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Circular Motion overview

Topic 2 of 5

Why acceleration points inward

A body can accelerate without speeding up. In uniform circular motion, its velocity changes direction continuously, so its acceleration is nonzero and points towards the centre.

Speed is the magnitude of velocity. The instantaneous velocity is tangent to the path. To find a change of velocity, subtract the vectors rather than their magnitudes; the vector method lets you translate an arrow without rotating or changing its length.

Average acceleration = Δv / Δt
Δv = vlater - vearlier

The symbols v here represent vectors. Place the two velocities at a common origin: the arrow from the earlier tip to the later tip is their difference.

Worked velocity change

Equal speeds, different velocity vectors

A body travels anticlockwise at 2.00 m/s on a circle of radius 1.00 m. Consider positions 10° below and 10° above the right-pointing radius, with x right and y up.

Equal speeds can have different velocity vectors

Use radius 1.00 m, speed 2.00 m/s and anticlockwise motion. The two positions are at -10° and +10° from the rightward radius. Right and up define the positive component directions.

Velocities are tangent at their own positions

Two equal-length tangent velocities on a short circular arcA true circle is shown with equal position scales. The first point is ten degrees below the rightward radius; its velocity components are positive 0.347296 and positive 1.969616 metres per second, pointing up and slightly right. The second point is ten degrees above the rightward radius; its velocity components are negative 0.347296 and positive 1.969616, pointing up and slightly left. Both blue velocity arrows are one hundred drawing units long. M is the middle of the short arc, at the rightmost point. A separate green instantaneous acceleration arrow at M points left towards the centre, with magnitude four metres per second squared. Its arrow has different units from the blue arrows and does not represent the finite-interval average.Ov1v2P1P2Ma at M: left

Both speeds are 2.00 m/s, but the directions differ. At M, the instantaneous acceleration is 4.00 m/s2 left, perpendicular to the upward velocity there.

Translate the vectors without rotating them

Velocity change points from the first velocity tip to the secondThe two original velocities are translated to the same origin without rotating them. Both use ninety drawing units per metre per second. The first tip is right of the vertical axis; the second tip is equally far left and at exactly the same height. The brown change arrow goes horizontally from the first tip to the second, pointing left. Thus delta v equals v two minus v one, with components negative 0.694593 and zero metres per second. Over the finite time interval 0.174533 seconds the average acceleration is approximately 3.97972 metres per second squared left. It is not labelled exactly equal to the four-metres-per-second-squared instantaneous acceleration at the middle of the arc.+x+yOv1v2Δv points left

The vertical velocity components cancel in the subtraction. Δv = (-0.694593, 0) m/s and Δt = 0.174533 s give average acceleration 3.97972 m/s2 left. A smaller angular interval brings this average closer to the instantaneous value, 4.00 m/s2.

The position view shows equal-length tangent velocities at two nearby points. The separate subtraction view translates those same vectors to a common origin. Its finite change points inward at the midpoint of the arc.

Keeping extra digits during the calculation, the velocity components in m/s are:

vearlier = (+0.347296, +1.969616)
vlater = (-0.347296, +1.969616)
Δv = (-0.694593, 0) m/s

The angular separation is 20° = π/9 rad. With ω = v/r = 2.00 rad/s, its duration is Δt = (π/9)/2 = 0.174533 s. Therefore:

aaverage = (-0.694593, 0)/0.174533
= (-3.97972, 0) m/s2

This finite-interval average points left, towards the centre from the arc's midpoint. The instantaneous acceleration at that midpoint is (-4.00, 0) m/s2. They are close but not identical. Reducing the angular interval makes the average approach the instantaneous acceleration.

Use the magnitude and give its direction separately

For a body travelling uniformly around a circle, the instantaneous centripetal, or inward, acceleration has magnitude:

a = v2/r = rω2

The second form follows by substituting v = rω into the first. Use radians per second for ω. These expressions give a magnitude, not a fixed x or y component. The inward direction itself changes as the body travels around the circle.

In the 1.00 m circle above, a = 2.002/1.00 = 4.00 m/s2. For the 0.400 m rotating marker with ω = π rad/s, a = 0.400π2 = 3.95 m/s2, inward.

A resultant perpendicular to velocity changes its direction without changing its speed, producing the curved path in uniform circular motion. The perpendicular force does no instantaneous work, consistent with constant kinetic energy.

If a body speeds up or slows down while following a curved path, it also has a tangential acceleration component. Its total acceleration and resultant force need not then be perpendicular to velocity. Do not apply the purely inward description to every changing-speed turn.

Optional check A body travels at constant speed around a circle. Which statement correctly describes its instantaneous velocity and acceleration?
A body travels at constant speed around a circle. Which statement correctly describes its instantaneous velocity and acceleration?

Next, identify which actual forces supply that inward resultant. A required acceleration does not introduce a new interaction by itself.