Centripetal force — what provides it in various situations: Worked examples

Before we start, one promise about symbols. We only use letters we have earned:

Definition The symbols we will reuse

$m$ ::: mass of the moving body, in kilograms (kg).
$v$ ::: speed = how fast, ignoring direction, in metres per second (m/s).
$r$ ::: radius of the circular path, in metres (m). Picture the string length, or the distance from the centre of the curve.
$ω$ (Greek "omega") ::: angular speed = how many radians of angle the body sweeps per second. Linked to speed by $v = ω r$ .
$a_{c} = \frac{v ^{2}}{r}$ ::: centripetal acceleration — how fast the velocity's direction is turning, in $m/s^{2}$ , always pointing toward the centre. It's the thing a force must produce (from Uniform Circular Motion).
$F_{c} = \frac{m v ^{2}}{r} = m a_{c}$ ::: the required inward force. This is a demand, a number the real forces must add up to. Not a force you draw.
$g = 9.8 m/s^{2}$ ::: strength of gravity near Earth's surface.
$T$ ::: tension — the inward pull a taut string or rope exerts, in newtons (N). Picture the string yanking the object back toward your hand.
$N$ ::: normal force — the push a surface gives, always perpendicular ("normal") to that surface, in newtons (N). Picture a road pushing straight up out of itself.
$f_{s}$ ::: static-friction force — the sideways grip between two surfaces that are not sliding, in newtons (N). Picture the tyres gripping the road.
$μ_{s}$ (Greek "mu", subscript $s$ ) ::: coefficient of static friction — a plain number (no units) saying how grippy two surfaces are. It sets friction's ceiling via $f_{s} \leq μ_{s} N$ .

If any of these feels shaky, revisit the parent note and Uniform Circular Motion first.

The scenario matrix

Every problem in this topic is one cell of this table. Our worked examples below are labelled by which cell they hit — together they touch all of them.

Cell	What makes it special	Provider of $F_{c}$	Covered by
A Single inward force, horizontal	tension = whole job	Tension	Ex 1
B Force splits into two components	vertical part holds weight, horizontal part turns you	Normal / Tension	Ex 2 (banking), Ex 3 (conical pendulum)
C Ceiling / limiting value	force has a maximum → a max speed exists	Static friction	Ex 4
D Both forces point inward together	gravity + contact both add up	Gravity + Tension/Normal	Ex 5 (top of loop)
E Sign flips around the loop	inward direction changes side-to-side	Gravity vs Tension	Ex 6 (bottom of loop)
F Inverse-square provider	force depends on $1/ r^{2}$	Gravity (orbit)	Ex 7
G Velocity-dependent provider	force itself grows with $v$	Magnetic force	Ex 8
H Degenerate / zero input	what happens as $v \to 0$ , $r \to \infty$ , $θ \to 0$	limiting logic	Ex 9

Example 1 — Cell A: stone on a string (single inward force)

Forecast: Will the tension be a few newtons, or hundreds? Guess a number, then check yourself.

Figure — Centripetal force — what provides it in various situations

What the figure shows: the faint circle is the stone's path, the dot in the middle is your hand (the centre). The orange arrow runs along the string from the stone straight in to the centre — that is the tension $T$ , the only force pulling inward. The teal arrow at the stone points sideways, tangent to the circle — that is the velocity $v$ , and notice it is at right angles to the string.

Steps.

Identify the real forces pointing inward. Why this step? The orange arrow in the figure lies exactly along the string, aimed at the centre: only the tension $T$ pulls the stone that way. No other arrow points along that inward line, so tension alone does the job.
Set net inward force = demand: $T = \frac{m v ^{2}}{r} .$ Why this step? Newton's 2nd law in the inward direction (Newton's Second Law) says the net inward force equals $m a_{c} = m v^{2} / r$ .
Put numbers in: $T = \frac{0.20 \times 4. 0 ^{2}}{0.50} = \frac{0.20 \times 16}{0.50} = 6.4 N .$

Verify: Units: $\frac{kg \cdot ( m/s ) ^{2}}{m} = \frac{kg \cdot m}{s ^{2}} = N$ . ✓ Sanity: double the speed → quadruples the tension (since $v^{2}$ ), which matches "faster whirl feels much harder." ✓

Example 2 — Cell B: car on a frictionless banked road

Forecast: A gentle 20° tilt — do you expect a walking pace, a city speed, or highway speed?

What the figure shows: the thick dark line is the tilted road surface, the plum square is the car sitting on it. The dark arrow pointing straight down is gravity $m g$ . The orange arrow leaving the car at right angles to the road is the normal force $N$ . The two teal dashed lines drop $N$ into its shadow on a vertical wall and a horizontal floor: the vertical dashed piece is $N cos θ$ (points up), and the horizontal dashed piece is $N sin θ$ (points in toward the curve's centre).

Steps.

Draw the two real forces: gravity $m g$ straight down, and the normal force $N$ perpendicular to the tilted road. Why this step? On a frictionless bank only these two exist — friction is switched off deliberately.
Split $N$ into pieces (follow the two teal dashed lines). The vertical piece $N cos θ$ holds the weight; the horizontal piece $N sin θ$ points toward the centre. $N cos θ = m g, N sin θ = \frac{m v ^{2}}{r} .$ Why this step? The car doesn't sink or fly, so vertical forces cancel; the leftover horizontal slice is the only thing available to turn the car — so it must equal the demand.
Divide the two equations to kill $N$ and $m$ : $tan θ = \frac{v ^{2}}{r g} \Rightarrow v = r g tan θ .$ Why this step? Dividing is the trick that removes the unknown $N$ — we only care about the ratio of the components, which is $tan θ$ (see Banking of Roads).
Numbers: $v = 80 \times 9.8 \times tan 2 0^{\circ} = 80 \times 9.8 \times 0.3640 = 285.3 \approx 16.9 m/s .$

Verify: $16.9 m/s \approx 61 km/h$ — a sensible highway-curve speed. ✓ Note $m$ cancelled: the design speed is the same for a truck and a scooter. ✓

Example 3 — Cell B: conical pendulum (tension splits)

Forecast: Same $tan θ = v^{2} / (r g)$ shape as banking — but here $r$ is hidden inside $L$ . Will you need it?

What the figure shows: the dark slanted line is the string running from the support down to the plum bob. The dotted vertical line drops from the support straight down, and the orange arc between it and the string marks the angle $θ$ . The teal dashed horizontal line from that vertical down to the bob is the circle's radius $r$ — it is the horizontal shadow of the string, which is why $r = L sin θ$ and not $L$ . The orange arrow up the string is the tension $T$ ; the dark arrow straight down is the weight $m g$ .

Steps.

The radius is not $L$ ! Reading the teal dashed line in the figure, the horizontal reach is $r = L sin θ$ . Why this step? The string is the slanted hypotenuse; the circle's radius is its horizontal shadow. $r = 1.2 sin 3 0^{\circ} = 1.2 \times 0.5 = 0.60 m .$
Split the tension: vertical part carries weight, horizontal part turns the bob. $T cos θ = m g, T sin θ = \frac{m v ^{2}}{r} .$ Why this step? Identical logic to banking — only the tilted force is now a string, not a road.
Divide the two equations to cancel the unknown $T$ (and the mass $m$ ), then solve for $v$ : $tan θ = \frac{v ^{2}}{r g} \Rightarrow v = r g tan θ .$ Why this step? We were never told $T$ or $m$ , so we must get rid of them. Dividing the horizontal equation by the vertical one leaves only the ratio $sin θ / cos θ = tan θ$ — an equation in known quantities, which we then rearrange to isolate $v$ . $v = 0.60 \times 9.8 \times tan 3 0^{\circ} = 0.60 \times 9.8 \times 0.5774 = 3.395 \approx 1.84 m/s .$

Verify: Units under the root: $m \cdot m/s^{2} = m^{2} / s^{2}$ , root gives m/s. ✓ If $θ \to 0$ (string nearly vertical), $tan θ \to 0$ so $v \to 0$ — a barely-swinging bob, correct. ✓

Example 4 — Cell C: flat curve, friction has a ceiling

Forecast: More friction than banking usually allows — so faster or slower than Example 2's 16.9 m/s?

What the figure shows: a bird's-eye (top-down) view of a flat curve. The faint arc is the road's edge; the plum square is the car. The orange arrow points inward toward the centre of the curve — that is the static friction $f_{s}$ , the only sideways grip turning the car. The teal arrow is the car's velocity $v$ , tangent to the arc. Off to the side a dashed orange line marks the ceiling height $μ_{s} m g$ : friction can grow up to that line and no further.

Steps.

On a flat road, the only inward force is static friction $f_{s}$ . Why this step? Gravity is vertical, normal is vertical; friction is the sole horizontal (inward) player (Friction) — exactly the orange arrow in the figure.
Friction cannot exceed its ceiling: $f_{s} \leq μ_{s} N = μ_{s} m g .$ Why this step? Static friction supplies exactly what's needed up to a maximum of $μ_{s} m g$ (the dashed ceiling line). Push past that and the tyres slip.
At maximum speed, demand = ceiling: $\frac{m v _{m a x}^{2}}{r} = μ_{s} m g \Rightarrow v_{m a x} = μ_{s} g r .$
Numbers: $v_{m a x} = 0.60 \times 9.8 \times 45 = 264.6 \approx 16.3 m/s .$

Verify: $m$ cancels again — a loaded truck skids at the same speed as a light car (surprising but true). ✓ Compare to Ex 2: this flat curve tops out at 16.3 m/s; banking the road would let you go faster with less friction wear. ✓

Example 5 — Cell D: top of a vertical loop (both forces inward)

Forecast: At the very top, gravity points down, which is toward the centre. Does that help or hurt the string?

What the figure shows: the faint circle is the vertical loop, the dot in the middle is its centre, and the plum dot sits at the very top. Both arrows leaving the ball point straight down: the orange arrow is the tension $T$ and the dark arrow is the weight $m g$ . The key visual is that at the top the centre is below the ball, so "inward" means "downward" — and both forces happen to line up that way.

Steps.

At the top, both gravity $m g$ and tension $T$ point downward = inward. Why this step? As the figure shows, the centre of the loop is below the ball, so "inward" is "down" here. Both real forces conveniently line up. $T + m g = \frac{m v ^{2}}{r} .$
Minimum speed is when the string is on the verge of going slack, $T = 0$ : $m g = \frac{m v _{m i n}^{2}}{r} \Rightarrow v_{m i n} = g r = 9.8 \times 1.0 \approx 3.13 m/s .$ Why this step? If gravity alone already exceeds the demand, the string can't push (strings only pull), so it slackens and the ball falls off the circle.
Now the tension at $v = 5.0$ : $T = \frac{m v ^{2}}{r} - m g = \frac{0.30 \times 5. 0 ^{2}}{1.0} - 0.30 \times 9.8 = 7.5 - 2.94 = 4.56 N .$

Verify: $T > 0$ ✓ (string is taut, consistent with $v = 5.0 > v_{m i n} = 3.13$ ). If we plugged $v = v_{m i n}$ , we'd get $T = 0$ exactly. ✓

Example 6 — Cell E: bottom of the loop (sign flips)

Forecast: At the bottom, does the string work with gravity or against it? Bigger or smaller tension than at the top?

What the figure shows: same faint loop and centre dot, but now the plum ball is at the very bottom. The centre is above the ball, so "inward" now means "upward." The orange arrow points up (tension $T$ , inward) while the dark arrow points down (weight $m g$ , outward) — the two forces now oppose each other, the opposite of Example 5.

Steps.

At the bottom, the centre is above the ball, so inward = up. As the figure shows, tension (orange) points up (inward, $+$ ); gravity (dark) points down (outward, $-$ ). Why this step? The inward direction flipped sides — this is the whole point of Cell E. $T - m g = \frac{m v ^{2}}{r} .$
Solve: $T = \frac{m v ^{2}}{r} + m g = 7.5 + 2.94 = 10.44 N .$ Why this step? Now the string must both supply the whole demand and hold up the weight, so tension is largest here.

Verify: Units: each term is $kg \cdot m/s^{2} = N$ . ✓ Sanity: $10.44 N$ (bottom) $> 4.56 N$ (top, Ex 5) — the string is most likely to snap at the bottom, which matches real experience with swings and buckets. ✓ The difference $10.44 - 4.56 = 5.88 N = 2 m g$ , exactly the sign flip of gravity ( $2 \times 0.30 \times 9.8$ ). ✓

Example 7 — Cell F: satellite (inverse-square provider)

Forecast: The ISS orbits at roughly this height. Do you expect a few hundred m/s, or several kilometres per second?

Steps.

