Answer to Question #103775 in Classical Mechanics for Ellen Sprot

Question

1. Suppose that an astronaut, wearing a space suit on Earth, can kick a ball a distance of 25 m. Using the datum gravitational acceleration is g = 1.6 m/s^2, estimate how far he could kick the ball on the moon, using the same action and kicking at the same angle. (Remember significant figures and neglect air resistance. Assume that the launch and landing heights are the same. Do not use exponent notation.)


2. This artist's conception imagines a 'wheel-shaped' space station. The idea is that the 'wheel' would turn on its axis at a rate such that the acceleration in the rim of the wheel would be 9.8 m/s^2. (This would make inhabitants feel as though they had their normal weight.) Suppose that the radius of the wheel is 1.0x10^2 m. What is the required period of rotation of the space station about its axis?
Period of rotation = _____ s. (Enter the number without using exponent notation.)

Answer 1

a) As per the given question,

Astronaut kick the ball at 25 m distance, on the surface of earth

we know that horizontal range of the projectile, when kicked at any angle is

$R=\dfrac{u^2 \sin2\theta}{g}$

we know that gravitational acceleration on the surface of the earth g=9.8 m/sec^2

as per the question, gravitational acceleration on the surface of the moon $g_m=1.6 m/sec^2$

As per the condition given in the question, let the range of the projectile at the surface of the moon is $R_1$ ,

taking the ratio of range on the surface of the moon tot he surface of the earth

$\dfrac{R}{R_1}=\dfrac{g_m}{g}$

$R_1=\dfrac{Rg}{g_m}=\dfrac{25\times9.8}{1.6}m$

$R_1=153.125 m$

b) As per the question,

acceleration on the wheel shaped space station (a) $= 9.8 m/sec^2$

and radius of the wheel shaped space station (R)=$1.0\times10^2 m$

we know that $a= R\omega^2$

where $\omega$ =angular speed of the space station.

$\omega=\sqrt{\dfrac{a}{R}}=\sqrt{\dfrac{9.8}{100}}=\sqrt{0.098}$

$\omega=0.313 m/sec$

We know that,

$\omega=\dfrac{2\pi}{T}$

where T is the time period

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{0.313}=20.07Sec$