Analyse Forces at Top and Bottom
- Calculate tension at the top and bottom of a vertical circle
- Apply Newton's second law with centripetal acceleration
Determine Minimum Speed
- Derive and apply the minimum speed to complete a vertical loop
- Calculate minimum entry speed at the bottom using energy conservation
Apply Conservation of Energy
- Use mechanical energy conservation for non-uniform vertical circular motion
- Find speed at any point in a vertical circle from initial conditions
On a roller coaster loop, where do you feel heaviest — at the top or the bottom? Why?
Vertical Circular Motion
Analyse objects moving in vertical circles. Calculate tension at the top and bottom of the loop, determine minimum speed to maintain contact, and use energy conservation for non-uniform vertical circular motion.
Forces at Top and Bottom
Applying Newton's second law to vertical circular motion
Vertical Circular Motion
Vertical Circle Detailed
In vertical circular motion, the speed is not constant — gravity speeds the object up on the way down and slows it on the way up. The forces providing centripetal acceleration also change. We analyse the top and bottom positions separately.
At the Top of the Circle
Both tension (or normal force) and weight point downward, toward the centre of the circle. Taking toward the centre as positive:
Forces at the Top
At the top, tension is minimum. If the speed is too low, tension becomes zero and the object falls away from the circular path. The minimum speed at the top occurs when $T = 0$:
Minimum Speed at Top
If $v < \sqrt{rg}$ at the top, the object cannot follow the circular path — it falls away (the string goes slack, or the vehicle leaves the track).
At the Bottom of the Circle
Tension points upward (toward the centre) and weight points downward (away from the centre). The net force toward the centre provides the centripetal acceleration:
Forces at the Bottom
Tension is always greater at the bottom than at the top for the same speed because the weight now opposes the centripetal force rather than assisting it. The difference is $T_{\text{bottom}} - T_{\text{top}} = 2mg$ when the speed is the same at both points.
At the top, gravity helps provide centripetal force, so tension can be smaller (or even zero). At the bottom, tension must overcome weight and provide centripetal force, so it is always larger. This is why you feel heaviest at the bottom of a roller coaster loop.
Worked Example — Tension at Top and Bottom
A 0.5 kg bob on a 0.8 m string is whirled in a vertical circle. At the top of the circle, its speed is 4 m/s. Find the tension in the string at the top and at the bottom.
$m = 0.5 \text{ kg}$, $r = 0.8 \text{ m}$, $v_{\text{top}} = 4 \text{ m/s}$
$T_{\text{top}}$ and $T_{\text{bottom}}$
First find tension at top using $T = \dfrac{mv^2}{r} - mg$. Then use energy conservation to find speed at bottom, and calculate tension at bottom.
Tension at top:
$T_{\text{top}} = \dfrac{0.5 \times (4)^2}{0.8} - (0.5 \times 9.8)$
$T_{\text{top}} = \dfrac{0.5 \times 16}{0.8} - 4.9 = \dfrac{8}{0.8} - 4.9 = 10 - 4.9$
$T_{\text{top}} = 5.1 \text{ N}$
Speed at bottom (energy conservation):
$\dfrac{1}{2}mv_{\text{bottom}}^2 = \dfrac{1}{2}mv_{\text{top}}^2 + mg(2r)$
$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr = 16 + 4 \times 9.8 \times 0.8 = 16 + 31.36 = 47.36$
$v_{\text{bottom}} = \sqrt{47.36} = 6.88 \text{ m/s}$
Tension at bottom:
$T_{\text{bottom}} = \dfrac{0.5 \times 47.36}{0.8} + (0.5 \times 9.8)$
$T_{\text{bottom}} = \dfrac{23.68}{0.8} + 4.9 = 29.6 + 4.9$
$T_{\text{bottom}} = 34.5 \text{ N}$
Minimum Speed to Complete the Loop
Finding the critical speed to maintain circular motion
For an object to complete a vertical circle, it must maintain contact with the track (or keep the string taut) at all points. The critical position is the top of the loop, where the normal force or tension is smallest.
Condition at the Top
At the top of the loop, the object is just about to lose contact when the normal force (or tension) becomes zero. At this point, gravity alone provides the centripetal force:
Critical Condition at Top
Relating Bottom Speed to Top Speed
Using conservation of mechanical energy, with the bottom of the loop as the reference level ($h = 0$):
$\dfrac{1}{2}mv_{\text{bottom}}^2 = \dfrac{1}{2}mv_{\text{top}}^2 + mg(2r)$
Substituting $v_{\text{top}}^2 = rg$ for the minimum case:
$v_{\text{bottom}}^2 = rg + 4gr = 5gr$
Minimum Speed at Bottom
This is a key result: to just complete a vertical loop of radius $r$, the speed at the bottom must be $\sqrt{5}$ times the minimum speed at the top. The factor of 5 arises because the object must gain enough kinetic energy to reach the top (converting $2mgr$ of kinetic energy to gravitational potential energy) and still have enough speed at the top to maintain contact ($\tfrac{1}{2}mrg$ of kinetic energy).
