Velocity of current-carrying cylinder

[steve B. 98]

Homework Statement

A cylindrical rod of length w carries a current I as shown(perpendicular to B) and is bathed in a field B perpendicular to the plane in which is a " shaped rail. The rod rolls without slipping on the rails, its length perpendicular to the two parallel rails and equal to the space between them. It starts at rest and rolls off after going a distance L. Show that its exit velocity is $v = \sqrt{ \frac{4BLIw}{3M}}$

Homework Equations

$F=B \times I w=BIw$
$\tau=r \times F$
$\tau=I\alpha$

The Attempt at a Solution

[/B]
I start by taking the torque
$\tau=r \times BIw$
Using the moment of inertia for a cylinder I get
$\tau=Mr^2\alpha /2$
$\alpha=\frac{2BIw}{Mr}$
since the problem was rolling without slipping we get.
$a=\frac{2BIw}{M}$
Using $v^2=0^2+2as$
$v=\sqrt{\frac{4BIwL}{M}}$
But this is off by a factor of $\frac{1}{\sqrt{3}}$.
What did I do wrong, and some hints toward the solution would be nice.

 

[AYPHY]

easy question pal i have solved it and i am placing the pictures.
First find net force which is force on wire by current and field and second is frictional force which will rotate it and hence net force is IwB-f which is Ma .
Then find torque which is fr=Iα when you will sove net acc. to be 2IwB/(3M) place it in v^2=2aL you will get the answer.

 

[AYPHY]

your acc was incorrect hence your answer came out wrong.

 

[AYPHY]

i think this might clear your doubt more then ask.