1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotating hoop (gyroscope?)

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Kleppner & Kolenkow 7.7
    A thin hoop of mass M and radius R is suspended from a string
    through a point on the rim of the hoop. If the support is turned with
    high angular velocity [tex]\omega[/tex], the hoop will spin as shown, with its plane nearly
    horizontal and its center nearly on the axis of the support. The string
    makes angle [tex]\alpha[/tex] with the vertical.
    Find, approximately, the small angle ([tex]\beta[/tex] between the plane of the
    hoop and the horizontal.

    2. Relevant equations

    3. The attempt at a solution

    torque = RMg cos[tex]\beta[/tex]
    angular momentum = I[tex]\omega[/tex]

    I don't know how to link these equations. I'm not really sure if this is a gyroscope problem to use its formula. Can anyone help me please?
    And one question, how do I determine if a system is precessing, means it's a gyroscopic problem or not?

    Attached Files:

  2. jcsd
  3. Dec 10, 2009 #2
    anyone please?
  4. Dec 10, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    Well, the torque on the loop (about a certain choice of origin) is just the time rate of change of the loops angular momentum (with respect to the same origin) isn't it?

    So, choose an appropriate reference point (like the point where the string attaches to the support) and calculate the toque about that reference point (it is not just RMg cos[itex]\beta [/itex]) and the angular momentum of the loop avout that same reference point and then set

    [itex]\mathbf{\tau}=\frac{d\textbf{L}}{dt}[/itex] and solve for [itex]\beta[/itex]....
  5. Apr 28, 2011 #4
    how do you find torque and angular momentum of hoop about that point??? the com of rope is not moving with respect to that point, so it seems that torque and angular momentum about that point is zero.
  6. Apr 28, 2011 #5
    http://books.google.com/books?id=Hm...ough a point on the rim of the hoop"&f=false"

    It says 'approximately' so i think what you're supposed to do is this.
    it tells that beta is very small which is an invitation for cos(beta)=1 sin(beta)=beta

    You just need to find the torque on the hoop caused by the rope (which is around the axis that goes through centre of the hoop and orthogonally to the axis of rotation and the line to rope attachment point), which is RMg, and which is equal to counter- torque caused by centrifugal force, which is an integral over the hoop.

    RMg = beta*M*omega2*(integral from a=0 to a=2pi (sin(a)*R)2*da )/(2pi) =
    beta*M*omega2*R2(integral from a=0 to a=2pi sin(a)2*da )/(2pi)
    the indefinite integral for sin(a)2*da is 0.5*(a-sin(a)*cos(a)) + const so the definite integral is pi so that neatly simplifies out into:
    RMg= beta* M*omega2*R2 / 2
    g = beta* omega2*R / 2
    beta = (2*g)/ omega2*R

    I have neglected the distance between centre of the hoop and axis of rotation (its said to be small). I'm not sure if it is 'approximately' good enough but judging by other problems in the book, its OK.

    On the second thought, it may be better to do the torque around the point where hoop is suspended, then you can easier calculate it without neglecting the distance from centre of mass and axis of rotation. I would guess though that the problem calls for neglecting that distance.

    Sorry, TEX does not refresh for me when I preview, do not know why, so no tex.
    Last edited by a moderator: Apr 25, 2017
  7. Apr 28, 2011 #6
    may i ask what is your education level? I'm 3rd year undergrad and these problems are very difficult for me, whereas you knocked it out with no problems.
    Last edited by a moderator: Apr 25, 2017
  8. Apr 28, 2011 #7
    Hmm I'd recommend you to doublecheck it first, I am rusty with this (and probably there's also some way that doesn't involve the integral, i just pretty much apply brute calculus to such stuff) and I have headache.
    For the precession, I didn't answer the question - I don't think there's anything complicated like that going on. The thing basically spins as a solid, around the axis, every tiny piece of mass simply spins around the axis in a horizontal circle, and all forces must balance out. It can be much more complicated if it was really dynamic, with the hoop bobbling.

    Actually, just highschool. But I got lucky to have PhD physics and math teachers in HS and they thought I was going to be some sort of next Feynman or something. Couldn't do that, after HS I had to move to another country and had to work for living, but got lucky to work with some PhD people. Currently living off royalties from this http://store.steampowered.com/app/67000
    (shameless self promotion)
    Last edited: Apr 28, 2011
  9. Oct 19, 2011 #8
    I didn't integrate anything and got the same answer as you.
    I used [itex] L=L_s cos(\beta) \hat{j} + L_s sin(\beta) (-\hat{r})[/itex]
    [itex]\tau=RT sin(\frac{\pi}{2} - \alpha + \beta) \hat{\theta}[/itex]
    (got T from F=ma in x and y)

    then used [itex]\frac{dL}{dt}=\tau[/itex]
    and got [itex]\beta=\frac{2g}{R\omega^2}[/itex]

    Can anyone else confirm that this is the answer ?
    Thank you.
  10. Nov 14, 2013 #9
    [tex]\text{elementary vectors } \hat{r},\hat{\theta},\hat{k}\\
    \vec{\omega} = \; <\omega \sin \beta\,,0,\omega\cos\beta\,>\\
    \vec{L} = \stackrel{\leftrightarrow }{I} \vec{\omega} = \frac{MR^2}{2}\omega \sin\beta\,\hat{r} + MR^2 \omega \cos\beta\,\hat{k}\\
    \text{Because for a disk, moments of inertia are: }I_\hat{k} = MR^2,\quad I_\hat{r}=\frac{MR^2}{2}\\
    \vec{\tau}=\frac{d}{dt}\vec{L} = \frac{MR^2}{2}\omega \sin\beta\,\dot{\hat{r}} + MR^2 \omega\cos \beta\,\dot{\hat{k}}\\
    \dot{\hat{r}}= \dot{\theta}\hat{\theta}= \omega\hat{\theta} \text{ (by geometry)},\; \dot{\hat{k}}=0\\
    \vec{\tau}= \frac{MR^2}{2}\omega^2\sin \beta\,\hat{\theta}\\
    = RMg\cos\beta\, \hat{\theta} \quad\text{ (torque due to gravity)}\\
    \text{Using small angle approximations }\sin\beta \approx \beta, \cos\beta \approx 1\text{, equating torques:}\\
    Last edited: Nov 14, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted