Roller Coaster Energy Reading Assignment
AP Physics C: Mechanics • Chapter 7: Energy & Circular Dynamics
The Vertical Loop: Physical Setup & Energy Conservation
Section 7.4
A roller coaster cart traversing a vertical circular loop represents one of the foundational systems in classical mechanics. Unlike horizontal circular motion where speed can remain constant, vertical loops subject the cart to a gravitational field that acts continuously in the downward direction. As the cart ascends, gravity decelerates it, trading kinetic energy for gravitational potential energy; as it descends, gravity accelerates it, replenishing kinetic energy.
Figure 7.1: Physical Geometry of the Roller Coaster Track & Loop Isolated System (\(W_{nc} = 0\))
Datum Reference Level: y = 0 (Lowest Point of Loop) Radius R R h y = 2R Release: v = 0 Bottom: vbottom Top: vtop Side: y = R
7.4.1 Work Done by Normal Force & Conservation of Energy
A critical question in AP Physics C is why the normal force (\(\vec{F}_N\)) does not alter the Total Mechanical Energy of the coaster car. At every infinitesimal displacement \(d\vec{r}\) along the track, the normal contact force is oriented strictly perpendicular to the surface. Calculating the infinitesimal work done by the track:
\[ dW_N = \vec{F}_N \cdot d\vec{r} = F_N \, |d\vec{r}| \cos(90^\circ) = 0 \]
Because the normal force does zero work throughout the entire ride, and assuming track friction and air resistance are negligible (\(W_{nc} = 0\)), the only force performing work is conservative gravity: \(W_{net} = W_g = -\Delta U_g\). By the Work-Energy Theorem, the Total Mechanical Energy \(E_{mech} = K + U_g\) remains strictly constant.
7.4.2 Exact Energy States at Key Track Locations
Let the lowest point of the loop (the track floor) be defined as the gravitational datum where elevation \(y = 0\) and potential energy \(U_g = 0\). We evaluate the mechanical energy at the three defining locations:
1. Release Hill (\(y = h\))
Released from rest (\(v_0 = 0\)):
\(E = mgh\)
100% potential energy. Sets total energy budget for the loop.
2. Bottom of Loop (\(y = 0\))
Lowest elevation (\(U_g = 0\)):
\(v_{bottom} = \sqrt{2gh}\)
All potential energy has converted into kinetic energy.
3. Top of Loop (\(y = 2R\))
Loop summit (\(U_g = 2mgR\)):
\(v_{top} = \sqrt{2g(h - 2R)}\)
Kinetic energy reduced by \(2mgR\). Requires \(h > 2R\).
The Invariant Velocity-Squared Difference: Subtracting the kinetic energy at the top from the bottom:
\(\frac{1}{2}m v_{bottom}^2 - \frac{1}{2}m v_{top}^2 = mg(2R) \implies v_{bottom}^2 - v_{top}^2 = 4gR\)
The quantity \(v_{bottom}^2 - v_{top}^2\) is a strict constant of the geometry. Regardless of release height, the square of the speed at the bottom will always exceed the square of the speed at the top by exactly \(4gR\).
Principles of Physics • AP Mechanics Edition Page 1 of 4
Chapter 7: Energy & Circular Dynamics
7.4.3 Forces and Free-Body Diagrams (Bottom vs. Top)
Radial Force Analysis
A common misconception is that "centripetal force" is an extra physical force acting on the coaster. In reality, centripetal force is merely a kinematic requirement: any mass moving in a circle of radius \(R\) at speed \(v\) must experience a net force toward the center equal to \(m v^2 / R\). In our system, this net radial force is created entirely by the vector interplay of gravity (\(\vec{F}_g\)) and track normal force (\(\vec{F}_N\)).
Figure 7.2: Vector Comparison of Forces at the Bottom and Top of the Loop
At the Bottom of the Loop (\(y = 0\)) + Radial Center FN, bottom Fg = mg
\[ \Sigma F_r = F_{N, bottom} - mg = m\frac{v_{bottom}^2}{R} \]
\(F_N\) and gravity oppose each other. \(F_N\) must overcome gravity and provide centripetal acceleration.
