An aircraft executes a horizontal loop of radius $1.00 \; km$ with a steady speed of $900 \; km/h$. Compare its centripetal acceleration with the acceleration due to gravity.
(N/A) Radius of the loop,$r = 1 \; km = 1000 \; m$.
Speed of the aircraft,$v = 900 \; km/h = 900 \times \frac{5}{18} = 250 \; m/s$.
Centripetal acceleration,$a_c = \frac{v^2}{r}$.
$a_c = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5 \; m/s^2$.
Acceleration due to gravity,$g = 9.8 \; m/s^2$.
Comparing the two,$\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38$.
Therefore,the centripetal acceleration is $6.38$ times the acceleration due to gravity,i.e.,$a_c = 6.38 \; g$.
Speed of the aircraft,$v = 900 \; km/h = 900 \times \frac{5}{18} = 250 \; m/s$.
Centripetal acceleration,$a_c = \frac{v^2}{r}$.
$a_c = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5 \; m/s^2$.
Acceleration due to gravity,$g = 9.8 \; m/s^2$.
Comparing the two,$\frac{a_c}{g} = \frac{62.5}{9.8} \approx 6.38$.
Therefore,the centripetal acceleration is $6.38$ times the acceleration due to gravity,i.e.,$a_c = 6.38 \; g$.
Explore More
Similar Questions
$A$ plane starts its flight in a direction $\theta$ with the runway. If the distance covered by it in both horizontal and vertical directions is $600\, m$,then find $\theta$. (in $^{\circ}$)
Medium
View Solution$A$ point moves in the $xy$-plane according to the following equations: $x = a \sin \omega t$ and $y = a(1 - \cos \omega t)$, where $a$ and $\omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.
If the radius of the circular path and the frequency of revolution of a particle of mass $m$ are doubled,then the change in its kinetic energy will be ($E_i$ and $E_f$ are the initial and final kinetic energies of the particle respectively). (in $E_i$)
DifficultMHT CET 2019
View Solution$A$ particle moves with a constant speed of $5 \, ms^{-1}$ as shown in the figure. What is the change in velocity during half a revolution in $ms^{-1}$?
Medium
View Solution$A$ particle crossing the origin at time $t=0$,moves in the $xy$-plane with a constant acceleration $a$ in the $y$-direction. If the equation of motion of the particle is $y = bx^2$ (where $b$ is a constant),then its velocity component in the $x$-direction is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo