Non-Inertial Frames & Pseudo-Forces

The bag that slides forward by itself

You are sitting in a bus with a shopping bag on the smooth floor beside you. The driver brakes hard. The bag slides forward, and nothing is touching it. No hand, no string, no wind. Newton says an object accelerates only when something pushes it. Something pushed it, then, but what?

Your friend on the pavement has no puzzle. She sees the bag carrying on at a steady velocity, exactly as an object with no horizontal force should, while the bus slows around it. Two observers, one event, and only one of them needs a mystery force.

That is the whole subject. Newton's second law is a law about the universe as measured from a frame that is not itself accelerating. Such a frame is called inertial. The pavement is close enough to one. The braking bus is not. You found the shape of a fluid at rest under gravity last lesson; here you will find the shape of one in a tank being dragged sideways, and a lot else, by learning what to do inside the wrong frame instead of climbing out of it.

You could refuse and work in the ground frame. Often that is horrible, because there the thing you care about is moving, rotating, or riding on a surface that is itself accelerating, and you end up with coupled equations.

So keep the accelerating frame and repair the law. Let the frame accelerate at $\vec{a}_f$ as seen from the ground, and let an object of mass $m$ have ground acceleration $\vec{a}$. Velocities and accelerations just subtract, so the acceleration you measure inside the frame is

$\vec{a}' = \vec{a} - \vec{a}_f.$

In the ground frame the real forces obey $\sum\vec{F} = m\vec{a}$. Substitute $\vec{a} = \vec{a}' + \vec{a}_f$ and move the extra term to the left:

$\sum\vec{F} - m\vec{a}_f = m\vec{a}'.$

Read that as an instruction. To write $\sum\vec{F} = m\vec{a}'$ inside the accelerating frame, add one extra force to every mass:

$\vec{F}_{\text{pseudo}} = -m\,\vec{a}_f,$

pointing opposite the frame's acceleration, with magnitude $ma_f$. The bus accelerates backward while braking, so the bag feels a forward pseudo-force. No mystery left.

Two features mark a pseudo-force out. It is exactly proportional to mass, like gravity, and it has no source object, so nothing anywhere feels a Newton's-third-law reaction to it. The first feature is the useful one: since $-m\vec{a}_f$ and real weight $m\vec{g}$ both scale with $m$, you can merge them into a single effective gravity

Inside a sealed accelerating cabin everything behaves as if gravity were tilted and stronger. Pendulums, floating blocks, water surfaces, the lot.

The pendulum accelerometer

Hang a bob from the ceiling of a cart accelerating forward at $a$ and work inside the cart. The bob hangs still, so it is in equilibrium under three forces: tension $T$ along the string, weight $mg$ down, pseudo-force $ma$ backward. The string must point along $\vec{g}_{\text{eff}}$, since tension has to cancel the other two. Resolve: horizontally $T\sin\theta = ma$, vertically $T\cos\theta = mg$. Divide the first by the second and $T$ disappears:

$\tan\theta = \frac{a}{g}.$

The string tilts backward from the true vertical by that angle, and the tension is $T = m\sqrt{g^2+a^2} = mg_{\text{eff}}$.

Example 1. A train accelerates at $a = 2.50\ \mathrm{m/s^2}$, with $g = 9.8\ \mathrm{m/s^2}$. Then $\tan\theta = 2.50/9.8 = 0.2551$, so $\theta = 14.31^\circ$ from the vertical, tilted toward the back.

That angle is where the bob settles. Start the cart suddenly and the bob is released from rest $\theta$ away from its new equilibrium, so it swings, and swings about the tilted direction, not the vertical. It is an ordinary pendulum living under $g_{\text{eff}}$ instead of $g$, so the small oscillation period is $T = 2\pi\sqrt{\ell/g_{\text{eff}}}$. For $\ell = 0.800\ \mathrm{m}$ in a cart at $a = 5.00\ \mathrm{m/s^2}$: $g_{\text{eff}} = \sqrt{9.8^2 + 5.00^2} = \sqrt{96.04+25.0} = \sqrt{121.04} = 11.00\ \mathrm{m/s^2}$, so $T = 2\pi\sqrt{0.800/11.00} = 2\pi\sqrt{0.07273} = 2\pi(0.2697) = 1.694\ \mathrm{s}$. Faster than at rest, because the effective gravity is stronger.

In an elevator $\vec a_f$ is vertical instead, so $g_{\text{eff}} = g \pm a$ and the scale reads $m(g+a)$ going up: $70.0 \times 11.8 = 826\ \mathrm{N}$ at $a = 2.00\ \mathrm{m/s^2}$. Accelerating down at $3.00\ \mathrm{m/s^2}$, $g_{\text{eff}} = 6.8$ and a $5.00\ \mathrm{kg}$ box presses with $34.0\ \mathrm{N}$.

A crate pressed against the inside back wall of a truck is the same idea with a twist. In the truck's frame it is static; horizontally the wall's normal force $N$ balances the pseudo-force, so $N = ma$, and vertically friction must carry the weight, $\mu N \ge mg$. The mass cancels: $a \ge g/\mu$. With $\mu = 0.400$, $a_{\min} = 9.8/0.400 = 24.5\ \mathrm{m/s^2}$.

Fluids in an accelerating tank

A free liquid surface always sits perpendicular to the local effective gravity. If it did not, some of $\vec g_{\text{eff}}$ would lie along the surface and the liquid would flow until it did. Tilt the gravity, tilt the surface. In a tank accelerating horizontally at $a$, the surface angle obeys the pendulum relation again:

$\tan\alpha = \frac{a}{g}.$

For a tanker at $3.00\ \mathrm{m/s^2}$, $\tan\alpha = 0.3061$ and $\alpha = 17.02^\circ$, fuel piled at the back. Over a tank of length $L$ the end-to-end height difference is $\Delta h = L\tan\alpha = La/g$: for $L = 1.20\ \mathrm{m}$ at $a = 2.00\ \mathrm{m/s^2}$, $\Delta h = 0.2449\ \mathrm{m}$.

The surface tilts about its midpoint, because the volume of liquid has not changed: the rear end rises $\Delta h/2$ and the front drops the same. That is what you need for pressure at the bottom of the rear wall, which is still just $\rho g$ times the depth there. Example 2. An open tank $1.00\ \mathrm{m}$ long, rest depth $0.400\ \mathrm{m}$, accelerating at $3.00\ \mathrm{m/s^2}$: $\Delta h = 1.00 \times 3.00/9.8 = 0.3061\ \mathrm{m}$, so the rear depth is $0.400 + 0.3061/2 = 0.5531\ \mathrm{m}$ and the gauge pressure is $1000 \times 9.8 \times 0.5531 = 5420\ \mathrm{Pa}$.

If the tank is sealed and completely full, there is no surface to tilt, but the pressure gradient is still the one that supplies the acceleration. Take a horizontal slab of fluid of area $A$ and length $L$, mass $\rho AL$. The only horizontal push on it is the pressure difference across its ends, so $(p_{\text{rear}} - p_{\text{front}})A = \rho A L a$, giving

$\Delta p = \rho a L.$

A sealed tank $2.00\ \mathrm{m}$ long, full of water ($\rho = 1000\ \mathrm{kg/m^3}$), at $4.00\ \mathrm{m/s^2}$ gives $\Delta p = 1000 \times 4.00 \times 2.00 = 8000\ \mathrm{Pa}$, rear wall higher. Note that $g$ never entered: at equal heights, gravity contributes nothing to the difference.

Blocks on wedges

Here the accelerating frame really earns its keep. A block sits on the frictionless face of a wedge of angle $\theta$, and you push the wedge horizontally so the block does not slide on the face. What acceleration do you need?

Work in the wedge's frame. The block is at rest there, so weight $mg$ down, normal force $N$ perpendicular to the face, and pseudo-force $ma$ backward must cancel. Resolve along the face: the direction the block would slide, and the one $N$ contributes nothing to. Weight pulls $mg\sin\theta$ down the slope; the pseudo-force pushes $ma\cos\theta$ up it. Setting them equal:

$ma\cos\theta = mg\sin\theta \quad\Longrightarrow\quad a = g\tan\theta.$

For $\theta = 25.0^\circ$: $a = 9.8 \times 0.4663 = 4.570\ \mathrm{m/s^2}$. Equivalently, the effective gravity must point straight into the face.