At the Top of the Loop (\(y = 2R\)) + Radial Center FN, top Fg = mg
\[ \Sigma F_r = F_{N, top} + mg = m\frac{v_{top}^2}{R} \]
\(F_N\) and gravity act in the same direction. Gravity actively assists the track in pulling the car downward toward the center.
Detailed Dynamics: At the Bottom of the Loop (\(y = 0\))
At the absolute bottom of the loop, the track surface curves underneath the car. The inward radial direction points straight up toward the center of curvature. The track pushes upward on the car with normal force \(F_{N, bottom}\), while gravity pulls downward with force \(mg\). Applying Newton's Second Law along the radial axis:
\[ F_{N, bottom} - mg = m\frac{v_{bottom}^2}{R} \implies F_{N, bottom} = m\left(g + \frac{v_{bottom}^2}{R}\right) \]
Apparent Weight: What a passenger physically "feels" is not gravity, but the normal contact force pressing against them. Because \(F_{N, bottom}\) must balance gravity plus provide the large centripetal acceleration demanded by the car's high speed at the bottom, riders feel dramatically heavier than normal.
Detailed Dynamics: At the Top of the Loop (\(y = 2R\))
At the apex of the loop, the cart is inverted beneath the rails. The center of curvature is directly below the cart, so the inward radial direction points straight down. Gravity pulls downward, and the rigid track can only push downward on the wheel assemblies:
\[ F_{N, top} + mg = m\frac{v_{top}^2}{R} \implies F_{N, top} = m\left(\frac{v_{top}^2}{R} - g\right) \]
The Role of Gravity: Because gravity already points toward the center, it contributes toward the required centripetal acceleration for free. The track only needs to supply whatever centripetal force gravity cannot cover. If the car is moving very fast, \(v_{top}^2/R > g\), and the track pushes down hard. But if the car slows down, the required track push diminishes.
Principles of Physics • AP Mechanics Edition Page 2 of 4
Chapter 7: Energy & Circular Dynamics
7.4.4 The Critical Velocity & Minimum Release Height (\(h_{\min} = 2.5R\))
Threshold Conditions
The Contact Boundary Condition (\(F_N \ge 0\))
On an idealized frictionless track without interlocking upstop safety wheels (such as a rolling marble or a classic coaster car resting on rails), the track surface can exert a compressive normal force but cannot exert a tensile force. The track can push the car, but it cannot pull it back if it attempts to separate. Therefore, the physical condition for the cart to successfully complete the circular path without falling off the rails is:
\(F_{N, top} \ge 0 \iff m\left(\frac{v_{top}^2}{R} - g\right) \ge 0 \implies \frac{v_{top}^2}{R} \ge g\)
Step 1: Critical Speed at the Top (\(v_{crit}\))
At impending separation, normal force drops to zero: \(F_{N, top} = 0\). Gravity alone supplies 100% of the centripetal acceleration:
\[ \Sigma F_r = mg = m\frac{v_{top, \min}^2}{R} \] \[ v_{top, \min} = \sqrt{gR} \]
Physical Sensation: At this speed, the car is in momentary free-fall at the peak, rendering passengers completely weightless (\(0\text{ g}\)). If \(v_{top} < \sqrt{gR}\), the car enters an inward parabolic projectile fall.
Step 2: Connecting to Release Height (\(h_{\min}\))
Apply Conservation of Energy from the release hill (\(y = h\)) to the loop summit (\(y = 2R\)) at critical speed:
\[ mgh_{\min} = mg(2R) + \frac{1}{2}m(\sqrt{gR})^2 \] \[ mgh_{\min} = 2mgR + 0.5mgR = 2.5mgR \] \[ h_{\min} = 2.5R = \frac{5}{2}R \]
Mass Independence: Mass \(m\) cancels entirely. The minimum release height is strictly 2.5 times the radius (or 1.25 times the loop diameter) for any point-mass cart.
The "6 g" Force Paradox at the Bottom of the Loop:
A critical engineering consequence emerges when evaluating the bottom of the loop for a cart released from this threshold height \(h = 2.5R\):
Speed at Bottom:
\(v_{bottom}^2 = 2g(2.5R) = 5gR\)
Normal Force at Bottom:
\(F_{N, bottom} = m\left(g + \frac{5gR}{R}\right) = 6mg\)
Engineering Reality: In order for the coaster to merely graze the top of the loop with zero contact force (\(0\text{ g}\)), the rider must survive a massive 6 g's of compressive apparent weight at the track entrance! This explains why real amusement park coasters avoid simple circular loops and use specialized geometry to mitigate passenger blackout risk.
Principles of Physics • AP Mechanics Edition Page 3 of 4
Chapter 7: Energy & Circular Dynamics
7.4.5 Acceleration Decomposition: Centripetal vs. Translational
2D Kinematics
In uniform circular motion, speed is constant, meaning the acceleration is purely centripetal. But a vertical loop is non-uniform circular motion: the velocity vector changes in both direction and magnitude simultaneously. Newton's Second Law resolves the net acceleration \(\vec{a}_{total}\) into two orthogonal vectors:
Centripetal (Radial) Accel. (\(\vec{a}_c\))
Changes velocity direction. Points perpendicular to the track toward the center:
\(a_c = \frac{v^2}{R} \quad \text{(Radially Inward)}\)
Translational (Tangential) Accel. (\(\vec{a}_t\))
Changes velocity magnitude (speed). Points parallel to the track:
\(a_t = \frac{dv}{dt} \quad \text{(Along Track Tangent)}\)
Figure 7.3: Resolving Total Acceleration \(\vec{a}_{total} = \vec{a}_c + \vec{a}_t\) Across the Loop Vector Triangles
Center ac ↑ Bottom: at = 0 • atotal = ac (Straight UP) ac ↓ Top: at = 0 • atotal = ac (Straight DOWN) ac at = g ↓ atotal Ascending (y = R) Slowing Down ac at = g ↓ atotal Descending (y = R) Speeding Up
| Track Position |
Elevation \(y\) |
Centripetal (\(a_c\)) |
Translational (\(a_t\)) |
Total Accel. (\(\vec{a}\)) Direction |
| Bottom of Loop |
\(0\) |
\(v_{bot}^2 / R\) (Max) |
\(0\) |
Straight UP (Purely Radial) |
| Ascending Side |
\(R\) |
\(v_{mid}^2 / R\) (Inward) |
\(g\) (Down) |
Inward & Downward |
| Top of Loop |
\(2R\) |
\(v_{top}^2 / R\) (Min) |
\(0\) |
Straight DOWN (Purely Radial) |
| Descending Side |
\(R\) |
\(v_{mid}^2 / R\) (Inward) |
\(g\) (Down) |
Inward & Downward |
Chapter 7 Complete Physical Synthesis
- Zero Work by Track: Because \(\vec{F}_N \perp d\vec{r}\), the track does no work; gravity alone exchanges energy: \(v_{bottom}^2 - v_{top}^2 = 4gR\).
- Bottom Dynamics: Normal force must exceed gravity to provide upward centripetal acceleration: \(F_{N, bottom} = m(g + v^2/R)\).
- Top Dynamics: Normal force and gravity both pull downward toward the center: \(F_{N, top} = m(v^2/R - g)\).
- Critical Apex Condition: Detachment threshold \(F_{N, top} = 0\) requires critical speed \(v_{top} = \sqrt{gR}\) and minimum release height \(h_{\min} = 2.5R\).
- Acceleration Directions: At the bottom and top, gravity is perpendicular to the track, so translational acceleration \(a_t = 0\). At mid-elevation (\(y = R\)), gravity is parallel to the track, generating \(a_t = g\) downward while \(a_c\) pulls horizontally inward.
Principles of Physics • AP Mechanics Edition Page 4 of 4
Friction Loop Dynamics Reading
AP Physics C: Mechanics • Chapter 7: Non-Conservative Dynamics
Vertical Loops with Track Friction & Thermal Dissipation
Section 7.5
In real-world roller coaster engineering, the idealized assumption of zero mechanical energy dissipation cannot hold. Contact friction between the wheel assemblies and the steel tubular rails, alongside wheel rotational resistance, generates a continuous kinetic friction force \(\vec{f}_k\). Because friction always opposes instantaneous displacement, it converts organized mechanical energy into disorganized internal thermal energy (\(E_{th}\)), fundamentally modifying the velocity profile and safety thresholds of the loop.
Figure 7.4: Energy Dissipation along a Dissipative Vertical Loop Non-Conservative System (\(W_{nc} = W_f < 0\))
Datum Reference Level: y = 0 h' y = 2R fk fk ↓ fk fk →
7.5.1 The Work-Energy Theorem for Non-Conservative Systems
When friction is present, the net work on the coaster car equals the work of conservative gravity plus the work of non-conservative friction: \(W_{net} = W_g + W_f\). Because \(W_{net} = \Delta K\) and \(W_g = -\Delta U_g\), rearranging gives the universal AP formulation:
\[ W_{nc} = W_f = \Delta E_{mech} = E_{final} - E_{initial} \iff E_{initial} + W_f = E_{final} \]
At every instant, the kinetic friction force is oriented exactly antiparallel to the cart's velocity vector: \(\vec{f}_k \cdot d\vec{r} = f_k \, |d\vec{r}| \cos(180^\circ) = -f_k \, ds\). Therefore, the work done by friction is strictly negative:
Total Dissipated Thermal Energy: \[ W_f = -\int_{release}^{apex} f_k \, ds = -\Delta E_{th} < 0 \] Mechanical energy is not conserved; it is monotonically drained along the entire travel path.
7.5.2 Modified Energy Balance between Release Hill & Loop Apex
Evaluating the energy accounting from release at height \(h'\) (released from rest, \(K_0 = 0\)) to the summit of the loop (\(y = 2R\)):
Energy at Release:
\(E_{initial} = mgh'\)
Initial gravitational potential energy.
Energy at Apex (\(y = 2R\)):
\(E_{final} = mg(2R) + \frac{1}{2}m v_{top}^2\)
Potential energy plus residual kinetic energy.
Master Energy Equation with Dissipation:
\[ mgh' - |W_f| = 2mgR + \frac{1}{2}m v_{top}^2 \]
Because \(|W_f| > 0\), achieving any specified velocity \(v_{top}\) requires a strictly higher release hill than in the frictionless ideal case.
Principles of Physics • AP Mechanics Edition Page 1 of 3
Chapter 7: Non-Conservative Dynamics
7.5.3 Free-Body Diagrams with Friction Across the Loop
2D Force Decomposition
When friction is introduced, three physical forces act simultaneously on the car: downward gravity (\(\vec{F}_g\)), inward track normal force (\(\vec{F}_N\)), and tangential friction (\(\vec{f}_k\)). Crucially, friction acts strictly parallel to the track, meaning it modifies the tangential acceleration but never provides radial force.
Figure 7.5: Free-Body Diagrams Including Tangential Friction Vectors
Bottom (\(y = 0\), Moving Right) FN ↑ Fg ← fk
Radial: \(F_N - mg = m\frac{v^2}{R}\) • Tangential: \(\Sigma F_t = -f_k = m a_t\)
Ascending Wall (\(y = R\), Moving Up) ← FN Fg fk
Radial: \(F_N = m\frac{v^2}{R}\) • Tangential: \(\Sigma F_t = -(mg + f_k) = m a_t\)
Apex Top (\(y = 2R\), Moving Left) FN ↓ Fg fk →
Radial: \(F_N + mg = m\frac{v^2}{R}\) • Tangential: \(\Sigma F_t = -f_k = m a_t\)
Descending Wall (\(y = R\), Moving Down) FN → Fg ↑ fk
Radial: \(F_N = m\frac{v^2}{R}\) • Tangential: \(\Sigma F_t = mg - f_k = m a_t\)
Physical Insights: Ascent vs. Descent Asymmetry
Notice the striking physical asymmetry between the ascending wall and descending wall:
During Ascent (Climbing):
Gravity and friction act in the same downward direction. Both forces act to decelerate the cart: \(a_t = -(g + f_k/m)\). The cart loses speed at an intensified rate.
