Circular Motion

A force that never speeds anything up

Tie a stone to a string and whirl it around your head at a steady rate. The stone's speed is not changing. Your hand is still pulling hard on the string, and you can feel it. Something has to be accelerating, because a force acts and nothing else is there to cancel it, but the speedometer reading of the stone, if it had one, would sit perfectly still.

That is the whole puzzle of this lesson, and the resolution is one line long: acceleration is the rate of change of velocity, and velocity is a vector. Turn the vector without lengthening it and you have still changed it. In the last lesson you watched a spring trade force for stretch along a single line. Here the force does something stranger: it bends the path and leaves the speed alone.

How big the inward acceleration is

Take an object going round a circle of radius $r$ at constant speed $v$. In a short time $\Delta t$ it moves along the arc through a small angle $\Delta\theta$. The velocity vector always points along the direction of travel, so as the object swings through $\Delta\theta$, the velocity vector swings through exactly the same angle $\Delta\theta$.

Draw the old velocity and the new velocity starting from the same point. Both have length $v$, and there is an angle $\Delta\theta$ between them. The change $\Delta \vec v$ is the little vector joining the tips. For a small angle that gap is an arc of a circle of radius $v$, so

$|\Delta \vec v| = v\,\Delta\theta$

Divide by $\Delta t$. The angle swept per second is the angular velocity $\omega = \Delta\theta/\Delta t$, measured in radians per second, so

$a = \frac{|\Delta \vec v|}{\Delta t} = v\,\omega$

And since the object covers arc length $r\,\Delta\theta$ in that same time, $v = \omega r$. Substituting either way round:

$a_c = \frac{v^2}{r} = \omega^2 r = \omega v$

Which direction does it point? $\Delta \vec v$ is perpendicular to $\vec v$ (it has to be, since the length of $\vec v$ is not changing) and it tilts toward the inside of the turn. So the acceleration points at the centre of the circle. That is what centripetal means: centre-seeking. The subscript $c$ is just a reminder of the direction.

Three more relations you will use constantly. If the period is $T$, one revolution of arc length $2\pi r$ takes $T$ seconds, so $v = 2\pi r/T$ and $\omega = 2\pi/T$. And revolutions per minute convert by dividing by 60 to get revolutions per second, then multiplying by $2\pi$.

Example 1. A point on a record sits 0.15 m from the centre and the record turns at 45 revolutions per minute. What is its centripetal acceleration?

Convert the rate first. $45$ rev/min $= 45/60 = 0.75$ rev/s, and each revolution is $2\pi$ radians, so

$\omega = 0.75 \times 2\pi = 4.712\ \mathrm{rad/s}$

Now use the form that already has $\omega$ in it, so you never need the speed:

$a_c = \omega^2 r = (4.712)^2 \times 0.15 = 22.21 \times 0.15 = 3.33\ \mathrm{m/s^2}$

There is no such thing as a centripetal force

This is the single habit that decides whether the rest of the topic is easy or miserable. $mv^2/r$ is not a force. It is a requirement. It is the answer the real forces have to add up to, in the inward direction, if the object is going to keep going round.

So the free body diagram never gets an arrow labelled "centripetal force". It gets the arrows it always got: gravity, the normal force, tension, friction. Then you pick the inward direction as positive, add up the components of those real forces along it, and set the total equal to $mv^2/r$:

$\sum F_{\text{inward}} = \frac{mv^2}{r}$

Perpendicular to that (along the direction of motion) the forces must sum to $ma_{\text{tangential}}$, which is zero when the speed is steady.

Cornering: when friction is the inward force

A car on a flat road turns because static friction from the road pushes it sideways, toward the centre of the curve. The vertical direction is boring ($N = mg$) so the most static friction the road can supply is $f_{\max} = \mu_s N = \mu_s m g$. Setting that equal to the requirement gives the fastest safe cornering speed:

$\mu_s m g = \frac{mv^2}{r} \quad\Longrightarrow\quad v_{\max} = \sqrt{\mu_s g r}$

The mass cancels, which is why a loaded truck and an empty one skid at roughly the same speed.

Friction is one vector, though, with one budget. If the car is also braking, the friction force has to do two jobs at once: point inward with magnitude $mv^2/r$ and point backward with magnitude $ma_{\text{brake}}$. Those two demands are perpendicular, so they add like the legs of a right triangle, and the total cannot exceed $\mu_s m g$.

Example 2. A car takes a flat curve of radius 100 m at 20 m/s on a road with $\mu_s = 0.80$. How hard can it brake at the same time without skidding?

The total friction available, per kilogram, is

$\frac{f_{\max}}{m} = \mu_s g = 0.80 \times 9.8 = 7.84\ \mathrm{m/s^2}$

The turning already spends

$a_c = \frac{v^2}{r} = \frac{20^2}{100} = \frac{400}{100} = 4.00\ \mathrm{m/s^2}$

These are at right angles, so what is left is not $7.84 - 4.00$. It is the other leg of the triangle:

$a_{\text{brake}} = \sqrt{(7.84)^2 - (4.00)^2} = \sqrt{61.47 - 16.00} = \sqrt{45.47} = 6.74\ \mathrm{m/s^2}$

Subtracting instead of using Pythagoras gives 3.84, and it is the most common wrong answer on this kind of problem.

Vertical circles and the speed you cannot go below

Now swing the stone in a vertical circle instead of a horizontal one. Gravity still points down, but "inward" now rotates as the stone goes round, so the component of gravity that counts keeps changing.

At the very top, the centre of the circle is directly below the stone. Both the tension $T$ and the weight $mg$ point down, which is inward, so

$T + mg = \frac{mv_{\text{top}}^2}{r}$

Rearranged, $T = mv_{\text{top}}^2/r - mg$. Slow the stone down and $T$ falls. A string can pull but it cannot push, so $T$ is not allowed to go negative. The slowest the stone can go while the string stays taut is the speed at which $T$ has just reached zero:

$0 + mg = \frac{mv_{\min}^2}{r} \quad\Longrightarrow\quad v_{\min} = \sqrt{gr}$

That is the famous condition, and the reason it matters is what it means physically: at that speed, gravity alone is exactly the force needed to bend the path into the circle. The string has nothing left to do. Go slower and gravity is more than enough, the stone falls inside the circle, and the string crumples. Notice the mass has dropped out: a bucket of water and a marble need the same minimum speed at the top of a 1.0 m circle, $\sqrt{9.8 \times 1.0} = 3.13$ m/s.

At the bottom, the centre is above, so inward is up:

$T - mg = \frac{mv_{\text{bot}}^2}{r} \quad\Longrightarrow\quad T = \frac{mv_{\text{bot}}^2}{r} + mg$

The plus sign here and the minus sign at the top come entirely from which way "inward" points. If you only remember one thing, remember to redraw the picture at the point in the circle the question asks about.

Linking the top speed to the bottom speed takes energy conservation, which you will meet properly soon but need here. Falling from the top of the circle to the bottom is a drop of $2r$, so $v_{\text{bot}}^2 = v_{\text{top}}^2 + 4gr$.

Example 3. A 0.25 kg ball is swung in a vertical circle at exactly the minimum speed at the top. What is the tension at the lowest point?

Here is the trick that makes this clean: you are not told $r$, so keep it symbolic and hope it cancels. At the top, minimum speed means $v_{\text{top}}^2 = gr$. Then

$v_{\text{bot}}^2 = gr + 4gr = 5gr$

Put that into the bottom equation:

$T = \frac{m(5gr)}{r} + mg = 5mg + mg = 6mg$

The radius cancelled. So the tension at the bottom of a critical loop is always six times the weight, whatever the size of the circle:

$T = 6 \times 0.25 \times 9.8 = 14.7\ \mathrm{N}$

When the support can push back

The $v_{\min} = \sqrt{gr}$ condition comes from the string's inability to push. Change the support and the condition changes with it.

