Grade by Grade
C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

Conservation of Mechanical Energy: Tracks, Pendulums and Loops

C2.3 C2.2About 75 minutes
Learning goals
  • I can explain why mechanical energy is conserved when only conservative forces do work.
  • I can use Ek₁ + Eg₁ = Ek₂ + Eg₂ to find speeds and heights on curved paths and for pendulums.
  • I can find the height of a pendulum bob from its length and angle.
  • I can combine energy conservation with circular motion to solve loop problems.

Words to know

mechanical energy
The sum of an object's kinetic and potential energies: Em = Ek + Eg (+ Ee).
law of conservation of mechanical energy
If only conservative forces do work on a system, its mechanical energy stays constant.
pendulum
A mass (the bob) on a string or rod that swings about a fixed pivot.
centripetal force
The net force toward the centre needed for circular motion, of size mv²/R (from the unit Dynamics: Forces and Motion in a Plane).

The roller coaster's secret

The first hill of a roller coaster is usually the tallest. A chain or a launch system does work to get the train up there; after that, the train runs on its own, trading height for speed and speed for height. If there were no friction or air resistance, it could keep doing that forever. That trade-off is the law of conservation of mechanical energy, and in Grade 12 you will use it for curved tracks, loops, pendulums and half-pipes.

Remember from Grade 11: SPH3U used Ek₁ + Eg₁ = Ek₂ + Eg₂ for falling objects, pendulums and roller coasters. This lesson shows why it works, when it applies and how to combine it with circular motion from the sibling unit Dynamics: Forces and Motion in a Plane.

Mechanical energy and why it is conserved

The mechanical energy of an object is the sum of its kinetic and potential energies:

Em = Ek + Eg (later in this unit you will add elastic potential energy, Ee)

Suppose only gravity does work on the object (other forces may act, but they do no work). Then:

  1. Work–energy theorem: W_net = ΔEk.
  2. Here W_net = W_g, and W_g = −ΔEg.
  3. So ΔEk = −ΔEg, which means ΔEk + ΔEg = 0.

When only conservative forces do work, mechanical energy is conserved:

Ek₁ + Eg₁ = Ek₂ + Eg₂, or ½mv₁² + mgh₁ = ½mv₂² + mgh₂

Forces that act perpendicular to the motion do no work, so they do not spoil conservation. The normal force of a smooth track and the tension in a pendulum string are always perpendicular to the motion. That is why a frictionless roller coaster on any curved track, and a pendulum, both conserve mechanical energy.

A (40 m, 3.0 m/s)C (25 m, 17 m/s)B (0 m, 28 m/s)ABCEgEksame total Em at A, B and C
A frictionless coaster (values per kilogram of train). At A it is almost all Eg, at B it is all Ek, and at C it is a mix, but the total height of each bar, the mechanical energy, never changes.
Worked example

Example 1: Speeds on a curved track

A roller coaster train passes point A, the top of a 40 m hill, at 3.0 m/s. Ignoring friction, find its speed at B (ground level) and at C (the top of a 25 m hill).

  1. Choose the ground as the reference level. The mass cancels: ½v₁² + gh₁ = ½v₂² + gh₂, so v₂² = v₁² + 2g(h₁ − h₂).
  2. At B: v² = 3.0² + 2(9.8)(40 − 0) = 9 + 784 = 793, so v_B = 28 m/s (28.2 m/s).
  3. At C: v² = 9 + 2(9.8)(40 − 25) = 9 + 294 = 303, so v_C = 17 m/s (17.4 m/s).
  4. The shape of the track between points does not matter, only the heights.

v_B ≈ 28 m/s, v_C ≈ 17 m/s.

θLL cos θhHeight of release above the lowest point:h = L − L cos θ = L(1 − cos θ)Speed at the bottom (no friction):mgh = ½mv², so v = √(2gh)
A pendulum of length L pulled back to angle θ is raised a height h = L(1 − cos θ) above its lowest point. The string's tension is always perpendicular to the motion, so it does no work.
Worked example

Example 2: A playground swing (a pendulum)

A swing with 1.5 m ropes is pulled back to 35° from the vertical and released from rest. How fast is the child moving at the bottom?

  1. Height of release above the bottom: h = L(1 − cos θ) = 1.5(1 − cos 35°) = 1.5(1 − 0.8192) = 0.271 m.
  2. Mechanical energy is conserved (tension does no work; ignore air resistance): mgh = ½mv².
  3. v = √(2gh) = √(2 × 9.8 × 0.271) = 2.3 m/s.

About 2.3 m/s, whatever the child's mass.

Worked example

Example 3: A half-pipe

A snowboarder drops into a half-pipe at 1.0 m/s from the lip, 4.5 m above the bottom. Ignore friction. (a) How fast is she going at the bottom? (b) If she rides up the other wall and leaves the lip moving straight up, how high above the lip does she rise?

  1. (a) v² = 1.0² + 2(9.8)(4.5) = 1 + 88.2 = 89.2, so v = 9.4 m/s.
  2. (b) She leaves the far lip (same height as the start) at the same speed she started with, 1.0 m/s.
  3. At the top of her flight v = 0: ½v² = gΔh, Δh = (1.0)² ÷ (2 × 9.8) = 0.051 m, about 5 cm above the lip.

(a) 9.4 m/s; (b) about 5 cm. To go higher, a rider must add energy by pumping (doing work with her legs).

