Free fall and circular motion

At the top of a vertical throw, a ball stops for an instant. A point on a uniformly rotating wheel keeps the same speed. Why is there acceleration in both cases? Let us examine changes in the magnitude and direction of velocity separately.

Release a compact heavy object and a compact light one together; they fall almost together. In the model without air resistance, gravity alone acts on both: this is free fall. Near Earth's surface, g\approx9{,}8\,\text{м/с}^2 points downwards and is independent of mass. This is an experimental finding; later, dynamics will connect it with gravitational force and inertia. A sheet of paper falling in air tests a different model.

The hand gives a ball its initial velocity during a throw, but stops pushing once it lets go. Gravity keeps changing the velocity downwards: it reduces the upward velocity on the way up and increases the downward velocity on the way down. Thus free fall includes upward motion when gravity is the only force. Free describes the conditions, not the direction of the trajectory.

Release a body without a push and point the axis downwards. Then v_0=0, a=g, and the familiar velocity relation becomes v=gt. With g=10 m/s², speed increases by 10 m/s each second. Doubling time doubles speed; reducing g reduces it for the same time. At t=0, the formula gives zero, as release from rest requires. Throwing the body initially would require a different initial condition.

During the first second the body is still slow; during the next it is faster. The average speed from release to time t is gt/2, so h=gt^2/2. Here h is the downward distance in metres. Doubling time quadruples this distance because both time and average speed double. The model requires constant g, an initial state of rest and negligible resistance; it cannot describe an entire parachute descent.

If the height is known, find how long the velocity has to build up. From h=gt^2/2 we get t=\sqrt{2h/g}, then v=\sqrt{2gh}. Choose the positive root: these are a duration and a speed. Quadrupling height doubles both time and speed. With smaller g, a body falls from the same height for longer but reaches a lower speed: the longer time does not fully compensate for the smaller acceleration.

Compare instants just before and after the top of a throw: the ball was moving up and will soon be moving down. Its velocity changes even as it passes through zero. At the top, v=0, but acceleration remains g downwards. Let H be the rise above the launch point and v_0 the initial speed. If acceleration also vanished and never returned, the ball would remain at rest at the top.

Point the axis upwards: a_x=-g, and at the top v_x=0. From v_x^2-v_{0x}^2=2a_xs_x we obtain H=v_0^2/(2g). Doubling launch speed quadruples the height: the ball starts faster and takes longer to lose its speed. Larger g reduces the rise. At v_0=0, there is no rise. We neglect air resistance and variation of g with height.

Mark a point on the rim of a uniformly rotating wheel. Over a short time its speed stays the same, but the velocity arrow turns. The difference between two nearby arrows points towards the centre, so acceleration points there too. Velocity itself is tangent to the circle: it is the instantaneous direction of motion. Radius R is measured in metres. Turning velocity is a change of vector even with a constant speedometer reading.

In time \Delta t, velocity turns through an angle of about v\Delta t/R. The change in the velocity arrow is then about v\times v\Delta t/R. Divide by time and take the limit of a short interval to obtain a_{\text{ц}}=v^2/R. One factor v comes from the arrow's length, the other from how quickly it turns. Doubling speed requires four times the acceleration; at the same speed, a larger radius makes the turn gentler.

Time one complete return of the mark on the wheel. This is period T, in seconds. Frequency \nu counts revolutions per second and is measured in hertz: \nu=1/T. If a revolution takes 2 s, half a revolution occurs per second: 0.5 Hz. Halving the period doubles frequency. Very slow rotation has a long period and a frequency close to zero; a stationary point has no finite revolution period.

In one period the mark travels the full circumference 2\pi R. Therefore v=2\pi R/T=2\pi R\nu. At fixed radius, halving the period doubles speed. At fixed period, a point at twice the radius covers twice the distance and moves twice as fast. To predict a change, state what is held constant: equal periods and equal linear speeds are different conditions.

Two points on the same wheel turn through the same angle each second but travel different distances. Angular velocity \omega gives a shared description: for uniform rotation, \omega=\varphi/t=2\pi\nu, in rad/s. Here \varphi is the angle swept from the chosen starting position in time t; a full revolution contains 2\pi radians. Radians link arc length to radius: arc length equals radius times angle in radians.

The arc travelled per second equals radius times angle per second: v=\omega R. Substitution gives a_{\text{ц}}=\omega^2R. A larger radius now seems to increase acceleration, whereas in v^2/R it decreased it. There is no contradiction: here \omega is fixed, so the outer point also moves faster. As R\to0 at fixed \omega, its linear speed and acceleration approach zero.

A point particle is released from rest. How far does it travel in 2 s without air resistance? Take g=10 м/с².

The conditions match release from rest with constant g.

We need distance from time, so choose h=gt²/2.

Substitute: h=10×2²/2=20 м. Final speed v=gt=20 м/с.

Check using average speed: (0+20)/2×2=20 м.

In 2 s, the body travels 20 m downwards and reaches a speed of 20 m/s. Its average speed was 10 m/s: it covered only 5 m in the first second, then 15 m in the second.

Zero velocity at one instant does not imply zero acceleration: gravity continues changing the ball's velocity at the top.

The formulas v = gt and h = gt²/2 describe release from rest with constant g and no resistance.

In circular motion, acceleration turns the velocity. Discuss the radius dependence together with what is held constant: v or ω.

Next we connect acceleration with the net force.