Add friction and the answer becomes a range. At the minimum acceleration the block is on the verge of sliding down, so static friction acts up the slope at full strength, $f = \mu N$. Along the slope, up-slope positive,

$ma\cos\theta + \mu N - mg\sin\theta = 0,$

and perpendicular to the slope it has no acceleration, so

$N = mg\cos\theta + ma\sin\theta.$

Substitute, cancel $m$, gather the $a$ terms on one side, and divide top and bottom by $\cos\theta$:

$a_{\min} = g\,\frac{\tan\theta - \mu}{1 + \mu\tan\theta}.$

Example 3. $\theta = 30.0^\circ$, $\mu = 0.250$. Then $\tan 30.0^\circ = 0.5774$. Numerator: $0.5774 - 0.250 = 0.3274$. Denominator: $1 + 0.250 \times 0.5774 = 1.1443$. So $a_{\min} = 9.8 \times 0.3274/1.1443 = 9.8 \times 0.2861 = 2.803\ \mathrm{m/s^2}$. If $\mu$ exceeded $\tan\theta$ the numerator would go negative: the frame telling you the block never needed help.

One harder variant: let the wedge, mass $M$, slide freely on a frictionless floor with the block, mass $m$, on its frictionless $\theta$ face. Nobody pushes the wedge; it recoils because the block leans on it. Its acceleration $A$ is unknown, and this is where the frame trick pays: treat the block in the wedge frame, the wedge in the ground frame.

In the wedge frame the block feels a pseudo-force $mA$ opposite the wedge's acceleration, and it stays on the face, so perpendicular to the face it has no acceleration at all. That one direction gives $N$ immediately: weight contributes $mg\cos\theta$ into the face, the pseudo-force $mA\sin\theta$ away from it, so

$N = mg\cos\theta - mA\sin\theta .$

Now the wedge, from the ground. The only horizontal force on it is the block's normal force pressing on the face, whose horizontal component is $N\sin\theta$:

$MA = N\sin\theta = (mg\cos\theta - mA\sin\theta)\sin\theta .$

Collect the $A$ terms, $MA + mA\sin^2\theta = mg\sin\theta\cos\theta$, and

$A = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}.$

For $m = 1.00$, $M = 3.00\ \mathrm{kg}$, $\theta = 40.0^\circ$: $\sin\theta\cos\theta = 0.6428 \times 0.7660 = 0.4924$, and $M + m\sin^2\theta = 3.00 + 0.4132 = 3.413$. So $A = 9.8 \times 0.4924/3.413 = 1.414\ \mathrm{m/s^2}$.

Rotating frames and the centrifugal term

A frame turning at constant angular speed $\omega$ is non-inertial too, because the direction of every velocity keeps changing. A point a distance $r$ from the axis has ground acceleration $\omega^2 r$ toward the axis, so $-m\vec a_f$ points outward:

$F_{\text{cf}} = m\omega^2 r,$

the centrifugal force. It grows with $r$, so it is not uniform, but at any one point you can still add it to gravity.

On a turntable at $\omega = 6.00\ \mathrm{rad/s}$, an object at $r = 0.500\ \mathrm{m}$ sits in a centrifugal field of $\omega^2 r = 36.0 \times 0.500 = 18.0\ \mathrm{m/s^2}$. A $60.0\ \mathrm{kg}$ passenger in a car rounding an $80.0\ \mathrm{m}$ curve at $20.0\ \mathrm{m/s}$ feels $mv^2/r = 60.0 \times 400/80.0 = 300\ \mathrm{N}$ pushing outward.

