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## 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=x

Right 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)