During Descent (Dropping):
Gravity acts downward to accelerate the cart, but friction acts upward to oppose motion: \(a_t = g - f_k/m\). The cart speeds up significantly slower than in free-fall.
Principles of Physics • AP Mechanics Edition Page 2 of 3
Chapter 7: Non-Conservative Dynamics
7.5.4 Critical Apex Speed & The Increased Release Height (\(h_{\min}'\))
Threshold Derivations
Why Does Critical Speed at the Apex Remain Invariant (\(v_{top, \min} = \sqrt{gR}\))?
A common student mistake is assuming friction changes the critical speed at the top of the loop. It does not! At the exact summit (\(y = 2R\)), the track is horizontal. The friction force \(\vec{f}_k\) acts purely horizontally (tangentially), having zero vertical/radial component. The non-detachment constraint remains strictly radial:
\[ \Sigma F_r = F_{N, top} + mg = m\frac{v_{top}^2}{R} \implies \text{Setting } F_{N, top} = 0 \implies v_{top, \min} = \sqrt{gR} \]
The threshold speed to prevent detachment depends solely on the local radius of curvature \(R\) and gravity \(g\). Friction does not alter the required velocity—it only robs the car of the energy needed to attain it!
Derivation of the Modified Minimum Release Height (\(h_{\min}'\))
Applying the Work-Energy Theorem with non-conservative thermal dissipation \(|W_f|\) between release from rest at height \(h_{\min}'\) and loop crest at critical speed \(v_{top} = \sqrt{gR}\):
\[ E_{release} - |W_f| = E_{apex} \] \[ mgh_{\min}' - |W_f| = mg(2R) + \frac{1}{2}m\left(\sqrt{gR}\right)^2 \] \[ mgh_{\min}' - |W_f| = 2mgR + 0.5mgR = 2.5mgR \] \[ h_{\min}' = 2.5R + \frac{|W_f|}{mg} \]
Physical Meaning: The coaster must be released from the theoretical frictionless height (\(2.5R\)) plus an additional elevation equal to the equivalent potential energy of the dissipated friction work: \(\Delta h = |W_f| / (mg)\).
The Danger of the Frictionless Height (\(h = 2.5R\)):
If a maintenance engineer releases a real cart from the theoretical frictionless minimum of \(h = 2.5R\):
\[ E_{top} = mg(2.5R) - |W_f| < 2.5mgR \implies v_{top} < \sqrt{gR} \]
The car runs out of sufficient kinetic energy before reaching the peak. As the car climbs into the upper hemisphere, normal force drops to zero prematurely (\(F_N = 0\) while \(y < 2R\)). The wheels lose contact with the rails, and the car falls inward along a catastrophic parabolic trajectory!
| Track Location |
Centripetal Accel. (\(a_c\)) |
Translational Accel. (\(a_t\)) |
Total Acceleration Vector (\(\vec{a}\)) |
| Bottom (\(y = 0\)) |
\(v^2 / R\) (Inward/Up) |
\(-f_k / m\) (Opposing \(\vec{v}\)) |
Tilted slightly backward from vertical |
| Ascending Wall (\(y = R\)) |
\(v^2 / R\) (Inward/Horizontal) |
\(-(g + f_k/m)\) (Down) |
Points down and inward |
| Apex Top (\(y = 2R\)) |
\(v^2 / R\) (Inward/Down) |
\(-f_k / m\) (Opposing \(\vec{v}\)) |
Tilted slightly backward from vertical |
| Descending Wall (\(y = R\)) |
\(v^2 / R\) (Inward/Horizontal) |
\(g - f_k/m\) (Down) |
Points down and inward |
Principles of Physics • AP Mechanics Edition Page 3 of 3
