Question

IIn traveling to the Moon, astronauts aboard the Apollo spacecraft put the spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine the angular acceleration, and the radial and tangential components of the linear acceleration of a point on the skin of the ship 6.0 min after it started this acceleration.

Answer 1

Answer:

α = 4.44*10^-5 rev/s^2

ar = 1.088*10^{-3}m/s^2

at = 1.88*10^{-4}m/s^2

Explanation:

T of ind the angular acceleration you can use the following formula from an accelerated harmonic motion:

$\omega=\omega_o+\alpha t\\\alpha=\frac{\omega-\omega_o}{t}$

w: angular velocity

w_o: initial angular velocity

t: time

α: angular acceleration

In this case you have that the cylinder increase its angular speed 1.0rev/min until 12min. The angular acceleration must be computed for 6 min. Hence, by replacing you obtain:

$\omega=1.0\frac{rev}{min}*\frac{1min}{60s}=0.016rev/s\\\alpha=\frac{0.016rev/s}{6(60s)}=4.44*10^{-5}\frac{rev}{s^2}$

hence, the angular acceleration is 4.44*10^-5 rev/s^2

the linear and tangential acceleration are calculated by using the formulas:

$a_t=\alpha r\\\\a_r=\frac{v^2}{r}$

where v is the speed of the border of the cylinder, that is for r=8.5m/2=4.25m. By replacing you obtain:

$a_t=(4.44*10^{-5}rev/s^2)(4.25m)=1.88*10^{-4}m/s^2\\\\v=\omega r=(0.016rev/s)(4.25m)=0.068m/s\\\\a_r=\frac{(0.068m/s)^2}{(4.25m)}=1.088*10^{-3}m/s^2$

hence, the radial acceleration is 1.088*10^{-3}m/s^2 and the tangential acceleration is 1.88*10^{-4}m/s^2