Loops: energy meets circular motion

In the sibling unit Dynamics: Forces and Motion in a Plane you learned that an object moving in a circle of radius R at speed v needs a net force of mv²/R toward the centre. At the top of a vertical loop, both gravity and the track's normal force point down, toward the centre:

mg + F_N = mv²/R

The track can push but not pull, so F_N ≥ 0. The slowest the car can go and still stay on the track is when F_N = 0, which gives v_top² = gR. Energy conservation then tells you how high the car must start.

start at resth = 2.5RRmgAt the top, at the minimum speed:F_N = 0, so mg = mv²/Rv_top² = gR
To get around a frictionless loop of radius R, a car starting from rest must begin at least 2.5R above the bottom. At the top it then has just enough speed that gravity alone supplies the centripetal force.
Worked example

Example 4: How high must the hill be?

A toy car starts from rest and coasts down a frictionless track into a vertical loop of radius R = 10 cm. What is the minimum starting height above the bottom of the loop?

  1. Condition at the top of the loop: v_top² = gR.
  2. Energy from the start (height h, at rest) to the top of the loop (height 2R): mgh = ½mv_top² + mg(2R).
  3. Substitute v_top² = gR: mgh = ½mgR + 2mgR = 2.5mgR, so h = 2.5R.
  4. For R = 10 cm: h = 25 cm.

At least 25 cm (2.5R). Real cars need more, because friction and rolling take energy.

Just for fun

Riddle: why did the roller coaster never get a big head?

Because after every high point, it remembered where it came from: all the way down.

Hands-on activity

Swing a pendulum and predict its speed

You need: a string about 1 m long, a heavy small object (a nut, a washer stack, or a small bag of coins taped shut), tape, a protractor, a ruler, and a phone that records slow-motion video. Safety: keep the swing clear of faces and breakables.

  1. Tape the string firmly to the top of a doorframe (or the edge of a high shelf) so the bob swings freely. Measure L from the pivot to the centre of the bob.
  2. Tape a ruler horizontally to a box or chair placed just behind the lowest point of the swing, at the height of the bob, as a scale.
  3. Pull the bob back to 30° (check with the protractor) and release it from rest. Film the bottom of the swing in slow motion.
  4. Step through the video frames. Measure how far the bob moves along the ruler between two frames and divide by the time between frames to estimate its speed.
  5. Predict the speed with v = √(2gL(1 − cos θ)) and compare.

What to look for: for L = 1.0 m and 30°, the prediction is about 1.6 m/s. Your measurement should be close; it will be a little lower if air resistance matters. Try 45° and 60° as well.

Your turn

1

A ball is dropped from rest 12 m above the ground. Ignoring air resistance, how fast is it moving when it lands?

2

A ball is thrown straight up at 14 m/s. How high does it rise?

3

A coaster car moves at 5.0 m/s at a height of 30 m. Find its speed at 10 m, ignoring friction.

4

A pendulum 0.80 m long is released from rest at 20° from the vertical. Find its speed at the bottom.

5

Why does the mass of the object cancel in most conservation-of-energy problems?

6

A child on a swing moves at 4.0 m/s at the lowest point. How high above the lowest point does she rise?

7

Is mechanical energy conserved for a skydiver falling at terminal velocity? Explain.

More practice: warm-up

1

Define mechanical energy.

2

When is mechanical energy conserved?

3

Does the normal force of a smooth track do work on a coaster car?

4

A 2.0 kg rock falls 5.0 m from rest. How much Eg turns into Ek?

5

Write the formula for the height of a pendulum bob released at angle θ.

More practice: on your own

1

A toboggan starts from rest at the top of an 8.0 m frictionless hill, then goes over a 3.0 m bump. How fast is it moving at the top of the bump?

2

Water flows over a 50 m waterfall with negligible speed at the top. Ignoring air resistance, how fast is it falling at the bottom?

3

A 2.0 m pendulum moves at 3.0 m/s at its lowest point. What is its maximum angle from the vertical?

4

A stone is thrown at 10 m/s, 30° above the horizontal, from a cliff 15 m above a lake. How fast is it moving when it is 5.0 m above the water?

5

A coaster car goes around a vertical loop of radius 8.0 m. What is its minimum speed at the top? From what minimum height above the bottom must it start from rest (no friction)?

Challenge

1

Show, with full reasoning, that a car released from rest needs a starting height of at least 2.5R to complete a frictionless vertical loop of radius R.

2

A 0.50 kg pendulum bob on a 1.2 m string is released from rest with the string horizontal. Find its speed and the string tension at the lowest point.

3

A child slides from rest off the very top of a smooth, dome-shaped snow mound of radius R. At what height above the centre of the dome does she leave the surface? (Hint: she leaves when the normal force reaches zero.)

Check your understanding

Quiz

1.

Mechanical energy is conserved when:

2.

A ball is dropped from 20 m. Ignoring air resistance, its speed at the ground is about:

3.

Why does the tension in a pendulum string not change the bob's mechanical energy?

4.

A pendulum of length 1.0 m is released at 60°. How high above its lowest point does it start?

5.

At the top of a vertical loop, a car is moving at the minimum speed to stay on the track. Which force(s) provide the centripetal force?

6.

Two frictionless slides start at the same height and end at the same height; one is steep and straight, the other long and curvy. Riders starting from rest reach the bottom with:

For parents and teachers

Big idea (C2.3): when only gravity (or another conservative force) does work, kinetic and potential energy trade places but their sum stays constant. Grade 12 students use this on curved tracks and combine it with circular motion for loops and pendulums.

How to help: at a playground, ask where a swing moves fastest and where it has the most potential energy. Do the pendulum activity together and compare the predicted speed with the video.

Common difficulties:

  • Using the length of the track instead of the change in height.
  • Forgetting the initial kinetic energy when the object does not start from rest.
  • For pendulums, using h = L instead of h = L(1 − cos θ).
  • Applying conservation when friction or air resistance clearly matters (that is next lesson).