gravitational force).","mcqOptions":[{"text":"Gravity (Gravitational force).","correct":true,"explanation":"The star’s gravitational attraction acts towards the star and supplies the inward resultant force needed to keep the planet in its orbit."},{"text":"Electrostatic force.","correct":false,"explanation":"Electrostatic attraction acts between charged objects, whereas the star and planet are treated as gravitationally interacting masses. Gravity supplies the required centripetal force."},{"text":"Magnetic force.","correct":false,"explanation":"Magnetic force does not maintain the orbit of a planet around a star. The inward force is supplied by the gravitational attraction between their masses."},{"text":"Strong nuclear force.","correct":false,"explanation":"The strong nuclear force acts only over extremely short subatomic distances. A planet’s orbital centripetal force is provided by the star’s gravitational attraction."}]},{"id":"84bf8dc1-fedb-4b35-b9c5-87bb8f1e6afe","cardType":"basic","points":1,"front":"What physical force provides the centripetal force for a car cornering on a flat road?","back":"Friction between the tyres and the road.","mcqOptions":[{"text":"Friction between the tyres and the road.","correct":true,"explanation":"Static friction at the tyre-road contact acts sideways towards the centre of the turn, providing the horizontal resultant force required for cornering."},{"text":"The normal reaction from the road.","correct":false,"explanation":"The normal reaction on a flat road acts vertically upwards and balances weight; it cannot provide the horizontal inward force. Tyre-road friction supplies that force."},{"text":"The driving force from the engine.","correct":false,"explanation":"Engine thrust primarily acts in the car’s forward direction rather than towards the turn’s centre. Sideways friction between the tyres and road provides the centripetal force."},{"text":"Air resistance acting on the car.","correct":false,"explanation":"Air resistance mainly opposes the car’s motion and is not the required sideways force. Friction at the tyre-road contact acts towards the turn’s centre."}]},{"id":"8d55eaa4-569a-4f8a-9cf9-20f8add83aa7","cardType":"basic","points":1,"front":"What physical force provides the centripetal force for an electron orbiting a nucleus?","back":"Electrostatic force.","mcqOptions":[{"text":"Electrostatic force.","correct":true,"explanation":"The negatively charged electron is attracted to the positively charged nucleus. In the orbital model, this inward Coulomb attraction supplies the required centripetal force."},{"text":"Strong nuclear force.","correct":false,"explanation":"The strong interaction binds quarks and helps bind nucleons, but it does not bind atomic electrons to nuclei. Electrostatic attraction supplies the inward force."},{"text":"Weak nuclear force.","correct":false,"explanation":"The weak interaction is responsible for processes such as beta decay, not ordinary atomic binding. The electron is attracted to the nucleus electrostatically."},{"text":"Gravitational force.","correct":false,"explanation":"Gravity between an electron and nucleus is negligibly small compared with their electrical attraction. The electrostatic force provides the centripetal force in this model."}]},{"id":"396912d6-26cc-48f9-863b-76bf06d024e7","cardType":"basic","points":2,"front":"If an object's linear speed doubles while the radius remains constant, what happens to the required centripetal force?","back":"The force quadruples ($F \\propto v^2$).","mcqOptions":[{"text":"The force quadruples ($F \\propto v^2$).","correct":true,"explanation":"At constant mass and radius, centripetal force is proportional to $v^2$. Replacing $v$ with $2v$ multiplies the force by $2^2 = 4$."},{"text":"The force doubles ($F \\propto v$).","correct":false,"explanation":"This treats force as directly proportional to speed rather than speed squared. Doubling $v$ multiplies $v^2$, and therefore the required force, by four."},{"text":"The force halves ($F \\propto \\frac{1}{v}$).","correct":false,"explanation":"This reverses the effect of increasing speed. Since $F = \\frac{mv^2}{r}$, doubling the speed makes the required centripetal force four times larger."},{"text":"The force remains unchanged ($F \\propto r$).","correct":false,"explanation":"This overlooks the speed dependence in $F = \\frac{mv^2}{r}$. With mass and radius fixed, doubling speed causes the required force to quadruple."}]},{"id":"7c0be125-7ffb-4595-9915-db1d6de46635","cardType":"basic","points":2,"front":"Explain why a car skids outwards if it travels too fast around a bend.","back":"The required centripetal force ($mv^2/r$) exceeds the maximum friction the tyres can provide.","mcqOptions":[{"text":"The required centripetal force ($mv^2/r$) exceeds the maximum friction the tyres can provide.","correct":true,"explanation":"The tyres’ friction must supply the centripetal force. Because this requirement increases with $v^2$, excessive speed makes the required force exceed the available friction, so the car cannot follow the bend."},{"text":"The required centripetal force ($mv^2/r$) acts outwards from the centre and pushes the tyres sideways.","correct":false,"explanation":"Centripetal force does not become an outward force at high speed; its required direction remains towards the bend’s centre. Skidding occurs when $mv^2/r$ exceeds the maximum available tyre friction."},{"text":"The available centripetal force ($mv^2/r$) decreases as the car’s speed increases around the bend.","correct":false,"explanation":"Friction is not necessarily reduced simply because the car travels faster. Instead, the required centripetal force rises with $v^2$ until the available friction cannot supply it."},{"text":"The car’s weight ($mg$) exceeds the normal contact force the road can provide.","correct":false,"explanation":"Weight and the normal contact force are primarily vertical on a level road, so their balance does not explain sideways skidding. The required centripetal force exceeds the maximum tyre friction."}]},{"id":"d06ca5ce-a3b2-4368-9dc6-da18377bd7cd","cardType":"basic","points":2,"front":"Why is centrifugal force considered a \"fictitious\" force?","back":"It is not a real force; it is the result of inertia (the body's tendency to continue in a straight line).","mcqOptions":[{"text":"It is not a real force; it is the result of inertia (the body's tendency to continue in a straight line).","correct":true,"explanation":"No physical interaction produces an outward force on the body in an inertial frame. The apparent effect arises because inertia makes the body tend to continue along a straight-line path."},{"text":"It is a real force; it is the reaction force opposing the inward centripetal force on the body.","correct":false,"explanation":"Centrifugal force is not the reaction force paired with centripetal force; Newton’s third-law pair acts on a different object. The apparent outward effect instead results from inertia."},{"text":"It is not a real force; it is the result of gravity pulling the body away from the centre.","correct":false,"explanation":"Gravity is a real interaction and is not generally responsible for the apparent outward effect in circular motion. Centrifugal force is fictitious because it represents inertia in a rotating frame."},{"text":"It is not a real force; it appears because the body has no acceleration while travelling around the circle.","correct":false,"explanation":"The body does accelerate towards the centre when undergoing circular motion, and this requires a real resultant force. The fictitious outward force instead arises from describing inertia in a rotating frame."}]},{"id":"d582c45e-bf87-41b8-b32a-33611d0d56e4","cardType":"cloze","points":null,"front":"When calculating circular motion, mass must be in [ ], distance in m, and time in s.","back":"When calculating circular motion, mass must be in kg, distance in m, and time in s.","mcqOptions":[{"text":"kg","correct":true,"explanation":"Kilograms are the SI base unit of mass. Using kg with metres and seconds makes circular-motion equations consistent with derived SI units such as newtons."},{"text":"g","correct":false,"explanation":"Grams measure mass, but circular-motion equations using standard SI quantities require kilograms. A value stated in grams must be divided by $1000$ before substitution."},{"text":"N","correct":false,"explanation":"Newtons measure force rather than mass. Mass must be entered in kilograms so that expressions such as $mv^2/r$ produce force in newtons."},{"text":"t","correct":false,"explanation":"Tonnes measure mass, but they are not the SI base unit required for direct substitution. Convert tonnes to kilograms by multiplying the value by $1000$."}]},{"id":"ab9f1995-c460-425a-8543-4e2c5dce6f19","cardType":"cloze","points":null,"front":"When calculating circular motion, mass must be in kg, distance in [ ], and time in s.","back":"When calculating circular motion, mass must be in kg, distance in m, and time in s.","mcqOptions":[{"text":"m","correct":true,"explanation":"Metres are the SI base unit of length. Radii, circumferences and other distances must be expressed in metres when using standard SI circular-motion equations."},{"text":"cm","correct":false,"explanation":"Centimetres are a valid length unit but are not the SI base unit used directly here. Divide centimetres by $100$ to convert the distance to metres."},{"text":"km","correct":false,"explanation":"Kilometres are unsuitable for direct substitution into standard SI equations. Multiply kilometres by $1000$ to express the circular path’s radius or distance in metres."},{"text":"m²","correct":false,"explanation":"Square metres measure area, not distance. Circular-motion equations require a radius or path length in metres, which is a one-dimensional quantity."}]},{"id":"a788abad-9fd6-4be8-9b08-31b47447dd68","cardType":"cloze","points":null,"front":"When calculating circular motion, mass must be in kg, distance in m, and time in [ ].","back":"When calculating circular motion, mass must be in kg, distance in m, and time in s.","mcqOptions":[{"text":"s","correct":true,"explanation":"Seconds are the SI base unit of time. Periods and other time measurements must be expressed in seconds when calculating angular speed, frequency or centripetal acceleration."},{"text":"ms","correct":false,"explanation":"Milliseconds measure time but must be converted before use in standard SI equations. Divide the millisecond value by $1000$ to obtain the time in seconds."},{"text":"min","correct":false,"explanation":"Minutes are not the SI base unit of time. Multiply a value in minutes by $60$ before substituting the corresponding time in seconds."},{"text":"Hz","correct":false,"explanation":"Hertz measures frequency, not time. Time must be measured in seconds; if frequency is known, the period is found using $T=1/f$."}]},{"id":"63198f46-5859-4036-98f3-5eb60ba09d4a","cardType":"basic","points":1,"front":"What is the direction of the velocity vector for an object in circular motion?","back":"Tangent to the circular path.","mcqOptions":[{"text":"Tangent to the circular path.","correct":true,"explanation":"At every point, instantaneous velocity points along the tangent because it gives the object’s direction of travel. The centripetal acceleration is perpendicular to it and points inwards."},{"text":"Radially inwards towards the centre.","correct":false,"explanation":"The inward radial direction belongs to centripetal acceleration and resultant force, not velocity. Instantaneous velocity points along the tangent to the circular path."},{"text":"Radially outwards from the centre.","correct":false,"explanation":"An outward radial direction may be mistakenly associated with centrifugal force, but it is not the direction of motion. Velocity is tangent to the circular path."},{"text":"Parallel to the circle’s diameter.","correct":false,"explanation":"Velocity does not point along a fixed line through the circle because its direction changes continuously. At each position, it is tangent to the circular path."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Circular motion (A-level only)","title":"Circular motion (A-level only) Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"177fb5c7-412c-5fee-9391-b88bf8c95d17","walled":"$L39"}],"$L3a","$L3b"]