Angular velocity of rod supported by yoyo

[timetraveller123]

Homework Statement

Homework Equations

The Attempt at a Solution

so since the rod and the floor is tangent to the circle then the tangent at external point theorem can be applied to find out that the two triangles are congruent
i assumed the the circle start out with tangent to the y-axis at first so the x distance to the centre at any time is vt +r this v is different from the v in the question
$\frac{r}{vt + r} = tan \frac{\theta}{2}\\ \frac{{sec \theta}^2}{2}\dot \theta = \frac{-r v}{(vt +r)^2}\\ \dot \theta = \frac{-2rv}{(vt + r)^2} = \frac{-2v tan\frac{\theta}{2}}{r (sec \theta)^2}$
where v is velocity of centre
how to relate the two velocities is my method even valid

 

[haruspex]

is my method even valid

Your method is basically ok, but what do you get for $\frac d{d\theta}\tan(\frac\theta 2)$?

 

[timetraveller123]

why is it with respect to theta i did it with respect to time which was $sec^2 \theta (\frac{1}{2}) \dot \theta$

 

[haruspex]

why is it with respect to theta i did it with respect to time which was $sec^2 \theta \frac{1}{2} \dot \theta$

You are, whether you realize it or not, using (actually, misusing) the chain rule. $\frac{df(y)}{dx}=f'(y)\frac{dy}{dx}$. So first do the derivative wrt theta. Better still, wrt ½θ

 

[timetraveller123]

no i understand i am using chain rule oh wait it is supposed to be

$sec^2 \frac{\theta}{2} (\frac{1}{2}) \dot \theta$
is it that -
$\frac{d f}{d \frac{\theta}{2}} \frac{d \frac{\theta}{2}}{d \theta}\frac{d \theta}{dt}$

 

[haruspex]

no i understand i am using chain rule oh wait it is supposed to be

$sec^2 \frac{\theta}{2} (\frac{1}{2}) \dot \theta$

Yes.

 

[timetraveller123]

then what about how to relate the v in the question to the v v of centre

 

[haruspex]

then what about how to relate the v in the question to the v v of centre

As a disc rolls, where is its instantaneous centre of rotation?

 

[timetraveller123]

the bottom of the cirlce so is it v/(R+r) * R

 

[haruspex]

the bottom of the cirlce so is it v/(R+r) * R

Yes.

 

[timetraveller123]

ok thanks for the help