MODULE 02 · C1 / CIRCULAR MOTION
Uniform Circular Motion
Constant speed, changing direction: why moving in a circle takes a force toward the centre.
TOP VIEW · PUCK ON A STRING · FRICTIONLESS TABLE
Learning objective
Conditions · locked for this activity
The experiment
ReadyA puck slides on a frictionless horizontal table, tied by a light string to a pin at the centre. Seen from above, the only horizontal force is the string’s pull toward the pin. Gravity and the table’s normal force are vertical and cancel.
Motion graphs
LIVEx and y are measured from the pin; dashed lines are the y components.
Key ideas
ALGEBRA| Period (time per lap) | T = 2πr / v | |
|---|---|---|
| Angular speed | ω = v / r = 2π / T | |
| Centripetal acceleration | a = v² / r | |
| Net force needed | F = m v² / r |
The speed never changes, but the direction of the velocity does, so the puck is accelerating. The acceleration and the net force point toward the centre. Nothing pushes the puck outward: cut the string and see.
Show the calculus
Turn on to see the position as a vector r(t), and the velocity and acceleration as its derivatives, with live values from the run.
r(t) = r (cos ωt, sin ωt)
v(t) = dr/dt = rω (−sin ωt, cos ωt), |v| = rω
a(t) = dv/dt = −ω² r(t), |a| = ω² r = v² / r
| Position r(t) | |
|---|---|
| Velocity v(t) | |
| r · v (perpendicular ⇔ 0) | |
| Acceleration a(t) | |
| −ω² r(t) |
d(|r|²)/dt = 2 r · v = 0 keeps the radius fixed, and d(|v|²)/dt = 2 v · a = 0 keeps the speed fixed: the acceleration only turns the velocity. On the graphs, the slope of x(t) is vₓ(t), and x and y are sinusoids a quarter period apart.
Results
Reflect on the evidence
Instructor-authored model answer
Top view, SI units. Frictionless table, light inextensible string, puck treated as a point. Motion is computed from the exact solution; after the cut the puck moves in a straight line until it leaves the 3.2 m table.