Irodov 1.84: The Middle of the Loop Needs Two Axes – Amit & Deepali Sinha Foundation

Problem Irodov 1.84

An aircraft flies a vertical circular loop of radius $R=500\text{ m}$ at the constant speed $v=360\text{ km/h}$ . A flyer of mass $m=70\text{ kg}$ rides in it. Find the flyer’s apparent weight, the magnitude of the force the seat exerts on the flyer, at three points of the loop: the lowest point, the highest point, and a point level with the center (halfway up, where the velocity is vertical).

The vertical loop, traveled at constant speed v, with its three points of interest: lower (bottom of the circle), upper (top), and middle (level with the center O, one quarter turn from either). — The vertical loop, traveled at constant speed $v$ , with its three points of interest: lower (bottom of the circle), upper (top), and middle (level with the center $O$ , one quarter turn from either).

Given

$R=500\text{ m}$
$v=360\text{ km/h}$
$m=70\text{ kg}$

Find

the seat force

N

at the lower, upper, and middle points

Solution

Step 1Convert to SI and find the centripetal acceleration

The speed converts to meters per second, and since the loop is flown at constant speed the acceleration everywhere on it is purely centripetal, of the same modulus.

\begin{align*}v &= 360\ \text{km/h} = 100\ \text{m/s} \\ a_c &= \frac{v^2}{R} = \frac{(100\ \text{m/s})^2}{500\ \text{m}}\end{align*}

a_c = \dfrac{v^2}{R} = 20\ \text{m/s}^2

(1)

Step 2Lower point: seat force and weight are collinear

At the lower point the center lies directly above the flyer, so the centripetal acceleration $a_c$ points straight up. The seat force $N$ (up) and gravity (down) are the only forces, and their sum along the vertical gives $a_c$ .

N_{\mathrm{low}}-mg = ma_c

Solving for the seat force and substituting $m=70\ \text{kg}$ , $g=9.8\ \text{m/s}^2$ and (1),

N_{\mathrm{low}} = m(g+a_c) = (70\ \text{kg})(9.8\ \text{m/s}^2+20\ \text{m/s}^2)

N_{\mathrm{low}} = m(g+a_c) = 2086\ \text{N} \approx 2.1\ \text{kN}

(2)

Step 3Upper point: both forces point toward the center

At the upper point the center lies directly below, so $a_c$ points straight down, the same sense as gravity. Both $N$ (now directed down, the seat pressing the flyer toward the center) and $mg$ point the same way.

N_{\mathrm{up}}+mg = ma_c

Solving for the seat force and substituting the same numbers,

N_{\mathrm{up}} = m(a_c-g) = (70\ \text{kg})(20\ \text{m/s}^2-9.8\ \text{m/s}^2)

N_{\mathrm{up}} = m(a_c-g) = 714\ \text{N} \approx 0.7\ \text{kN}

(3)

Step 4Middle point: resolve the seat force onto two axes

Free body at the middle point: the seat force $\vect N$ , horizontal toward the center, and the weight $m\vect g$ , vertical. Unlike at the lower and upper points, the two do not share a line.

Here the center is horizontal from the flyer, so $a_c$ has no vertical component: gravity alone is unbalanced vertically, and the seat force alone supplies the whole horizontal, center-directed pull (figure above). Newton’s second law along each axis,

\begin{align*}N_{\mathrm{mid},x} &= ma_c \\ N_{\mathrm{mid},y} &= mg\end{align*}

gives the seat force as the vector sum of a purely horizontal and a purely vertical part, so its modulus follows from Pythagoras.

N_{\mathrm{mid}} = \sqrt{N_{\mathrm{mid},x}^2+N_{\mathrm{mid},y}^2} = m\sqrt{a_c^2+g^2}

Substituting the same numbers once more,

N_{\mathrm{mid}} = (70\ \text{kg})\sqrt{(20\ \text{m/s}^2)^2+(9.8\ \text{m/s}^2)^2}

N_{\mathrm{mid}} = m\sqrt{a_c^2+g^2} = 1559\ \text{N} \approx 1.56\ \text{kN}

(4)

Check

Dimensionally each boxed result is a mass times an acceleration, hence a force, and $N_{\mathrm{up}}<N_{\mathrm{mid}}<N_{\mathrm{low}}$ as it must: gravity works against the seat at the top, with it at the bottom, and at a right angle to it in between, so the middle value lands strictly between the two extremes. The book’s printed key rounds these to $2.1$ , $0.7$ and $1.5$ kN; the third is a loose down-rounding of the exact $1559\ \text{N}$ found above, since the same nearest- $0.1$ -kN convention that gives the other two ( $2086\to2.1$ , $714\to0.7$ ) would round $1559$ to $1.6$ , not $1.5$ . The exact value, not either rounded form, is what this solution carries.

Remark

The middle point’s force is not the average of the other two, nor does it split the difference in any simple way: it is the hypotenuse of a right triangle whose legs are $ma_c$ (horizontal, the same at every point of the loop) and $mg$ (vertical, always present). The lower and upper points are just the two special cases where that triangle degenerates to a single line, once with the legs adding and once with them subtracting.