In traveling to the Moon, astronauts aboard the Apollo spacecraft put themselves into a slow rotation to distribute the sun's energy evenly. At the start of their trip they accelerated from no rotation to 1revolution every minute during a 12 min time interval. THe spacecrafy can be thought of as a cylindar with a diameter of 8.5m. Determin the angular acceleration and the radial and tengential components of the linear acceleration of a point on the skin of the ship 5 min after it started this acceleration.
2006-10-31 06:55:49 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
This is just simple arithmetic. Get out your calculator and get to work.
2006-10-31 06:59:55 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Angular acceleration is the change in angular velocity divided by the number of seconds required to affect that change.
Tangential acceleration is the angular acceleration times the radius.
Radial acceleration is 2 times the angular acceleration times the radius.
Doug
2006-10-31 15:07:20 · answer #2 · answered by doug_donaghue 7 · 1⤊ 0⤋
First change all quantities into SI units
Initial angular velocity = 0
final angular velocity = 1 rpm = 2pi/60 rad/s
time = 12 min= 720s
angular acceleration = ?
final angular velocity = Initial angular velocity + angular accn*time
2 pi/60=0 +ang accn*720
ang accn= 0.000145444104 rad/s/s
Angular accn is constant so:
Tangential linear acceleration = Radius* ang accn
Tangential linear acceleration = 8.5* 0.000145444104
=0.00123627488
=1.23E-3 m/s/s
Five minutes into the accn
Initial angular velocity = 0
final angular velocity = ?
time = 5 min= 300s
angular acceleration = 0.000145444104 rad/s/s
final angular velocity = Initial angular velocity + ang accn*time
=0 + 0.000145444104 *300
Final angular velocity = 0.0436332312 rad/s
The radial acceleration uses the centripetal formula
a = V^2/R
As linear velocity= radius * angular velocity
V=8.5*0.0436332312
v=0.370882465m/s
a = V^2/R
a= (0.370882465)^2/8.5
=0.0161828003m/s/s
a (radial) = 0.0162m/s/s toward the centre of the circle.
2006-10-31 15:22:33 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Hi. Interesting, I didn't know that. (I knew they rotated but not that it took 12 minutes to develop the rotation.) Assuming the thrusters were turned on for the whole acceleration period (not likely) then they would have sped up 1/12 rev/min or about .083 r/m. (Divide by 60 and we have about .0014. rev/sec.) .083 * 5=.415 rev/min after 5 minutes.
2006-10-31 15:03:12 · answer #4 · answered by Cirric 7 · 1⤊ 0⤋
I'll get right on that after I floss my dog's teeth.
2006-10-31 14:57:17 · answer #5 · answered by martin h 6 · 1⤊ 1⤋