Conduct investigations to explain and evaluate, for objects executing uniform circular motion, the relationships that exist between centripetal force, mass, speed and radius, and solve problems using the relationships a_c = v^2 / r, v = 2 pi r / T, F_c = m v^2 / r and omega = delta theta / delta t

Centripetal force is the net force directed toward the centre of the circular path.

$F_c = \frac{m v^2}{r} = \frac{1200 \times 18^2}{80} = \frac{1200 \times 324}{80} = 4860$ N.

The centripetal force is provided by the friction between the tyres and the road. On a flat (unbanked) bend, friction is the only horizontal force available to push the car toward the centre of the curve. If friction is insufficient (for example, on a wet road), the car cannot complete the turn and skids outward.

Markers reward the correct numerical answer with units, explicit identification of friction as the source of the centripetal force, and the link between the force and the curved path.

Acceleration is the rate of change of velocity, not speed. Velocity is a vector with both magnitude and direction.

In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of the velocity changes continuously because the object follows a curved path. A change in direction is a change in velocity, and therefore the object is accelerating.

The acceleration is directed toward the centre of the circle (centripetal acceleration), with magnitude $a_c = \frac{v^2}{r}$. By Newton's second law, this acceleration requires a net inward force.

Markers reward the distinction between speed and velocity, the explicit mention of the direction change, and the link to centripetal acceleration.

Reading the graph. The line is straight through the origin with gradient $= \dfrac{F_c}{v^2}$. Taking the marked point at $v^2 = 16\ \text{m}^2\text{s}^{-2}$, $F_c = 8.0\ \text{N}$ gives gradient $= \dfrac{8.0}{16} = 0.50\ \text{kg m}^{-1}$, which is $m/r$.

Direct calculation. $F_c = \dfrac{mv^2}{r} = \dfrac{0.40 \times 4.0^2}{0.80} = \dfrac{0.40 \times 16}{0.80} = 8.0\ \text{N}$.

This matches the graph's data point at $v^2 = 16\ \text{m}^2\text{s}^{-2}$, and $m/r = 0.40/0.80 = 0.50\ \text{kg m}^{-1}$ matches the gradient read off the graph.

Marks: one for correctly reading the gradient from the graph as $F_c/v^2$, one for the correct substitution into $F_c = mv^2/r$, one for $F_c = 8.0$ N with the unit and the consistency check against the graph.

$a_c = \dfrac{v^2}{r} = \dfrac{(7.6 \times 10^3)^2}{6.8 \times 10^6} = \dfrac{5.78 \times 10^7}{6.8 \times 10^6} = 8.5\ \text{m/s}^2$.

Marks: one for the correct formula with values substituted, one for the answer stated to two significant figures with the unit $\text{m/s}^2$.

$\omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.25} = 25\ \text{rad/s}$.

Marks: one for using $\omega = 2\pi/T$ (or $\Delta\theta/\Delta t$ with $\Delta\theta = 2\pi$), one for the answer $\omega = 25\ \text{rad/s}$ (2 s.f.) with the correct unit (radians per second, not revolutions or degrees).

(a) Derivation. With no friction, only gravity ($mg$, down) and the normal force $N$ (perpendicular to the road) act on the car. Resolving vertically, $N\cos\theta = mg$. Resolving horizontally, the only inward component is $N\sin\theta$, which supplies the centripetal force: $N\sin\theta = \dfrac{mv^2}{r}$.

Dividing the horizontal equation by the vertical equation cancels $N$: $\tan\theta = \dfrac{v^2}{rg}$, so $v = \sqrt{rg\tan\theta}$.

(b) Numerical answer. $v = \sqrt{60 \times 9.8 \times \tan 20^{\circ}} = \sqrt{60 \times 9.8 \times 0.364} = \sqrt{214} = 15\ \text{m/s}$.

Marks: one for correctly resolving $N$ into vertical and horizontal components, one for dividing the two equations to eliminate $N$ and $m$, one for correctly rearranging to $v = \sqrt{rg\tan\theta}$, one for the numerical answer $v = 15\ \text{m/s}$ (2 s.f.) with the unit.

(a) Radius

$r = L\sin\theta = 0.90 \times \sin 30^{\circ} = 0.90 \times 0.50 = 0.45\ \text{m}$.

(b) Tension

Vertically, the tension's vertical component balances gravity: $T\cos\theta = mg$, so $T = \dfrac{mg}{\cos\theta} = \dfrac{0.60 \times 9.8}{\cos 30^{\circ}} = \dfrac{5.88}{0.866} = 6.8\ \text{N}$.

(c) Speed

The horizontal component of tension is the centripetal force: $T\sin\theta = \dfrac{mv^2}{r}$, so $v = \sqrt{\dfrac{Tr\sin\theta}{m}} = \sqrt{\dfrac{6.8 \times 0.45 \times 0.50}{0.60}} = \sqrt{2.55} = 1.6\ \text{m/s}$.

Marks: one for $r = 0.45$ m, one for correctly resolving tension vertically ($T\cos\theta = mg$), one for $T = 6.8$ N with the unit, one for correctly resolving tension horizontally to find $v$, one for $v = 1.6$ m/s (2 s.f.) with the unit.

Friction supplies the centripetal force, up to its maximum value $f_{\max} = \mu_s N = \mu_s mg$ (on a flat track, $N = mg$). At the maximum speed, $\mu_s mg = \dfrac{mv_{\max}^2}{r}$, and $m$ cancels: $v_{\max} = \sqrt{\mu_s g r}$.

$v_{\max} = \sqrt{0.60 \times 9.8 \times 18} = \sqrt{105.8} = 10\ \text{m/s}$.

Marks: one for equating maximum friction to the required centripetal force ($\mu_s mg = mv^2/r$), one for correctly cancelling $m$ and rearranging to $v_{\max} = \sqrt{\mu_s g r}$, one for the answer $v_{\max} = 10\ \text{m/s}$ (2 s.f.) with the unit.

Band-6 plan. (1) State that the centripetal force is constant in magnitude (same $v$, same $r$) but always directed toward the centre, so its direction relative to gravity changes at top versus bottom. (2) Draw the free-body reasoning at each point: identify the two real forces (gravity, normal force) and apply Newton's second law along the radial direction. (3) Solve for $N$ at the bottom and at the top. (4) Explain physically why $N$ differs (the "heavier at the bottom, lighter at the top" sensation) and note the special case if $N$ reached zero.

Model answer. The centripetal force needed is $F_c = \dfrac{mv^2}{r} = \dfrac{70 \times 3.0^2}{15} = 42\ \text{N}$, directed toward the centre of the wheel at every point of the ride, since $m$, $v$ and $r$ do not change.

At the bottom of the wheel, the centre is directly above the passenger, so the inward (upward) direction is positive. The two forces acting are the normal force $N$ (up) and gravity $mg$ (down). Newton's second law toward the centre gives $N - mg = F_c$, so $N = mg + F_c = (70 \times 9.8) + 42 = 686 + 42 = 728\ \text{N}$.

At the top of the wheel, the centre is directly below the passenger, so the inward (downward) direction is now positive. Both gravity and, this time, the deficit of normal force act toward the centre: $mg - N = F_c$, so $N = mg - F_c = 686 - 42 = 644\ \text{N}$.

The normal force is larger at the bottom ($728\ \text{N}$) than at the top ($644\ \text{N}$) because the direction of the required centripetal force is the same (always toward the centre) but the direction of gravity relative to that centre reverses between the two points. At the bottom, the seat must push up hard enough both to support the passenger's weight and to supply the extra inward force, so $N > mg$ and the passenger feels heavier. At the top, gravity itself supplies part or all of the centripetal force, so the seat can push less, $N < mg$, and the passenger feels lighter. If the wheel spun fast enough that $F_c \geq mg$, $N$ at the top would fall to zero and the passenger would momentarily leave the seat.

Marker's note: the top band sets up Newton's second law separately at the top and bottom with correctly signed equations (not just quoting a memorised "add/subtract $mg$" rule), computes both normal forces correctly, and explains the physical reason for the difference (the fixed centripetal force direction versus the reversing direction of gravity) rather than only describing the sensation of feeling heavier or lighter.

Band-6 plan. (1) State the claim's core error: centripetal force is a role, not a distinct force, then define it correctly. (2) Justify with Newton's second law and $a_c = v^2/r$: any object moving on a curved path must have a net inward force, which must be supplied by some real force. (3) Give at least two worked examples identifying the real force each time (friction on a bend; tension in a conical pendulum or on a string; gravity for an orbit; normal force on a banked track). (4) Reach an evaluative judgement on the claim, explicitly rejecting or qualifying it with reasoning.

Model answer. The claim is not correct. Centripetal force is not a new, separate physical force alongside gravity, friction, tension and the normal force; it is simply the name given to whatever net force happens to point toward the centre of a circular path. This follows directly from Newton's second law: any object executing uniform circular motion has velocity that is constantly changing direction, so by $a_c = v^2/r$ it has a non-zero acceleration directed toward the centre, and by $F = ma$ there must be a real net force $F_c = mv^2/r$ producing that acceleration. The equation $F_c = mv^2/r$ describes the size of the required net force; it does not describe a new source of force.

Two examples illustrate this. First, a car rounding a flat, unbanked bend at speed $v$ and radius $r$ needs an inward force $F_c = mv^2/r$; the only horizontal force available is static friction between the tyres and the road, so friction plays the centripetal role, and the car skids if $\mu_s mg < mv^2/r$. Second, a bob on a conical pendulum swinging at constant angle $\theta$ needs the same inward force $F_c = mv^2/r$; here the horizontal component of the string's tension, $T\sin\theta$, supplies it, while the vertical component $T\cos\theta$ separately balances gravity. In an orbiting satellite, gravity alone supplies $F_c$; on a banked, frictionless curve, a component of the normal force supplies it. In every case, the "centripetal force" is a different real force depending on the physical setup, not a fifth fundamental force.

The claim therefore over-states what centripetal force is: it correctly recognises that circular motion always requires a net inward force of magnitude $mv^2/r$, but it is wrong to treat this as a distinct force in its own right rather than as the net effect, or a component, of gravity, friction, tension or the normal force acting in a given situation.

Marker's note: the top band explicitly rejects or heavily qualifies the claim (not a fence-sitting "it depends" with no conclusion), derives the need for a net inward force from $a_c = v^2/r$ and Newton's second law rather than asserting it, and correctly names the real force in two distinct, clearly explained examples. Confusing centripetal with centrifugal force, or describing centripetal force as "the force that pushes objects outward," is a serious error that caps the response well below Band 6.