Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
8
Rotational Motion
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8-1: Angular Quantities
8-2 and 8-3: Constant Angular Acceleration; Rolling
8-4: Torque
8-5 and 8-6: Rotational Dynamics
8-7: Rotational Kinetic Energy
8-8: Angular Momentum
8-9: Angular Quantities as Vectors

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 15
Q

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 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

  1. the angular acceleration, and
  2. 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.
A
  1. 1.5×104 rad/s21.5 \times 10^{-4} \textrm{ rad/s}^2
  2. aT=6.2×104 m/s2aR=1.2×102 m/s2a_T = 6.2 \times 10^{-4} \textrm{ m/s}^2 \textrm{, } a_R = 1.2 \times 10^{-2} \textrm{ m/s}^2
Giancoli 7th Edition, Chapter 8, Problem 15 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The angular acceleration of this Apollo spacecraft is gonna be the final angular velocity minus the initial angular velocity divided by the time it takes to make that change. So we need to know what the angular velocities are in radians per second so we'll convert this one revolution per minute into radians per second by multiplying it by 1 minute for every 60 seconds and then multiply by 2π radians for every revolution leaving us with radians per second that's 0.1047 radians per second. So we plug that in for ω f and then minus 0 because it starts with no angular velocity and divide that by 12 minutes converted into seconds by multiplying by 60 seconds per minute and we end up with 1.5 times 10 to the minus 4 radians per second every second is the angular acceleration. And for the tangental and radial components of the acceleration of a point on the skin of the ship, we'll find the tangental component of the acceleration by going the radius of the ship multiplied by the angular acceleration. So the 8.5 meter diameter has to get divided by 2 to get radius multiplied by the 1.45444 times 10 to the minus 4 radians per second squared that we found in part (a). I didn't write out the full number there unrounded in this part here but in using a value in a subsequent calculation, you should always use the unrounded answer to avoid intermediate rounding error— you don't want to use 1.5 times 10 to the minus 4 here for example. So we get 6.2 times 10 to the minus 4 meters per second squared is the tangental component of acceleration of a point on the skin of the ship. And then the radial component of its acceleration: we need to take this final angular velocity after 6 mintues so we take the initial angular velocity of 0 plus the angular acceleration that we found multiply by 6 minutes converted into seconds and we end up with 0.0523599 radians per second. And then square that, multiply by the radius and we get 1.2 times 10 to the minus 2 meters per second squared is the radial component of the ship's acceleration.

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