Vertical circular motion — minimum speed conditions: Worked examples

This page is the drill deck for Vertical circular motion — minimum speed conditions. The parent note built the three master results; here we hammer them against every case a problem can throw at you — every position on the circle, both constraint types (string vs rod), the degenerate slow case, the fast case, energy links, and exam twists.

Before we start, one reminder in plain words. A vertical circle just means a loop standing up in the room, like a Ferris wheel, so height changes as you go around and gravity ( $m g$ , the downward pull on a mass $m$ ) matters differently at each point. The letter $r$ is the radius (distance from center to the ball), $v$ is the speed (how fast it moves along the circle), and $g \approx 9.8 m/s^{2}$ is the strength of gravity. The symbol $\frac{m v ^{2}}{r}$ is the required inward force — not a new push, just the total that real forces (gravity + tension + normal) must add up to, so the ball keeps curving instead of flying straight.

The scenario matrix

Every vertical-circle problem is one (or a combo) of these cells. Each worked example below is tagged with the cell(s) it lands on.

Cell	What varies	Danger / trick it tests
A — Top, string	Position = top, constraint pulls only	$T \geq 0$ ; slack when too slow
B — Bottom, string	Position = bottom	tension adds to the load ( $6 m g$ )
C — Side, height $= r$	Position = 90°	gravity is tangent, not central
D — Arbitrary angle $θ$	General point on loop	resolve gravity into radial + tangential
E — Degenerate slow	$v_{top} < g r$	ball leaves circle → projectile
F — Rigid rod	Constraint can push ( $T < 0$ allowed)	min top speed $= 0$
G — Real-world word	bucket / coaster / bike	translate words to a cell
H — Exam twist	asks a ratio or a proof	algebra, no numbers
I — Fast case	speed well above minimum	tension grows, never slack

The examples below are numbered to cover every cell: Ex1→A, Ex2→B, Ex3→C, Ex4→D, Ex5→E, Ex6→F, Ex7→G, Ex8→H, Ex9→I. Figures accompany the geometric cases (A–D).

Example 1 — Top of a string loop (Cell A)

Figure — Vertical circular motion — minimum speed conditions

Step 1 — Locate the center; assign directions. In figure s01 the red dot is the center and the white dot is the ball, sitting directly above it. So "toward center" points straight down. The blue arrow is the string tension $T$ and the yellow arrow is gravity $m g$ — trace them with your eye: both point downward, into the center. The green note flags the payoff: because both are inward, they add. Why this step? Direction of the net inward force is the #1 thing students get wrong — always find the center first, then read the arrows off the figure.

Step 2 — Write Newton's second law toward the center. $T + m g = \frac{m v _{top}^{2}}{r}$ Why this step? This is Newton's Second Law — net force form applied along the radius: the sum of real inward forces (the two downward arrows in s01) equals the required $\frac{m v ^{2}}{r}$ .

Step 3 — Impose the physical edge $T = 0$ . A string can only pull, so $T \geq 0$ . The slowest circular motion is when tension just vanishes (the blue arrow in s01 shrinks to nothing): $m g = \frac{m v _{top,min}^{2}}{r} \Rightarrow v_{top,min} = g r = 9.8 \times 0.80 = 2.80 m/s$ Why this step? At $T = 0$ gravity (the yellow arrow) alone supplies the whole turning force — the defining condition of "minimum".

Step 4 — Tension at that minimum speed. By construction $T = 0$ .

Verify: $9.8 \times 0.8 = 7.84 = 2.80$ ✓. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ ✓. Tension $0 N$ is exactly the minimum-condition answer, not "almost zero".

Example 2 — Bottom of the same loop (Cell B)

Step 1 — Get $v_{bot}$ from energy. Start with full Conservation of mechanical energy between bottom and top, keeping the mass $m$ visible. Kinetic energy is $\frac{1}{2} m v^{2}$ and gravitational potential energy is $m g h$ ; the climb from bottom to top is a height $Δ h = 2 r$ . Tension is always perpendicular to motion, so it does no work: $KE bottom \frac{1}{2} m v_{bot}^{2} = KE top \frac{1}{2} m v_{top}^{2} + PE gained m g (2 r)$ Now every term has the same factor $m$ , so divide the whole equation by $m$ — the mass cancels and never affects the speeds: $\frac{1}{2} v_{bot}^{2} = \frac{1}{2} v_{top}^{2} + g (2 r) \Rightarrow v_{bot}^{2} = v_{top}^{2} + 4 g r = g r + 4 g r = 5 g r$ $v_{bot} = 5 g r = 5 \times 9.8 \times 0.8 = 39.2 = 6.26 m/s$ Why this step? The bottom must be faster to "pay" for the height it will climb; writing $m g h$ first (not the shortcut) shows exactly why $m$ drops out, so nobody wonders where the mass went.

Step 2 — Directions at the bottom. In figure s02 the center (red dot) is now above the ball. Toward center = up. The blue arrow (tension) points up toward the center; the yellow arrow (gravity) points down. Reading those two off the figure: $T_{bot} - m g = \frac{m v _{bot}^{2}}{r}$ Why this step? Opposite geometry to the top (compare arrows in s01 vs s02) — that's the whole point of Cell B: here tension and gravity fight instead of adding.

Step 3 — Solve. $T_{bot} = m g + \frac{m ( 5 g r )}{r} = m g + 5 m g = 6 m g = 6 (0.20) (9.8) = 11.76 N$ Why this step? We substitute the known $v_{bot}^{2} = 5 g r$ into the radial equation and isolate $T_{bot}$ — this converts the general force balance into the single number the problem asked for.

Verify: $6 m g = 6 \times 0.2 \times 9.8 = 11.76 N$ ✓. Sanity: it's exactly six times the weight $m g = 1.96 N$ ✓. And $T_{bot} - T_{top} = 11.76 - 0 = 11.76 = 6 m g$ ✓.

Example 3 — At the side, height $= r$ (Cell C)

Step 1 — Speed at the side by energy. Same per-unit-mass energy balance as Example 2, but the climb from bottom is only $Δ h = r$ , so the $v^{2}$ drop is $2 g r$ : $v_{side}^{2} = v_{bot}^{2} - 2 g r = 5 g r - 2 g r = 3 g r = 3 \times 9.8 \times 0.8 = 23.52$ $v_{side} = 23.52 = 4.85 m/s$ Why this step? Same energy bookkeeping, just a smaller climb.

Step 2 — Directions at the side. In figure s03 the center (red dot) is horizontal from the ball, so the blue arrow (tension, toward center) is horizontal. Gravity — the yellow arrow — points straight down, which here is tangent to the circle (along the direction of motion). The green note spells it out: gravity changes the speed, not the turning, so only tension acts inward: $T_{side} = \frac{m v _{side}^{2}}{r}$ Why this step? This is the key lesson of Cell C — gravity's whole job here is tangential (yellow arrow perpendicular to the blue radial arrow), so it drops out of the radial equation.

Step 3 — Solve. $T_{side} = \frac{m ( 3 g r )}{r} = 3 m g = 3 (0.20) (9.8) = 5.88 N$ Why this step? Plugging $v_{side}^{2} = 3 g r$ into the radial equation collapses it to a clean multiple of the weight $m g$ , giving the number the problem wants and exposing the $0, 3 m g, 6 m g$ pattern.

Verify: $3 m g = 5.88 N$ ✓. Pattern check: tensions read $0, 3 m g, 6 m g$ (top, side, bottom) — matching the $1 : 3 : 5$ speed-squared ladder shifted by the changing gravity component ✓.

Example 4 — General angle $θ$ (Cell D)

Step 1 — Read the angle off the figure. Look at s04: the yellow arc at the center is the angle $θ$ , measured from the downward vertical (the line from center to the bottom) round to the dashed blue radius pointing at the ball. So $θ$ is literally "how far round from the bottom" the ball has travelled. This is the anchor that fixes all our signs. Why this step? Every sign below depends on where $θ$ is measured from; the figure pins it down so we can't get the projection backwards.

Step 2 — Resolve gravity into radial + tangential. Gravity (the yellow down-arrow at the ball in s04) is straight down. The dashed blue radius makes angle $θ$ with that downward vertical (that's exactly the arc we just read). The piece of gravity lying along the radius, pointing outward (away from center) is the green arrow, of size $m g cos θ$ . The inward radial equation is therefore: $T - m g cos θ = \frac{m v ^{2}}{r}$ Why this step? Only the radial slice of gravity competes with tension; the leftover tangential slice $m g sin θ$ merely speeds up/slows the ball. We use $cos θ$ because $θ$ is the angle between the down-direction and the radius, and the cosine projects gravity onto that radius (adjacent-side rule).

Step 3 — Check the three landmark angles.

$θ = 0$ (bottom): $cos 0 = 1 \Rightarrow T - m g = \frac{m v ^{2}}{r}$ ✓ (Cell B).
$θ = 9 0^{\circ}$ (side): $cos 9 0^{\circ} = 0 \Rightarrow T = \frac{m v ^{2}}{r}$ ✓ (Cell C).
$θ = 18 0^{\circ}$ (top): $cos 18 0^{\circ} = - 1 \Rightarrow T + m g = \frac{m v ^{2}}{r}$ ✓ (Cell A).

Why this step? A general formula is only trustworthy if it collapses to the special cases you already proved — the sign of $cos θ$ flips exactly as the geometry in s01–s03 demands.

Step 4 — Numbers at $θ = 6 0^{\circ}$ . Radial gravity component $= m g cos 6 0^{\circ} = m g (0.5) = 0.5 m g$ . So gravity supplies half its weight along the radius here. Why this step? Evaluating a concrete angle turns the abstract $cos θ$ into a number you can feel — at $6 0^{\circ}$ exactly half of gravity fights the tension, which is the intuition the Forecast asked you to guess.

Verify: $cos 6 0^{\circ} = 0.5$ ✓, and the formula gives $T = \frac{m v ^{2}}{r} + 0.5 m g$ . At $θ = 0, 90, 18 0^{\circ}$ it returns the bottom/side/top laws exactly ✓.

Example 5 — Too slow at the top: it leaves the circle (Cell E)

Step 1 — Compute the tension the circle would demand. $T = \frac{m v ^{2}}{r} - m g = m (\frac{v ^{2}}{r} - g) = m (\frac{2. 0 ^{2}}{1.0} - 9.8) = m (4.0 - 9.8) = - 5.8 m$ So the equation asks for a negative tension of $- 5.8 m$ newtons (per kg). Why this step? A negative $T$ means the string would need to push — a string can't push (recall the sign convention: negative $T$ = push outward).

Step 2 — Interpret physically. Since the string cannot supply the demanded inward force, it goes slack ( $T = 0$ ). The ball is now a free projectile under gravity alone — it falls inside the circle and the string re-tightens later on the way down. It never completed the loop as a circle. Why this step? This is the real-world outcome of every "too slow" problem: not a smaller circle, but departure from the constraint.

Step 3 — Threshold check. The break-even is exactly $v^{2} / r = g$ , i.e. $v = g r = 3.13 m/s$ . Below it, $T < 0$ (slack); at it, $T = 0$ ; above it, $T > 0$ (taut). Why this step? Pinning down the exact break-even speed tells you the precise boundary between "circular" and "falls off", so you can classify any given top speed at a glance instead of re-solving each time.

Verify: $\frac{2. 0 ^{2}}{1.0} - 9.8 = 4.0 - 9.8 = - 5.8 < 0$ ✓ (negative → slack). Threshold $9.8 = 3.13 m/s$ ✓.

Example 6 — Rigid rod vs string (Cell F)

Step 1 — Drop the $T \geq 0$ rule. A rod can pull or push, so $T$ may be negative (recall the sign convention: negative $T$ = rod in compression, pushing outward). The only requirement to "reach the top" is $v_{top} \geq 0$ . Why this step? The $g r$ result was a string condition (it came entirely from demanding $T \geq 0$ ); remove that constraint and the physics of "minimum" changes completely.

Step 2 — (a) Min top speed and (b) min bottom speed. $v_{top,min} = 0.$ Just enough energy to arrive at the top with zero speed, using the same per-unit-mass energy balance as Example 2: $v_{bot}^{2} = v_{top}^{2} + 4 g r = 0 + 4 g r = 4 g r$ $v_{bot,min} = 4 g r = 2 g r = 2 9.8 \times 0.8 = 2 \times 2.80 = 5.60 m/s$ Why this step? With no slack condition, the only thing the rod must guarantee is that the ball arrives at the top ( $v_{top} \geq 0$ ); the cheapest way is $v_{top} = 0$ , and energy conservation over the climb $2 r$ then converts that into the required bottom speed.

Step 3 — (c) Rod force at the top when $v_{top} = 0$ . Use the top radial equation with $v = 0$ : $T + m g = \frac{m v ^{2}}{r} = \frac{m ( 0 )}{r} = 0 \Rightarrow T = - m g$ The negative sign means the rod is in compression: it pushes the ball outward (upward at the top) with force $m g$ , exactly holding the ball up against gravity so it can hang there motionless. A string, which cannot push, would simply drop the ball — this is precisely why a rod completes the loop where a string fails. Why this step? It shows concretely, with the sign convention, what "the rod can push" buys you: the missing support force $m g$ at the top.

Verify: $2 9.8 \times 0.8 = 2 7.84 = 2 \times 2.8 = 5.60 m/s$ ✓, which is less than the string's $5 g r = 6.26 m/s$ ✓ (rod is easier). Rod force $= - m g$ at top (compression) ✓.

Example 7 — Bucket of water (real-world word problem, Cell G)

Step 1 — (a) Translate the words into a cell. This is the modelling step every word problem needs. Water "spills" only if it tends to fall away from the bucket bottom — i.e. if the bucket bottom (the surface pushing on the water) would need to pull the water inward to keep it circling. The bottom exerts a normal force $N$ , and a surface can only push, so $N \geq 0$ — the exact same maths as a string in Cell A. The "ball" is the water, the "string" is the bucket bottom. So the minimum: $v_{top,min} = g r = 9.8 \times 1.2 = 11.76 = 3.43 m/s$ Why this step? Half of every real-world problem is spotting which cell it is; naming the constraint ( $N \geq 0$ ) reduces the bucket to Cell A.

Step 2 — (b) Effective weight at $v = 4.0 m/s$ . Draw the free body of the water at the top: gravity $m g$ down, normal force $N$ down (bucket bottom pushes water toward the center, which is below). For water of mass $m$ the top radial equation reads: $N + m g = \frac{m v ^{2}}{r} \Rightarrow N = m (\frac{v ^{2}}{r} - g) = m (\frac{16.0}{1.2} - 9.8) = m (13.33 - 9.8) = 3.53 m$ So the water presses on the bucket with $N = 3.53 m$ newtons per kg — this is its "effective weight" at the top. Why this step? By Newton's third law the force the bucket pushes on the water ( $N$ ) equals the force the water pushes back on the bucket, so solving the radial equation for $N$ is the "effective weight" the question asks for — and its sign tells us whether the water stays in ( $N > 0$ ) or spills ( $N < 0$ ).

Verify: (a) $9.8 \times 1.2 = 11.76 = 3.43 m/s$ ✓. (b) $\frac{16}{1.2} - 9.8 = 13.333 - 9.8 = 3.533 > 0$ ✓ (positive → water stays, presses harder than free-fall). Since $4.0 > 3.43$ , no spill ✓.

Example 8 — Exam twist: prove $T_{bot} - T_{top} = 6 m g$ for ANY speed (Cell H)

Step 1 — Write both radial equations. $T_{bot} - m g = \frac{m v _{bot}^{2}}{r}, T_{top} + m g = \frac{m v _{top}^{2}}{r}$ Why this step? Directions differ (center below at top, above at bottom), so the $m g$ signs flip — bottom subtracts $m g$ , top adds it. These are just Cells B and A written side by side with symbols instead of numbers.

Step 2 — Subtract the top equation from the bottom. $T_{bot} - T_{top} = \frac{m}{r} (v_{bot}^{2} - v_{top}^{2}) + 2 m g$ Why this step? The two $m g$ terms combine into $+ 2 m g$ (subtracting the top's $+ m g$ gives $- (+ m g)$ on the left, i.e. $+ m g$ moved over, added to the bottom's $- (- m g)$ ); the speeds gather into one bracket, ready for the energy substitution.

