Lesson 8 · 40 min
Energy on Curved Paths
On a curved path, energy tells you how fast the particle is going, but not how hard the track or the string must push or pull to bend its path. That takes Newton's second law along the normal. Put the two together and you can find the forces on a roller coaster, the tension in a pendulum's string, and the point where a particle leaves a smooth surface.
Learning objectives
- Combine conservation of energy (or the general form) with \(\sum F_n = mv^2/\rho\) to find normal forces and cord tensions.
- Find the smallest height or speed that keeps a particle on a loop, or a cord taut, at the top.
- Find where a particle leaves a smooth convex surface.
- Explain the difference between a cord or a one-sided track, which can only pull or push one way, and a rod or rails.
Two equations, two unknowns
Forces on a curved path
\[ T_1 + V_1 + \textstyle\sum\left(U_{1\to2}\right)_{nc} = T_2 + V_2 \qquad\text{and}\qquad \sum F_n = m\,\dfrac{v^2}{\rho} \]The energy equation gives \(v^2\) at the point of interest. The normal equation, with \(n\) pointing toward the center of curvature, then gives the normal force or tension there.
The normal force does no work, so it never appears in the energy equation. The energy equation never needs the acceleration, but the normal equation does: \(a_n = v^2/\rho\), which is why you need the speed first. On a circle of radius \(R\), \(\rho = R\).
The loop-the-loop
Example 8.1 — Minimum height, and the forces at the top and bottom
A small cart is released from rest at a height \(h\) above the bottom \(B\) of a smooth track with a vertical circular loop of radius \(R\). (a) Find the smallest \(h\) for which the cart stays on the track at the top of the loop, \(C\). (b) For \(h = 3R\), find the normal force on the cart at \(C\) and at \(B\) as multiples of its weight. (c) A \(70\ \text{kg}\) rider goes round a loop with \(R = 10\ \text{m}\) and \(h = 3R\). What normal force acts on the rider at the bottom?
Show solution
(a) At \(C\) the center of the loop is below the cart, so \(n\) points down, and both the weight and the track's push point down:
\[ \textstyle\sum F_n = m a_n: \qquad N_C + mg = m\,\dfrac{v_C^2}{R} \]The track can only push, \(N_C \ge 0\), so the cart needs \(v_C^2 \ge gR\). Energy from the start to \(C\), with the datum at \(B\):
\[ 0 + mgh = \tfrac12 mv_C^2 + mg(2R) \quad\Rightarrow\quad v_C^2 = 2g(h - 2R) \] \[ 2g(h - 2R) \ge gR \quad\Rightarrow\quad h_{\min} = \tfrac52 R \](b) With \(h = 3R\): \(v_C^2 = 2gR\), so \(N_C = m(2gR)/R - mg = mg\). At \(B\), \(v_B^2 = 2g(3R) = 6gR\), and \(n\) points up, toward the center, against the weight:
\[ N_B - mg = m\,\frac{v_B^2}{R} = 6mg \quad\Rightarrow\quad N_B = 7mg \](c) \(N_B = 7(70)(9.81) = 4.81\ \text{kN}\): seven times the rider's weight.
Pendulums and vertical circles
Example 8.2 — The tension in a pendulum's string
A bob of mass \(m\) on a string of length \(L\) is released from rest at an angle \(\theta_0\) from the vertical. Find the string's tension when the string makes an angle \(\theta\) with the vertical, and evaluate it at the bottom for a \(2\ \text{kg}\) bob released from horizontal.
Show solution
Energy (the string does no work), with the bob \(L(\cos\theta - \cos\theta_0)\) lower than at release:
\[ v^2 = 2gL(\cos\theta - \cos\theta_0) \]Normal direction (toward the pivot): the tension pulls toward it, and the weight has a component \(mg\cos\theta\) away from it:
\[ T_s - mg\cos\theta = m\,\frac{v^2}{L} \quad\Rightarrow\quad T_s = mg\left(3\cos\theta - 2\cos\theta_0\right) \]At the bottom, released from horizontal (\(\theta = 0\), \(\theta_0 = 90^\circ\)): \(T_s = 3mg = 3(2)(9.81) = 58.9\ \text{N}\), three times the bob's weight, whatever the length of the string.
| The particle is held by | It can | Condition at the top | Speed needed at the bottom |
|---|---|---|---|
| A string or cord | only pull | \(T \ge 0\): \(v_\text{top}^2 \ge gR\) | \(v_\text{bottom}^2 \ge 5gR\) |
| The inside of a track | only push outward | \(N \ge 0\): \(v_\text{top}^2 \ge gR\) | \(v_\text{bottom}^2 \ge 5gR\) |
| A rod, or rails it is locked to | push or pull | it only has to arrive: \(v_\text{top} \ge 0\) | \(v_\text{bottom}^2 \ge 4gR\) |
Leaving a smooth surface
On the outside of a curve, the surface can only push away from the center. As the particle speeds up going over, the push it needs to stay on the curve keeps shrinking, until at some point the surface would have to pull. There the particle leaves the surface and flies off as a projectile.
Example 8.3 — Where does it leave?
A particle starts from rest at the top of a smooth hemisphere of radius \(R\) and slides down. At what angle \(\theta\) from the vertical does it leave the surface?
Show solution
Energy: at angle \(\theta\) the particle has dropped \(R(1 - \cos\theta)\), so \(v^2 = 2gR(1 - \cos\theta)\).
Normal direction (toward the center, below the particle): \(mg\cos\theta - N = mv^2/R\). It leaves when \(N = 0\):
\[ mg\cos\theta = m\,\frac{2gR(1 - \cos\theta)}{R} \quad\Rightarrow\quad 3\cos\theta = 2 \quad\Rightarrow\quad \theta = \cos^{-1}\tfrac23 = 48.2^\circ \]It leaves at a height \(\tfrac23 R\) above the center, having dropped only a third of the radius. With a starting speed \(v_0\), the same steps give \(\cos\theta = \tfrac23 + v_0^2/(3gR)\): a particle with \(v_0^2 \ge gR\) leaves at once.
Check your understanding
Key takeaways
- On a curved path: energy for \(v^2\), then \(\sum F_n = mv^2/\rho\) with \(n\) toward the center of curvature for the force.
- A smooth loop needs \(h \ge \tfrac52 R\); then \(N_\text{top} = mg\) and \(N_\text{bottom} = 7mg\) for \(h = 3R\).
- "Just makes it" means the force at the top reaches zero, not the speed.
- Pendulum: \(T_s = mg(3\cos\theta - 2\cos\theta_0)\). Dome from rest: it leaves at \(\cos\theta = \tfrac23\).
- You have finished the lessons. Practise in the Practice Lab, explore your own tracks in the Energy Explorer, then try the Self-Check Quiz.