Solving Homework Equations: Your Step-by-Step Guide

[rado5]

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi rado5!

(have an alpha: α and an omega: ω and a tau: τ )

to clarify: is this a mass m on a string of length l and tension T rotating at angle α with angular velocity ω?

or is there also a horizontal cable with tension F?

assuming the former, you've calculated τ (about the vertical axis) using the wrong force …

τ is r x -mgz, not r x F

(T and F don't count because they go through the vertical axis; F also doesn't count if there's no horizontal cable, because then you just invented F)

 

[rado5]

Hi tiny-tim!

Thank you very much for your kind help.

Yes, this is a mass m on a string of length l and tension T rotating at angle α with angular velocity ω.

 

[tiny-tim]

hi rado5!

i'm getting confused … it would be easier if you typed your answer, instead of giving us a photo of your handwriting

you're trying to do τ = r x F_net = dL/dt

but about the vertical axis, L is constant, and τ is zero

we solve this sort of problem with the ordinary linear F = ma equation​

 

[rado5]

Hi tiny-tim!

you're trying to do τ = r x F_net = dL/dt

but about the vertical axis, L is constant, and τ is zero

we solve this sort of problem with the ordinary linear F = ma equation​

There are two problems in my book. The first one is an example with its solution. As I told you before the solution is \vec{\tau}=mglsin\alpha \vec{e_{\theta}} for the problem \vec{\tau}= \vec{r} \times \vec{F}. The second one is a problem with no solution which asks "Is \vec{\tau}= \frac{d\vec{L}}{dt} correct for \vec{\tau}= \vec{r} \times \vec{F} in the first example". I have been trying to show that it must be correct. I mean I have to show that \vec{\tau}= \vec{r} \times \vec{F} = \frac{d\vec{L}}{dt}.

 

[tiny-tim]

oh i see now!

the confusion is that you're not treating τ and L as vectors

your τ is calculated about the top of the string, and is r x -mgz,

with (as you say) magnitude mglsinα, and direction tangential

your L (calculated about the top of the string) is r x v, which is sticking up diagonally outward

L's vertical component is constant, so you need only bother with d/dt of its horizontal component …

that should give you the required τ = dL/dt

 

[rado5]

Hi tiny-tim!

Thank you very much for your kind help.

I think my book solved the first example in a bad way!

I was very naive about \vec{L}, because I wrote \vec{L} = mlr \omega \vec{e_{r}} which is wrong! Because \vec{r} = lsin \alpha \vec{e_{r}} - lcos \alpha \vec{k} and \vec{v} = r \omega \vec{e_{\theta}} so \vec{L} = \vec{r} \times mv = mlr \omega cos \alpha \vec{e_{r}} + mlr \omega sin \alpha \vec{k}.

Now we have \vec{\tau}= \frac{d \vec{L}}{dt} = mlr \omega ^{2} cos \alpha \vec{e_{\theta}} = lFcos \alpha \vec{e_{\theta}}.

\vec{\tau} = lmgtan \alpha cos \alpha \vec{e_{\theta}}

\vec{\tau} = mglsin \alpha \vec{e_{\theta}} which is the right answer!

 

[tiny-tim]

you got it!