Step 3 — Kill the speeds with energy conservation. Over the height $2 r$ from bottom to top, the per-unit-mass energy balance gives a fixed speed-squared drop, independent of how fast you swing: $v_{bot}^{2} - v_{top}^{2} = 4 g r$ Substitute: $T_{bot} - T_{top} = \frac{m}{r} (4 g r) + 2 m g = 4 m g + 2 m g = 6 m g$ Why this step? The speed cancels because the difference in $v^{2}$ is set by geometry (the $2 r$ climb) alone, not by the launch speed — that's exactly why the identity is universal, which is what the problem asked us to prove.

Verify: $\frac{m}{r} (4 g r) + 2 m g = 4 m g + 2 m g = 6 m g$ , and the result contains no $v$ ✓ — it holds for every taut speed.

Example 9 — The fast case: tension keeps growing (Cell I)

Step 1 — Tension at the top. $T_{top} = \frac{m v _{top}^{2}}{r} - m g = \frac{0.20 \times 6. 0 ^{2}}{0.80} - 0.20 \times 9.8 = 9.0 - 1.96 = 7.04 N$ Why this step? Same top radial equation as Cell A, but now $v_{top}$ is large, so $\frac{m v ^{2}}{r}$ far exceeds $m g$ and the string is firmly taut ( $T > 0$ ) — the opposite of the slack case in Example 5.

Step 2 — Bottom speed by energy. $v_{bot}^{2} = v_{top}^{2} + 4 g r = 36.0 + 4 (9.8) (0.80) = 36.0 + 31.36 = 67.36 m^{2} / s^{2}$ Why this step? The same per-unit-mass energy link as Example 2 — the fixed $4 g r$ boost — carries the fast top speed down to the bottom.

Step 3 — Tension at the bottom. $T_{bot} = \frac{m v _{bot}^{2}}{r} + m g = \frac{0.20 \times 67.36}{0.80} + 1.96 = 16.84 + 1.96 = 18.80 N$ Why this step? Bottom radial equation from Cell B; substituting the larger $v_{bot}^{2}$ shows the tension scales up with speed and never threatens to go slack.

Step 4 — Check the universal identity. $T_{bot} - T_{top} = 18.80 - 7.04 = 11.76 N = 6 m g$ Why this step? Confirms Example 8: even far from the minimum, the difference is locked at $6 m g$ — only the individual tensions grow.

Verify: (a) $\frac{0.2 \times 36}{0.8} - 0.2 \times 9.8 = 9.0 - 1.96 = 7.04 N$ ✓. (b) $\frac{0.2 \times 67.36}{0.8} + 1.96 = 16.84 + 1.96 = 18.80 N$ ✓. Difference $18.80 - 7.04 = 11.76 = 6 m g$ ✓ — identity holds, both tensions positive (never slack) ✓.

Recall

Recall Which cell tests "gravity is tangent, not central"?

Cell C — the side point (height $r$ ), where tension alone is inward and $T = 3 m g$ for the minimum loop.

Recall At

$θ$ from the bottom, what is the radial equation? $T - m g cos θ = \frac{m v ^{2}}{r}$ ; it gives the bottom/side/top laws at $θ = 0/90/18 0^{\circ}$ .

Recall Why does the too-slow ball NOT just make a smaller circle?

The string can't push, so $T = 0$ (slack) and the ball becomes a free-falling projectile inside the loop.

Recall Is the rod's minimum bottom speed larger or smaller than the string's?

Smaller: $2 g r$ vs $5 g r$ , because the rod only needs to reach the top, not stay taut.

Recall In the fast case (well above minimum), what stays constant and what grows?

Both individual tensions grow with speed, but the difference $T_{bot} - T_{top}$ stays locked at $6 m g$ .

Connections

Centripetal force and acceleration
Newton's Second Law — net force form
Conservation of mechanical energy
Tension and constraint forces
Normal force in circular tracks
Banked curves and horizontal circular motion
Free-body diagrams