A rigid rod can push as well as pull. Balanced at the top and nudged, a ball on a rod can start with essentially zero speed, because the rod will hold it up on the way round. So for a rod of length $\ell = 0.70$ m, the speed at the bottom comes straight from the $2\ell$ drop with $v_{\text{top}} \approx 0$:

$v_{\text{bot}} = \sqrt{4g\ell} = \sqrt{4 \times 9.8 \times 0.70} = \sqrt{27.44} = 5.24\ \mathrm{m/s}$

A surface can push but not pull, which flips the logic the other way. A block sliding over the outside of a dome of radius $R$ has the centre of the circle below it, so at an angle $\theta$ from the top the inward equation is

$mg\cos\theta - N = \frac{mv^2}{R}$

The block leaves the surface when $N$ falls to zero, so it leaves when $v^2 = gR\cos\theta$. Meanwhile energy gives $v^2 = 2gR(1 - \cos\theta)$ for a block starting from rest at the top. Setting those equal:

$gR\cos\theta = 2gR(1-\cos\theta) \quad\Longrightarrow\quad 3\cos\theta = 2 \quad\Longrightarrow\quad \cos\theta = \tfrac{2}{3}$

The height above the centre of the dome is $R\cos\theta = \tfrac{2}{3}R$, so a dome of radius 1.5 m releases the block at $\tfrac{2}{3}(1.5) = 1.0$ m. The angle is the same for every dome; only the height scales.

Two more sources of the inward force

A rotating string carrying several masses. Cut the string anywhere in your mind and ask what the piece beyond the cut needs. The inner segment has to supply the inward force for everything outside it. For two 0.30 kg beads at 0.40 m and 0.80 m rotating at 5.0 rad/s, the inner segment carries both:

$T = m\omega^2 r_1 + m\omega^2 r_2 = 0.30 \times 25 \times (0.40 + 0.80) = 7.5 \times 1.2 = 9.0\ \mathrm{N}$

For a continuous chain the same logic holds with the mass of the outer piece placed at its own centre of mass. Tension is largest at the axis and zero at the free tip.

Gravity, for an orbit. A satellite in a circular orbit is held in by gravity and nothing else, so $GMm/r^2 = mv^2/r$. The satellite's mass cancels. Writing $v = 2\pi r/T$ and rearranging gives $T^2 \propto r^3$, so $T \propto r^{3/2}$. Four times the radius means $4^{3/2} = 8$ times the period.

A banked track uses the horizontal component of the normal force. With no friction, the normal force's vertical part carries the weight and its horizontal part does the turning, and dividing one by the other gives $\tan\theta = v^2/(gr)$. For $r = 50$ m, $v = 20$ m/s and $g = 10$ m/s², $\tan\theta = 400/500 = 0.8$, so $\theta = 38.7°$. The next lesson takes that apart properly, along with what happens when friction joins in.

Recognising it

Reach for this technique when the problem contains any of these:

• The words circle, loop, curve, corner, orbit, revolution, rpm, or a period for something going round.
• A radius together with a speed, or a radius together with an angular rate.
• A question about the minimum or maximum speed for something to stay on a track, in a bucket, or on a string.
• A normal force or tension that you are asked to find at a specific point of a curved path.

Then the procedure is always the same four steps. Draw the object at the exact moment asked about. Mark which way the centre is from there: that is "inward". Add up the real forces along that direction, with signs, and set the sum equal to $mv^2/r$. If two different points of the path are involved, link their speeds with energy conservation before you solve.

The one extra question to ask, every time: can the thing supplying the force only pull, only push, or both? A string only pulls, a surface only pushes, a rod does either. That single answer is what sets the critical condition, and it is where most of the hard problems in this topic hide.