At minimum speed, the total energy at the bottom is $\tfrac{1}{2}m(5gr) = 2.5mgr$. At the top, the kinetic energy is $\tfrac{1}{2}m(rg) = 0.5mgr$ and the potential energy is $mg(2r) = 2mgr$. The sum is $0.5mgr + 2mgr = 2.5mgr$ — energy is conserved.
Worked Example — Loop-the-Loop Entry Speed
A toy car must complete a vertical loop-the-loop of radius 2.5 m. What is the minimum speed the car must have at the bottom of the loop?
$r = 2.5 \text{ m}$, $g = 9.8 \text{ m/s}^2$
$v_{\text{min(bottom)}}$
Use $v_{\text{min(bottom)}} = \sqrt{5gr}$
$v_{\text{min(bottom)}} = \sqrt{5 \times 9.8 \times 2.5}$
$v_{\text{min(bottom)}} = \sqrt{122.5}$
$v_{\text{min(bottom)}} = 11.1 \text{ m/s}$
Energy Method for Non-Uniform Motion
Using conservation of mechanical energy to find speed at any point
Unlike uniform circular motion, vertical circular motion has a speed that changes continuously as the object rises and falls. The speed is maximum at the bottom and minimum at the top (assuming no external driving force). We use conservation of mechanical energy to relate speeds at different heights.
Setting Up the Energy Equation
Taking the bottom of the circle as the reference level ($h = 0$), mechanical energy is conserved:
Energy Conservation (No Friction)
At the bottom of the loop: $h = 0$, so $PE = 0$ and $KE = \tfrac{1}{2}mv_{\text{bottom}}^2$
At the top of the loop: $h = 2r$, so $PE = mg(2r)$ and $KE = \tfrac{1}{2}mv_{\text{top}}^2$
For a roller coaster entering a loop at known speed, we can find the speed at any height using energy conservation, then calculate the normal force at that point using $N - mg\cos\theta = \dfrac{mv^2}{r}$ (for points at angle $\theta$ from the bottom).
Worked Example — Roller Coaster in a Loop
A roller coaster car of mass 300 kg starts from rest at a height of 15 m above the ground and enters a vertical loop of radius 4 m. Find (a) the speed of the car at the top of the loop, and (b) the normal force on the car at the top of the loop.
$m = 300 \text{ kg}$, $h_{\text{start}} = 15 \text{ m}$, $r = 4 \text{ m}$, $v_{\text{start}} = 0$
Height at top of loop: $h_{\text{top}} = 15 - 2r = 15 - 8 = 7 \text{ m}$ above the bottom of the loop (or $h_{\text{drop}} = 15 - 8 = 7 \text{ m}$ from start)
(a) $v_{\text{top}}$ (b) $N_{\text{top}}$
Use energy conservation to find speed at top, then apply Newton's second law at the top: $N + mg = \dfrac{mv^2}{r}$
(a) Speed at top:
$mgh_{\text{start}} = \dfrac{1}{2}mv_{\text{top}}^2 + mgh_{\text{top}}$
$300 \times 9.8 \times 15 = \dfrac{1}{2} \times 300 \times v_{\text{top}}^2 + 300 \times 9.8 \times 7$
$44100 = 150v_{\text{top}}^2 + 20580$
$150v_{\text{top}}^2 = 23520$
$v_{\text{top}}^2 = 156.8$
$v_{\text{top}} = 12.5 \text{ m/s}$
(b) Normal force at top:
$N + mg = \dfrac{mv^2}{r}$
$N = \dfrac{300 \times 156.8}{4} - (300 \times 9.8)$
$N = 11760 - 2940$
$N = 8820 \text{ N}$
Essential Formulae — Vertical Circular Motion
Common Misconceptions
"The normal force is always greater at the top" — No. The normal force (or tension) is always greater at the bottom than at the top. The difference is $T_{\text{bottom}} - T_{\text{top}} = 2mg$ for the same speed, and in practice the speed at the bottom is also greater, making the disparity even larger.
"You feel weightless at the bottom of a loop" — No. You feel heaviest at the bottom because the seat must push up with force $N = \dfrac{mv^2}{r} + mg$, which is greater than your weight alone. Weightlessness occurs at the top when $v = \sqrt{rg}$.
The Clothoid Loop
The loop on a modern roller coaster is not a perfect circle — it is a clothoid shape (also called an Euler spiral). This design reduces dangerous g-force variation and improves rider experience.
In a circular loop with $v = 15 \text{ m/s}$ and $r = 5 \text{ m}$ at the bottom, the centripetal acceleration would be:
$a_c = \dfrac{v^2}{r} = \dfrac{(15)^2}{5} = \dfrac{225}{5} = 45 \text{ m/s}^2 \approx 4.6g$
The rider feels $N = ma_c + mg = m(4.6g + g) = 5.6g$. Most people experience greyout above $5g$, which is why clothoid loops are used — they have a larger radius at the bottom (reducing peak g-force) and a smaller radius at the top (ensuring sufficient speed to maintain contact). The continuously varying radius keeps g-forces in a comfortable range of approximately 2–3.5g throughout the loop.
Tension at the Top
1 A 0.3 kg ball on a 0.6 m string moves in a vertical circle. At the top, its speed is 3.5 m/s. Find the tension in the string at the top.
Minimum Entry Speed
2 Find the minimum speed required at the bottom of a vertical loop of radius 3.0 m to complete the loop.
Roller Coaster Forces
3 A 500 kg roller coaster car enters a vertical loop of radius 5.0 m at a speed of 12 m/s. Calculate the normal force on the car at the bottom of the loop. Determine whether the car has sufficient speed to reach the top of the loop.
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▾Key Definitions
- Vertical circular motion: motion in a circle where gravity causes speed to vary with height
- Minimum speed at top: $v_{\text{min}} = \sqrt{rg}$ — speed where tension/normal force becomes zero
- Clothoid loop: a non-circular loop shape with varying radius to control g-forces
Force Equations
- Top: $T + mg = \dfrac{mv^2}{r}$
- Top rearranged: $T = \dfrac{mv^2}{r} - mg$
- Bottom: $T - mg = \dfrac{mv^2}{r}$
- Bottom rearranged: $T = \dfrac{mv^2}{r} + mg$
Energy Conservation
- $\dfrac{1}{2}mv_A^2 + mgh_A = \dfrac{1}{2}mv_B^2 + mgh_B$
- $v_{\text{min(bottom)}} = \sqrt{5gr}$
- $v_{\text{min(top)}} = \sqrt{rg}$
Key Principles
- Tension is always greater at the bottom than at the top
- Minimum speed at top requires $T = 0$
- You feel heaviest at the bottom of a loop
- Modern coasters use clothoid loops to reduce peak g-force
Revisit Your Prediction
You predicted where you would feel heaviest on a roller coaster loop. Now you know: at the bottom of the loop, the normal force is $N = \dfrac{mv^2}{r} + mg$, which is greater than your weight. At the top, $N = \dfrac{mv^2}{r} - mg$, which can even be zero at the minimum speed. You feel heaviest at the bottom and lightest at the top.
Has your understanding changed? Write a revised explanation:
1. At the top of a vertical circle, tension is minimum when:
2. For a roller coaster at the bottom of a loop:
3. To just complete a vertical loop of radius $r$, the minimum speed at the bottom must be:
4. A bucket of water is whirled in a vertical circle. The water stays in at the top because:
5. Comparing top and bottom of a vertical loop at the same speed $v$:
A 0.4 kg stone on a 0.5 m string is whirled in a vertical circle. At the top, the tension is 3.2 N. Calculate the speed of the stone at the top of the circle.
A roller coaster car of mass 250 kg approaches a vertical loop of radius 6.0 m. Calculate the minimum speed required at the bottom of the loop for the car to complete the loop. Calculate the normal force on the car at the bottom at this minimum speed.
Evaluate the design of a circular roller coaster loop versus a clothoid-shaped loop. Use physics principles to explain why modern roller coasters use non-circular loops and the effect on rider experience.
Model Answers
▾Question 1 (2 marks)
At the top: $T + mg = \dfrac{mv^2}{r}$
$3.2 + (0.4 \times 9.8) = \dfrac{0.4 \times v^2}{0.5}$
$3.2 + 3.92 = \dfrac{0.4v^2}{0.5}$
$7.12 = 0.8v^2$
$v^2 = 8.9$
$v = 2.98 \text{ m/s}$ (2 marks — 1 for method, 1 for answer)
Question 2 (3 marks)
Minimum speed at bottom: $v_{\text{min(bottom)}} = \sqrt{5gr}$
$v_{\text{min(bottom)}} = \sqrt{5 \times 9.8 \times 6} = \sqrt{294} = 17.1 \text{ m/s}$ (1 mark)
At the bottom: $N - mg = \dfrac{mv^2}{r}$
$N = \dfrac{250 \times 294}{6} + (250 \times 9.8)$ (1 mark for method)
$N = 12250 + 2450 = 14700 \text{ N}$ (1 mark)
This equals $6mg$, which is the normal force at the bottom for minimum entry speed.
Question 3 (4 marks)
In a circular loop, $a_c = \dfrac{v^2}{r}$ depends on both speed and radius (1 mark). For vertical circular motion, speed varies with height due to energy conservation — the speed is maximum at the bottom and minimum at the top (1 mark).
At the bottom of a circular loop, the speed is highest and the fixed radius would produce dangerously high centripetal acceleration. At the top, the speed is lowest and the same radius may not provide sufficient centripetal acceleration to maintain contact (1 mark).
A clothoid loop has continuously varying radius — large at the bottom (reducing $a_c = v^2/r$ and peak g-force) and small at the top (ensuring sufficient $a_c$ even at reduced speed). This reduces maximum g-force from approximately 6g (circular) to approximately 3.5g (clothoid), making the ride safer and more comfortable for riders (1 mark).
Mark Lesson 9 Complete
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