Ball rolling down slope (with wedge-shaped groove)

1. The problem statement, all variables and given/known data[/b]

The question is B7 here: http://ift.tt/1gpD2wC

I managed to derive the acceleration required in the first part, but the second part is giving me trouble.

2. Relevant equations

3. The attempt at a solution

I have calculated the moment of inertia of the ball about the axis it is instantaneously rotating around (I don’t quite know what is meant by an ‘instantaneous axis’ – but anyway), I arrived at…

[tex]{I_{AB}} = \frac{2}{5}m{a^2} + m{a^2}{\sin ^2}(\frac{1}{2}\phi ) = m{a^2}(\frac{2}{5} + {\sin ^2}(\frac{1}{2}\phi ))[/tex]
1. The problem statement, all variables and given/known data

Now, I have an equation for the linear motion of the centre of mass of the ball, and it is… [tex]mg\sin \theta – 2F = m\frac{{dv}}{{dt}}[/tex] where F is the frictional force acting in the direction of the slope, providing the moment (two points of contact so I wrote it as 2F)

The rotational motion is confusing me, I tried to write…

[tex]G = I\frac{{d\omega }}{{dt}} = \frac{I}{a}\frac{{dv}}{{dt}} = ma(\frac{2}{5} + {\sin ^2}(\frac{1}{2}\phi ))\frac{{dv}}{{dt}}[/tex]

but how do I find G, the total external moment, and which axis do I take the moment about – (because if I were to take the moment about AB I’d have cos terms in my result)? Which I am guessing is supplied by the two frictional forces? Am I thinking about this all wrong?

http://ift.tt/1poxO3R

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons