In traveling to the Moon, astronauts aboard the Apollo spacecr...

Question

In 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 for 12 minutes from no rotation to 1.0 revolution per minute which they then maintained. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine 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.

Accepted Answer

Welcome back everyone in this problem to stabilize its orientation. For a communication link. A satellite initiates a maneuver to rotate about its axis initially stationary. The satellite accelerates to achieve a rotational speed of two revolutions per minute in just 10 seconds, assume the satellite is cylindrical with a radius of 1 m rotating about its longitudinal axis. Calculate both the radial and tangential components of the linear acceleration of a point on the outer surface of the satellite. Five seconds after it begins this maneuver for our answer choices. A says the radial acceleration is 1.1 multiplied by 10 to the negative 2 m per second squared. And the tangential acceleration is 2.1 multiplied by 10 to the negative 2 m per second. Squared. B says they are 2.2 multiplied by 10 to the negative two and 1.1 multiplied by 10 to the negative 2 m per second squared respectively. C says they are 4.2 multiplied by 10 to the negative two and 1.1 multiplied by 10 to the negative 1 m per second squared respectively. And the D says they are 1.1 multiplied by 10 to the negative one and 2.1 multiplied by 10 to the negative 1 m per second squared respectively. Now, if we're going to figure out the radial and tangential components of the linear acceleration at our point, let's first make note of all the information that we already have. So so far, we know that the satellite accelerates to achieve a rotational speed of two revolutions per minute. So let's convert that to SI units here. Let's convert that for our angular velocity omega. So this is telling us that our final o final angular velocity omega F equals two revolutions per minute, which we can multiply it by two pi radians divided by 60 seconds to convert it to SI units. Which when we do that, we get it to be 0.209 radiance per second. OK. So that's our final angular velocity. Next, we know that the final time. OK. It achieves this velocity in is 10 seconds. We also know that our cylindrical satellite has a radius of 1 m and we want to figure out the radial and tangential components five seconds after it begins the maneuver. So t actual t equals five seconds while our final time is 10 seconds. So if we're going to figure out our radial acceleration and our tangential acceleration, let's first ask ourselves, what do we know about acceleration for this particular motion? Well, recall, OK. If you think about acceleration, recall that our angular acceleration is going to be equal to the change in angular velocity divided by the change in time. In this case, we have our final angular velocity and our final time. So that means the angular acceleration is going to be uh 0.209 radiance per second. Our final velocity minus or rather I should leave. So this has the change in w because it's going to be the final minus the initial. So our final is 0.209 radiance per second. And our final is our initial sorry, angular velocity is zero because it was initially stationary. So we subtract zero from our change in time which is 10 seconds minus zero, which is just 10 seconds. Now, when we do that, we're dividing 0.209 by 10. So we get our angular acceleration to be 0.0209 radiance per square second. Now that we have our angular acceleration, how can we relate it to our radial and tangential acceleration at that point? Well, starting with our radial acceleration, recall that radar acceleration is going to be equal to or it's going to be equal to our angular velocity squared divided by R at that point at t being equal to five seconds. We know what our final angular velocity is, but we don't know what it is that T equals five seconds. However, we can figure that out using our angular acceleration because if it's accelerating at a rate of 0.0209 radiance per square seconds. Ok. Then at five seconds at five seconds, its angular velocity is going to be equal to 0.0209 radiance per second squared multiplied by five seconds. And again, that makes sense because speed is defined as acceleration multiplied by time. OK. In other words, our velocity is gaining 0.0209 radiance per second for every second. So when we do that, then we can find Omega at T equals five seconds to be 0.104 radiance per second. Now that we have the value for omega, that means our radar acceleration A is going to be equal to 0.104 radiance per second squared divided by R which is 1 m. And when we do that, OK, when we do that, we should find that we get it to be equal to 0.0108 m per second squared which when rounded to or when written in standard form, rounded to two significant figures, we can write it as 1.1 multiplied by 10 to the negative 2 m per second. So that's our radial acceleration. Now, what about our tangential acceleration? Well, for a tangential acceleration recall that the tangential acceleration is equal to alpha multiplied by R OK. Or angular acceleration multiplied by the radius, we know our angular acceleration is zero point 029. Ok. So that's 0.029 radius per square seconds. And our radius is 1 m. So that means our tangential acceleration is therefore equal to 0.0209 m per second squared. Again, if we write that uh to two significant figures, then it's going to be 2.1 multiplied by 10 to the negative 2 m per square second. And likewise our previous radar acceleration, my apologies, this unit should have been in square seconds. Therefore, our radial acceleration is 1.1 multiplied by 10 to the negative 2 m per second squared. And our tangential acceleration is 2.1 multiplied by 10 to the negative 2 m per second squared. If we go back to our answer choices, that means a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Angular Velocity

Centripetal Acceleration

Tangential Acceleration

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