Dynamics: Doors, Wheels, and Coast-Down Tests

Dynamics is two equations used carefully: F = ma for things that slide and T = Iα for things that spin. A modeling course like MIT’s 2.003 dresses them up with coordinates and free-body diagrams ¹, but the everyday job is the same: what force or torque does this motion need, and what happens when it stops? Two pieces of hardware cover most of it: a sliding barn door and a bandsaw.

Starting a barn door

An 80 kg door on a track. You want it moving at 1 m/s within half a second of pushing.

• Acceleration: a = Δv / t = 1 / 0.5 = 2 m/s².
• Force: F = ma = 80 × 2 = 160 N, about 36 lb, plus whatever the rollers resist.

Move the same door at the same speed but take a full second to get it going, and the push halves to 80 N. The force depends on how quickly you change the speed, not on the speed itself ².

Stopping is the hard part

At 1 m/s the door carries kinetic energy ½mv² = ½ × 80 × 1² = 40 J. Whatever stops it has to soak up those 40 joules, and the average force is the energy divided by the distance it stops over ²:

Same door, same energy, a fifty-fold difference in force. That’s what tears out track bolts and cracks hangers, and it’s why a damper is worth the money on a heavy door. The same logic applies to a drawer, a gate, a falling tool, or a conveyor carriage hitting its end stop.

Spinning: the bandsaw

Rotation uses the same ideas with new names: torque T for force, angular acceleration α for acceleration, and moment of inertia I for mass. I depends on how far the mass sits from the axis: for a solid disk, I = ½mr²; for a thin rim, I = mr² ². Mass at the rim counts double.

Take a 14-inch bandsaw. Treat each cast-iron wheel as a 9 kg disk with a 0.178 m radius:

• One wheel: I = ½ × 9 × 0.178² ≈ 0.14 kg·m². Both wheels: ≈ 0.29 kg·m². Real spoked wheels carry more of their mass at the rim, so the true value is a bit higher. These are estimates to show the method.
• A 1,725 rpm motor drives the lower wheel through a belt with about a 2.1:1 reduction, so the wheels turn about 820 rpm (86 rad/s). The blade speed is wheel circumference × rpm = 3.67 ft × 820 ≈ 3,000 ft/min.
• Stored energy at speed: ½Iω² = ½ × 0.29 × 86² ≈ 1.1 kJ. That’s why a bandsaw keeps running for a long time after you hit the switch.

Seen through the belt

The motor doesn’t feel 0.29 kg·m². Through a speed reduction, the load’s inertia as seen by the motor divides by the ratio squared: 0.29 / 2.1² ≈ 0.065 kg·m² ¹. One factor of the ratio comes from the belt multiplying the motor’s torque; the other comes from the load turning slower, so it needs less angular acceleration.

A 1 hp motor at 1,725 rpm (181 rad/s) has a rated torque of P / ω = 746 / 181 ≈ 4.1 N·m. Induction motors deliver roughly double that while starting, so call it 8 N·m. The motor’s acceleration is then α = T / I = 8 / 0.068 ≈ 120 rad/s² (adding a little for the motor’s own rotor), and it reaches 181 rad/s in about 1.5 seconds. That’s about what you hear when you switch one on. Storing 1.1 kJ in 1.5 s is an average of about 730 W, close to the motor’s rating, which is a good sign the numbers hang together.

The same arithmetic explains why a big flywheel or a heavy lathe chuck on a small motor trips breakers or spins up slowly: the starting current lasts until the inertia is up to speed.

The coast-down test

Friction is the hardest thing in a machine to calculate and the easiest to measure. Switch the saw off at full speed and time it to a stop. Say it takes 40 seconds.

If friction torque is roughly constant, the wheels decelerate steadily, and

T_friction = I × ω / t = 0.29 × 86 / 40 ≈ 0.6 N·m

At full speed that’s 0.6 × 86 ≈ 55 W going into bearings, tire flex, blade bending, and air drag. Do the test again after changing something (new bearings, less blade tension, a dried-out tire) and the change in coast-down time tells you how much friction you added or removed. If the saw coasts noticeably shorter than it used to, something is dragging.

Real coast-down curves fall faster at the start, because air drag and other speed-dependent losses are highest then, and slower near the end, where dry friction takes over. For a first estimate the straight line is fine.

Checking your answers

• Units: every term in an equation should come out in newtons, or newton-meters, or joules.
• Limits: double the mass and the acceleration should halve; take away the force and nothing should speed up.
• Energy: the energy a motor puts in during a move has to cover the kinetic energy gained plus the friction losses. If it doesn’t, a number is wrong.

Try it

Time the coast-down of a machine in your shop: a bandsaw, a lathe with the chuck on, a grinder. Estimate the rotating parts’ inertia from their weight and radius, compute the friction torque and the power lost at speed, and write the coast-down time on the machine. Next year, time it again.