The provider is gravity, whose strength falls as $1/ r^{2}$ (Gravitation and Orbits): $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} .$ Why this step? Gravity is the only inward force in orbit — set it equal to the demand.
Cancel one $r$ and the mass $m$ : $v = \frac{GM}{r} .$ Why this step? The satellite's own mass drops out — orbit speed depends only on where you are, not on how heavy the satellite is.
Numbers: $v = \frac{3.986 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} = 5.694 \times 1 0^{7} \approx 7.55 \times 1 0^{3} m/s .$

Verify: About $7.5 km/s$ — the well-known low-Earth-orbit speed. ✓ Larger $r$ → smaller $v$ : far orbits are slower, as observed for the Moon vs the ISS. ✓

Example 8 — Cell G: charge in a magnetic field (velocity-dependent provider)

Forecast: Microscopic mass, decent speed — will the radius be centimetres, or kilometres?

Steps.

The provider is the magnetic force $q v B$ , always perpendicular to velocity (Magnetic Force on Moving Charges), so it curves the path without changing speed — a perfect centripetal force. $q v B = \frac{m v ^{2}}{r} .$ Why this step? A force always at right angles to motion does no work (no speed change) but constantly turns the direction — exactly what a circle needs.
Cancel one $v$ and solve: $r = \frac{m v}{q B} .$ Why this step? Notice the force itself grew with $v$ , so one factor of $v$ cancels — radius rises only linearly with speed, not as $v^{2}$ .
Numbers: $r = \frac{1.67 \times 1 0 ^{- 27} \times 2.0 \times 1 0 ^{6}}{1.60 \times 1 0 ^{- 19} \times 0.50} = \frac{3.34 \times 1 0 ^{- 21}}{8.0 \times 1 0 ^{- 20}} \approx 0.0418 m .$

Verify: About $4.2 cm$ — a plausible cyclotron-scale radius. ✓ Units: $\frac{kg \cdot m/s}{C \cdot T} = m$ ✓ (since $T = kg / (C \cdot s)$ ).

Example 9 — Cell H: degenerate & limiting cases

Forecast: These are the "what if it's basically not a circle?" checks. Predict each before reading.

Steps.

$v \to 0$ : demand $F_{c} = m v^{2} / r \to 0$ . Why this step? A body barely moving needs almost no inward force — a parked car needs zero friction to "turn." The formula agrees: no motion, no requirement.
$r \to \infty$ : $F_{c} = m v^{2} / r \to 0$ . Why this step? An infinitely large circle is a straight line, and a straight line needs no inward force at all (Newton's 1st law). The formula smoothly returns the "no force" answer.
$θ \to 0$ (flat bank): the frictionless design speed is $v = r g tan θ$ . As $θ \to 0$ , $tan θ \to 0$ , so $v \to 0$ . Why this step? A perfectly flat road contributes zero horizontal normal component ( $N sin 0 = 0$ ), so with friction switched off the only "safe" frictionless speed is standing still. This is exactly why real flat curves cannot rely on tilt alone and must borrow friction (which is Cell C, Example 4). The banking picture and the friction picture meet cleanly at $θ = 0$ .

Verify: All three limits collapse to "no inward force needed / available," each consistent with straight-line motion. No formula blows up or gives a nonsense negative, and case (c) hands the job back to friction — no gap between the models. ✓

Recall Quick self-test

Which cell is each of these? (i) bucket of water at top of a swing. (ii) coin on a spinning turntable not sliding. (iii) Moon around Earth. (i) ::: Cell D — gravity and tension/normal both inward (like Ex 5). (ii) ::: Cell C — static friction provides $F_{c}$ with a ceiling (like Ex 4). (iii) ::: Cell F — gravity, inverse-square provider (like Ex 7).

Connections

Newton's Second Law — the inward projection $\sum F_{in} = m a_{c}$ used in every example.
Uniform Circular Motion — where $v = ω r$ and $a_{c} = v^{2} / r$ come from.
Banking of Roads — Ex 2's geometry in detail.
Friction — the $μ_{s} N$ ceiling behind Ex 4.
Gravitation and Orbits — Ex 7's provider.
Magnetic Force on Moving Charges — Ex 8's provider.
Pseudo-forces and Non-inertial Frames — why "centrifugal" never appears in these ground-frame solutions.
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