An aircraft executes a horizontal loop of radius $1.00\ \text{km}$ with a steady speed of $900\ \text{km/h}$. Compare its centripetal acceleration with the acceleration due to gravity.

An aircraft executes a horizontal loop of radius 1.00km with a steady speed of 900km/h. Compare its centripeta

Exercises · 3.15

1.00\ \text{km}

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Concept understanding — Uniform Circular Motion

Uniform Circular Motion

Uniform circular motion is motion along a circular path at constant speed. Although the speed stays the same, the velocity does not — its direction keeps changing at every instant — so the motion has an acceleration even though the speed never changes. This concept covers the kinematics of that motion; the force that causes it (centripetal force) belongs to the Laws of Motion unit.

1. Angular Quantities

As the body sweeps through an angle $\theta$ , its angular velocity is

\omega = \frac{d\theta}{dt}

For one full revolution in a period $T$ , at frequency $f$ :

\omega = \frac{2\pi}{T} = 2\pi f, \qquad T = \frac{1}{f}

To convert rpm to rad/s:

\omega = \frac{2\pi \cdot (\text{rpm})}{60} = \frac{\pi}{30}\,(\text{rpm})

The linear (rim) speed $v$ and the angular speed $\omega$ are related by

v = \omega r

2. Centripetal Acceleration

Even though the speed is constant, the velocity vector keeps turning — this produces an acceleration directed toward the centre of the circle:

a_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r

Use whichever form matches the data you are given. This acceleration is often expressed as a multiple of $g$ (as $a_c/g$ , taking $g = 10\ \text{m/s}^2$ ).

3. Directions

The velocity is always tangential (along the direction of motion); the acceleration is centripetal — radial, pointing inward, and perpendicular to the velocity.

Because the direction of the velocity keeps changing, the change in the velocity vector over an angle $\Delta\theta$ has magnitude

|\Delta v| = 2v\sin\left(\frac{\Delta\theta}{2}\right)

So a quarter turn gives $|\Delta v| = v\sqrt{2}$ , a half turn gives $2v$ , and a full turn gives $0$ .

The average acceleration over an arc is $|\Delta v|/\Delta t$ — this is smaller in magnitude than the instantaneous centripetal acceleration $v^2/r$ , and its direction lies along the perpendicular bisector of the chord joining the two points (which, for a circle, always passes through the centre) rather than being radially inward from either endpoint's own position.

4. Points on a Rotating Body

Every point on a rigid rotating body (a wheel, a disc, a clock hand, the Earth) shares the same $\omega$ , but the rim speed $v = \omega r$ grows with the radius.

Clock hands: the second hand turns at $\omega = \dfrac{2\pi}{60}$ rad/s, the minute hand at $\dfrac{2\pi}{3600}$ rad/s, the hour hand at $\dfrac{2\pi}{43200}$ rad/s.
The Earth spins with $\omega = \dfrac{2\pi}{86400} \approx 7.3\times10^{-5}$ rad/s. A point on the equator moves at $\omega R \approx 465$ m/s, and a point at latitude $\lambda$ moves at $\omega R\cos\lambda$ (since the circle of latitude has radius $R\cos\lambda$ ).

5. Connected Systems

Wheels joined by a belt (or two gears in mesh) share the same rim speed, so $\omega_1 r_1 = \omega_2 r_2$ — the larger wheel turns with the smaller $\omega$ .
Wheels on the same axle (concentric) share the same $\omega$ , so the outer rim moves faster ( $v \propto r$ ).

6. Non-Uniform Circular Motion

If the speed also changes, there is a tangential acceleration

a_t = \frac{dv}{dt} = r\alpha

(from the angular acceleration $\alpha = d\omega/dt$ ), in addition to the radial $a_c = v^2/r$ . These two are perpendicular, so the total acceleration is

$a = \sqrt{a_c^2 + a_t^2}, \qquad \tan\beta = \frac{a_t}{a_c}$ …

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The centripetal acceleration of the aircraft is about 6.38 times the acceleration due to gravity. This is found by converting speed to m/s, using $a_c = v^2/r$ , and dividing by $g = 9.8\ \text{m/s}^2$ .

