Question

A plane is pulling out for a dive at a speed $720\text{km/h}$. Assuming its path to be vertical circle of radius $1000\text{m}$ and its mass to be $15000\text{kg}$, find the force exerted by the air on it at the lowest point.
(Take $g=10{\text{m/s}}^2$)

• A. $500\text{kN}$
• B. $750\text{kN}$
• C. $147\text{kN}$
• D. $600\text{kN}$

Answer 1

The correct option is B $750\text{kN}$

Given that,
Mass of plane, $m=15000\text{kg}$
Radius of circle, $R=1000\text{m}$
Speed of plane, $v=720\text{km/h}=720\times \frac{5}{18}=200\text{m/s}$


Let $F$ be the force exerted by the air on the plane at the lowest point.
At the lowest point of the journey, the centripetal acceleration is vertically upwards (towards the centre) and its magnitude is $v^2/R$
$\therefore$ Newton's second law of motion gives
$F-mg=m\frac{v^2}{R}$
$F=mg+\frac{m{v}^2}{R}=m(g+\frac{v^2}{R})$

$=15000(10+\frac{200\times 200}{1000})=750\text{kN}$

Why this question? Tips: At high speed, the force exerted by air is considerable.