Rod with current and magnetic field

[gsquare567]

Homework Statement

A rod of mass m and radius R rests on two parallel rails that are a distance d apart and have a length L. The rod carries a current I and rolls along the rails without slipping. A uniform magnetic field B is directed perpendicular to the road and the rails, pointing downwards. If it starts from rest, what is the speed of the rod as it leaves the rails.

Homework Equations

(1) F = I L x B = ma
(2) torque = r x F

The Attempt at a Solution

By subbing in the given constants into (1), I get a = IdB/m. However, I think that I need to use equation 2 instead. I am wondering how to relate the torque on the rod to the velocity of the rod. I don't think the the acceleration I got applies to the rod, because it receives its motion due to torque, rather than just the magnetic force.

Thanks for any help!

 

[Delphi51]

It could be done with force and torque, but I think the easier way is to
- calculate the force on the rod and work done over the distance L
- work = linear kinetic energy + rotational energy
- solve for v, which will appear in both energy terms

 

[gsquare567]

ok, let me see if i got it straight:

W = Krotational + Ktranslational = 1/2Iw^2 + 1/2mv^2 = 1/2(1/2mR^2)(v/R)^2 + 1/2mv^2 = 3/4mv^2
and
W = F * d = (IdB)(L)
so
3/4mv^2 = IdBL
v = 2/sqrt(3) sqrt(IdBL/m)
now my question is, why doesn't the radius of the rod, R, matter? did i use the right moment of inertia and angular velocity?

 

[Delphi51]

It looks correct to me. The calculation itself "explains" why R doesn't appear in the answer I guess. A wee bit surprising, though.

 

[gsquare567]

Interesting. Thanks for all your help!

 

[Delphi51]

Most welcome.