The problem asks you to compare two accelerations — one from circular motion, one from gravity. The comparison is just a ratio, but the real insight is that centripetal acceleration depends only on speed and radius, not on mass. So a plane, a car, or a satellite on the same circular path would all feel the same inward pull per unit mass.

The key formula is the centripetal acceleration:

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

where $v$ is the speed along the circular path and $r$ is the radius. The acceleration due to gravity near Earth’s surface is $g = 9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ if an approximation is allowed, but we’ll use the standard value).

Now, the trap most students fall into: mixing units. The radius is given in kilometres, the speed in km/h. You cannot plug these directly into $v^2/r$ and expect to get $\text{m/s}^2$ . You must convert everything to SI units (metres and seconds) first.

Let’s work through it step by step.

Convert the radius to metres.

$r = 1.00\ \text{km} = 1000\ \text{m}$ .
Convert the speed from km/h to m/s.

The conversion factor: $1\ \text{km/h} = \frac{1000\ \text{m}}{3600\ \text{s}} = \frac{5}{18}\ \text{m/s}$ .

So $v = 900\ \text{km/h} = 900 \times \frac{5}{18}\ \text{m/s}$ .

$900 \div 18 = 50$ , then $50 \times 5 = 250$ .

Hence $v = 250\ \text{m/s}$ .

Tip

A quick mental shortcut: to convert km/h to m/s, multiply by $5/18$ . For 900, just do $900/18 = 50$ , then $50 \times 5 = 250$ . No calculator needed.
Calculate the centripetal acceleration.

a_c = \frac{v^2}{r} = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5\ \text{m/s}^2.

Compare with $g$ . …

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Concept: Centripetal Acceleration via the Period of the Loop

Method: Find the Loop's Time Period $T$ First, Then Use $a_c=\omega^2 r$ with $\omega=2\pi/T$ (Not the Direct $v^2/r$ Substitution)

The existing solutions substitute the converted speed and radius directly into $a_c=v^2/r$ . This method instead finds how long the aircraft takes to complete one full loop (its period $T$ ), converts that into an angular speed $\omega$ , and only then computes $a_c=\omega^2r$ — a route through time rather than straight through speed.

Step 1 — Convert to SI units (unavoidable first step either way)

r = 1.00\ \text{km} = 1000\ \text{m}, \qquad v = 900\ \text{km/h} = 900\times\frac{5}{18} = 250\ \text{m/s}

Step 2 — Find the period: how long one full loop takes

The aircraft covers the circle's entire circumference, $2\pi r$ , once every period $T$ :

T = \frac{2\pi r}{v} = \frac{2\pi(1000)}{250} = 8\pi\ \text{s} \approx 25.13\ \text{s}

Step 3 — Convert the period into an angular speed

\omega = \frac{2\pi}{T} = \frac{2\pi}{8\pi} = \frac{1}{4} = 0.25\ \text{rad/s}

(This is consistent with $\omega=v/r=250/1000=0.25\ \text{rad/s}$ computed the direct way — a useful cross-check that the period was found correctly.)

Step 4 — Centripetal acceleration from $\omega$

a_c = \omega^2 r = (0.25)^2(1000) = 0.0625\times1000 = 62.5\ \text{m/s}^2

Step 5 — Compare with $g$

$\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38$ …

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Q.An aircraft executes a horizontal loop of radius 1.00 km1.00\ \text{km}1.00 km with a steady speed of 900 km/h900\ \text{km/h}900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

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Q.An aircraft executes a horizontal loop of radius $1.00\ \text{km}$ with a steady speed of $900\ \text{km/h}$ . Compare its centripetal acceleration with the acceleration due to gravity.

Unlock everything free for 14 days