How Does Angular Momentum Affect Torque in a Suspended Hoop?

[decerto]

Homework Statement

This is problem 7.7 from [kleppners mechanics book]http://books.google.ie/books?id=Hmq...rough a point on the rim of the hoop"&f=false but its also a general question on understanding this type of problem.

Homework Equations

L=Iω
dL/dt=ΩL=rxF

The Attempt at a Solution

The standard approach appears to be compute the angular momentum from Iω and find how it changes with respect to time, generally its ΩL where Ω is the precession about the axis, then compute the relevant torque and set them equal and solve for whatever your looking for.My problems arises in 7.7 as the force creating the torque(the tension in the string) is at an angle to axis and the r from r x F is also at angle to the axis. So I'm not quite sure how I get the relevant force, I'm thinking it's as simple as r x (perp component of T) but I'm not 100%.

I'm also not sure since, I'm breaking down ω as a vector into components, perp and parallel to the hoop, then computing the spin angular momentum from the perp ω and ignoring the parallel ω one. But I'm not sure if you can just do that in this problem because the way the torque is setup, maybe I need to include the other component of angular momentum into it?.

Help would be much appreciated.

As they say solve it approximately with small angles, I assume that means cosx=1 and sinx=xRight now for dL/dt i.e ΩL I have MR^2 ω^2 β
and for the torque I have with T=Mg torque=RMg(1+αβ)

 

[voko]

Google does not want to show that part of book anymore. You will have to reproduce the problem here to get any help.

 

[decerto]

Google does not want to show that part of book anymore. You will have to reproduce the problem here to get any help.

Really? It shows me it just fine when I click on the link

http://www.maths.tcd.ie/~kovacs/Teaching/Mechanics/Kleppner-Kolenkow.pdf

page 352 question 7.7

 

[voko]

I have looked through the whole chapter in the book, and I have a question: how is the hoop different from a gyroscope tilted down?

 

[Nickie]

Hello,

Thank you for your question. In this problem, we are dealing with a rigid body rotation, which means that the object is rotating around a fixed axis. The key concept here is that the sum of torques acting on the object must be equal to the rate of change of angular momentum.

In this case, the torque is created by the tension in the string, which is acting at an angle to the axis of rotation. To find the relevant force, you can use the component method as you suggested, where you take the perpendicular component of the tension force and use that in your calculation of torque.

As for the components of angular momentum, you are correct in separating the perpendicular and parallel components. The perpendicular component will contribute to the spin angular momentum, while the parallel component will contribute to the precession angular momentum. In this problem, we are only concerned with the spin angular momentum, so we can ignore the parallel component.

To solve for the tension force, you can set the torque equal to ΩL and solve for T. You can then use this value of T in your calculation of the precession angular momentum.

I hope this helps. Let me know if you have any further questions.