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Homework Help: Gyroscope- Angular Speed, Rotational Kinematics, Precession

  1. Nov 1, 2008 #1
    Gyroscope-- Angular Speed, Rotational Kinematics, Precession

    1. The problem statement, all variables and given/known data

    A Gyro Stabilizer. The stabilizing gyroscope of a ship is a solid disk with a mass of 6.50×10^4 kg; its radius is 2.30m, and it rotates about a vertical axis with an angular speed of 500 rev/min.

    1) How much time is required to bring it up to speed, starting from rest, with a constant power input of 7.45×10^4 W?

    2)Find the torque needed to cause the axis to precess in a vertical fore-and-aft plane at an angular rate of 1.00 degree/sec.



    2. Relevant equations
    I may be incorrect as I apparently keep coming up with the wrong answer. However, the moment of inertia for a solid cylinder I=1/2mr^2. P=power, I=moment of inertia, w=angular speed, a=angular acceleration, T=torque.

    P=Tw. Ia=T. W=at.

    For the second question, I'm a lot more lost. I'm guessing it's Torque/angular momentum but I continue to get incorrect answers.

    3. The attempt at a solution

    Ok, perhaps it's an issue with me actually understanding what is given because I would assume you would need a length from a pivot in order to work with a gyroscope. Since there is no such info in the problem, I assume that the disk is rotating directly on the pivot point.

    I started with finding I. I=1/2(6.5*10^4)(2.3^2). I get 171925. Next, I use the Power equation to find the torque, using the given power and target angular speed. 7.45*10^4=T(500(rev/min). I convert 500 Rev/min to rads per second using 500*(2pi/60) to get 52.3599.

    Then, solving for T, (7.45*10^4)/52.3599 to get 1422.84.

    T=Ia. Using 1422.84=171925a, we find alpha to be .008276. Here is where I think there is something I may not be accounting for with torque due to the weight of the disc, but I assumed that since it is acting at the center of mass there is no torque. I tried the solution with torque being equal to mg + 1422.84, and the answer is still wrong.

    Anyway, now that we have a, and given that the gyro starts from rest, w=at. So, 52.3599/.008276 = t, which =6327.

    This answer is incorrect. Am I making incorrect assumptions somewhere as to using equations that wouldn't apply in this situation or something I am missing completely? Any help is much appreciated.

    As for the precession part, that can wait until after I've figured out the first part.
     
  2. jcsd
  3. Nov 3, 2008 #2
    Re: Gyroscope-- Angular Speed, Rotational Kinematics, Precession

    Any ideas?
     
  4. Nov 3, 2008 #3
    Re: Gyroscope-- Angular Speed, Rotational Kinematics, Precession

    Figured it out. My approach went off on a tangent because the angular velocity is not constant. Instead, you use the energy approach K=1/2Iw^2 and know that the work is equal to the change in kinetic energy, and Power is work over time. So, Power * change in time =W.

    1/2Iw^2=P*t, divide by the given power.
     
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