Uniform Circular Motion Simulator

A dark dot travels around a dashed circle at constant speed. The teal arrow on it is the velocity, always tangent to the path. The coral arrow is the net (centripetal) force and acceleration, always aimed at the center point.

The readout shows $a_c = v^2/r$ in m/s², $F_c = m a_c$ in newtons, the period T in seconds and the angular speed ω in rad/s, all to two decimal places. The coral arrow's length follows the acceleration, so changing only the mass leaves the arrow alone even though the force readout changes. The arrow also stops growing once $a_c$ passes about 60 m/s², so use the numbers, not the arrow, at high speeds and small radii.

Two scaling checks are worth doing. Double the speed from 3 to 6 m/s at a fixed radius: $a_c$ becomes four times larger. Then double the radius from 2 to 4 m at a fixed speed: $a_c$ halves, and the period doubles because the object has twice as far to go.

An object moving at constant speed in a circle is still accelerating, because the direction of its velocity keeps changing. That acceleration points toward the center and has magnitude $$a_c = \frac{v^2}{r}$$

Newton's second law in the radial direction says the net force toward the center must supply it: $$F_{net,c} = m a_c = \frac{mv^2}{r}$$ “Centripetal force” is not a new kind of force. It is the name for whatever net force points inward: tension in a string, friction on tires, gravity on a satellite, or a combination such as the normal force and weight on a car going over a hill.

The time for one lap and the angular speed are $$T = \frac{2\pi r}{v} \qquad \omega = \frac{v}{r} = \frac{2\pi}{T}$$ If the inward net force suddenly disappeared, the object would leave along the tangent line in the direction of its velocity at that instant, not fly straight outward.

A 2.5 kg ball on a string moves in a horizontal circle of radius 4 m at a constant 8 m/s on a frictionless table. Find its acceleration, the tension, the period and the angular speed.

Acceleration. $a_c = (8)^2/4 = 16$ m/s², pointing toward the center.

Tension. The string's tension is the only horizontal force, so it is the net inward force: $T_{string} = (2.5)(16) = 40$ N. (The weight and the normal force from the table are vertical and cancel.)

Period and angular speed. $T = 2\pi(4)/8 = 3.14$ s and $\omega = 8/4 = 2$ rad/s.

What if? If the string can hold at most 50 N, the fastest safe speed is $v = \sqrt{F r/m} = \sqrt{(50)(4)/2.5} = 8.9$ m/s.

Set v = 8, r = 4 and m = 2.5 in the simulator. The readout should show $a_c$ = 16 m/s², $F_c$ = 40 N, T = 3.14 s and ω = 2 rad/s. Keep the speed at 8 m/s and shrink the radius to 2 m: the acceleration and force double to 32 m/s² and 80 N.

Circular motion is Topic 2.9 in Unit 2, Force and Translational Dynamics. Typical questions include the tension in a swinging pendulum at its lowest point, the normal force on a car at the top of a hill or the bottom of a dip, the friction needed for a car to round a flat curve, and orbits, where gravity provides the inward force. Each one starts with a free-body diagram and the radial form of Newton's second law.