Question

A spaceship captain lands on an unknown planet. Before venturing forth, he needs to find out the acceleration due to gravity on that planet. All he has available to him is some thin light string, a stopwatch, and a small 2.75-kg metal ball (it was a rough landing). So he lets the ball swing from a 1.5-m length of the string, starting at rest, and measures that it takes 1.9 s for it to swing from the place where he released it to the place where it first stops as it reverses direction. What is the acceleration due to gravity on this planet

Answer 1

g = 4,10 m/s

Step-by-step explanation:

Este es un ejercicio de péndulo simple, donde la velocidad angular esta dada por

w =
`\sqrt{ (g)/(L) }`

la velocidad angular esta relacionada con el periodo

w =2π/T

sustituimos

`(2\pi )/(T) = \sqrt{ (g)/(L) }`2pi/T = Ra g/L

T² =
`4\pi ^2 ( L)/(g)`

g =
`4\pi ^2 \ (L)/(T^2)`

el periodo es el tiempo de una oscilación completa, osea desde el punto donde se suelta hasta que regresa al mismo punto, en este caso tenemos medio periodo, desde donde se suelta hasta donde se detiene

T/2 = 1,9

T = 2 1,9

T = 3,8 s

calculemos

g = 4π² 1,5 /3,8²

g = 4,10 m/s