A ring space station of radius $90.0\ \mathrm{m}$ wanting $\omega^2 R = 9.8\ \mathrm{m/s^2}$ at the rim needs $\omega = \sqrt{9.8/90.0} = 0.330\ \mathrm{rad/s}$; the crew stand on the inside of the rim with the centrifugal force playing the part of weight. A centrifuge at $4000\ \mathrm{rev/min} = 4000\times2\pi/60 = 418.9\ \mathrm{rad/s}$ gives a sample at $r = 0.0800\ \mathrm{m}$ an acceleration $418.9^2 \times 0.0800 = 1.404\times10^{4}\ \mathrm{m/s^2}$, about 1400 $g$.

Every centripetal-force problem becomes a statics problem. A coin on a turntable at $r = 0.120\ \mathrm{m}$ with $\mu = 0.350$ stays put while friction can balance the centrifugal term: $\mu m g \ge m\omega^2 r$, so $\omega_{\max} = \sqrt{\mu g/r} = \sqrt{0.350\times 9.8/0.120} = \sqrt{28.58} = 5.346\ \mathrm{rad/s}$. A banked curve with no friction needs the effective gravity perpendicular to the road, i.e. $\tan\beta = v^2/(rg)$, so $v = \sqrt{rg\tan\beta} = \sqrt{120 \times 9.8 \times \tan 12.0^\circ} = \sqrt{250.0} = 15.81\ \mathrm{m/s}$. A conical pendulum of length $\ell$ at half-angle $\theta$ has $r = \ell\sin\theta$ and $\tan\theta = \omega^2 r/g$, giving $\omega = \sqrt{g/(\ell\cos\theta)} = \sqrt{9.8/(1.20\times 0.8660)} = 3.071\ \mathrm{rad/s}$.

Swing a bucket of water in a vertical circle of radius $r$. At the top, in the bucket's frame, the water feels $mg$ down and $mv^2/r$ up. It stays in as long as the bucket does not have to pull, so the limit is $mv^2/r = mg$ and $v_{\min} = \sqrt{gr} = \sqrt{9.8} = 3.13\ \mathrm{m/s}$ for $r = 1.00$ m.

Coriolis: the second rotating pseudo-force

Centrifugal force is the whole story only while the object sits still in the rotating frame. Start it moving and a second pseudo-force appears. Roll a ball outward from the centre of a turntable: from above it travels in a straight line, but the floor beneath it is turning, so in turntable coordinates its track curves sideways. Nothing pushed it; the frame swung out from under it.

That apparent sideways push is the Coriolis force. For a velocity $\vec v\,'$ measured in a frame rotating at $\vec\Omega$, it is

$\vec F_{\text{Cor}} = -2m\,\vec\Omega \times \vec v\,',$

with magnitude $2m\Omega v'\sin\phi$, where $\phi$ is the angle between $\vec\Omega$ and $\vec v\,'$, directed perpendicular to both. It vanishes for anything at rest in the frame, which is why everything above could ignore it.

An aircraft at $v = 200\ \mathrm{m/s}$ flying horizontally over the North Pole moves perpendicular to Earth's spin axis, so $\sin\phi = 1$ and the Coriolis acceleration is $2\Omega v = 2 \times 7.292\times10^{-5} \times 200 = 0.0292\ \mathrm{m/s^2}$. Tiny next to $g$, but it acts sideways for hours, and that is what spins hurricanes.

The rotating bucket

Spin a cylinder of water about its axis and the surface curves up at the rim. Use the surface rule again: the free surface is perpendicular to $\vec g_{\text{eff}}$, which at radius $r$ is $g$ down and $\omega^2 r$ outward. The slope of the surface is therefore

A parabola. The rise from the axis out to radius $r$ is $\omega^2 r^2/2g$: at $\omega = 8.00\ \mathrm{rad/s}$ and $r = 0.100\ \mathrm{m}$, that is $64.0 \times 0.0100/19.6 = 0.0327\ \mathrm{m}$.

To locate $z_0$ you need volume conservation, and one geometric fact settles it: a paraboloid fills exactly half the cylinder containing it, so the mean surface height is still the rest depth $H$. With total rise $h = \omega^2R^2/2g$ across the radius, the centre sits at $H - h/2$ and the rim at $H + h/2$.

Example 4. A bucket of radius $0.150\ \mathrm{m}$ holds water $0.200\ \mathrm{m}$ deep. At what $\omega$ does the surface just touch the centre of the bottom?

Touching the bottom means the centre has dropped the full depth: $h/2 = 0.200\ \mathrm{m}$, so $h = 0.400\ \mathrm{m}$. Then $\omega^2 R^2/(2g) = 0.400$, so $\omega^2 = 0.400 \times 2 \times 9.8/(0.150)^2 = 7.84/0.0225 = 348.4$, and $\omega = 18.67\ \mathrm{rad/s}$.

Spin faster and a disc of the bottom goes dry. The half-volume trick fails (the paraboloid is truncated) so integrate. With dry radius $r_0$ the surface is $z = \omega^2(r^2-r_0^2)/2g$ for $r \ge r_0$, and the water volume is

so $R^2 - r_0^2 = 2R\sqrt{gH}/\omega$. For $R = 0.120\ \mathrm{m}$, $H = 0.100\ \mathrm{m}$, $\omega = 25.0\ \mathrm{rad/s}$: $2 \times 0.120 \times \sqrt{0.980}/25.0 = 0.240\times0.9899/25.0 = 0.009504$. Then $r_0^2 = 0.0144 - 0.009504 = 0.004896$ and $r_0 = 0.0700\ \mathrm{m}$.

Two traps, and the Earth itself

The first trap is double counting. Centripetal force is not a new force; it is the name for the net inward force in the ground frame, while centrifugal force is a pseudo-force in the rotating frame. Write both into one equation and you have counted the same physics twice with opposite signs. Pick a frame, name it out loud, stay there.

The second trap is asking whether a pseudo-force is "real". Inside the frame it is as real as anything: it bends the water surface and it reads on the scale. It just has no source.

A last example. The Earth rotates, so a plumb bob does not point at the centre. Let true gravity be $g$ toward the centre and the latitude $\lambda$. The distance to the spin axis is $R\cos\lambda$, so the centrifugal term is $\Omega^2 R\cos\lambda$ away from the axis. Split it: $\Omega^2R\cos^2\lambda$ straight up, weakening apparent gravity, and $\Omega^2R\sin\lambda\cos\lambda$ horizontal toward the equator, tilting the plumb line by

$\tan\varepsilon = \frac{\Omega^2 R\sin\lambda\cos\lambda}{g - \Omega^2R\cos^2\lambda}.$

At $\lambda = 45.0^\circ$ with $\Omega = 7.292\times10^{-5}\ \mathrm{rad/s}$ and $R = 6.371\times10^6\ \mathrm{m}$: $\Omega^2 R = 0.03388\ \mathrm{m/s^2}$. Both $\sin45^\circ\cos45^\circ$ and $\cos^2 45^\circ$ are $0.500$, so the numerator is $0.01694$ and the denominator $9.8 - 0.01694 = 9.783$, giving $\tan\varepsilon = 0.001731$ and $\varepsilon = 0.0992^\circ$. At the equator the tilt vanishes and the whole term points up: $g_{\text{eff}} = 9.8 - 0.03388 = 9.766\ \mathrm{m/s^2}$.

Recognising the technique

Reach for a non-inertial frame when the problem hands you a container and asks about what is inside it: "inside the accelerating cart", "relative to the wedge", "the surface in the rotating bucket", "what does the scale read". Anything at rest with respect to something that is accelerating is a statics problem in disguise. Three questions:

1. Is there an object that stays put relative to an accelerating or rotating body? Go to that body's frame; the object is then in equilibrium.
2. Is the frame's acceleration uniform? If so, fold it into $\vec g_{\text{eff}}$ and treat the whole problem as ordinary statics under a tilted, stronger gravity: surfaces perpendicular to it, pendulums along it.
3. Is the frame rotating? Then the pseudo-force is $m\omega^2 r$ outward and grows with $r$, so write it as a function of position before you balance anything, and add the Coriolis term $2m\Omega v'$ if the object is also moving in that frame.

If none of those fit, the ground frame is probably still the better bet. The trick is a tool, not an obligation.

Next you leave mechanics behind and start on the force that runs the rest of the course, in Coulomb's Law and superposition.