If it starts from rest, what is the speed of the rod as it leaves the rails?

+5 votes

A rod of mass m and radius R rests on two parallel rails (see figure below) that are a distance d apart have a length L. The rod carries a current I in the direction shown and rolls along the rails without slipping. A uniform magnetic field B is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails? (Use any variable or symbol stated above as necessary.)

v =                m/s

asked Feb 19, 2012 in Physics by Joey33 ~Expert~ (1,216 points)

1 Answer

+4 votes
Best answer

The magnetic force acting on the rod can be described by:

Magnetic force = I(d X B)  (*where I=current, d=distance, B=magnetic field, and X=cross product)

Using the work-energy theorem, which is:

(K trans + K rot)initial + magnetic Force * L = (K trans + K rot)final  (*where K = kinetic energy, trans=translational, rot=rotational)

Since it starts from rest, the initial Kinetic energy = 0 and hence the equation simplifies to:

 magnetic Force = (K trans + K rot)final


K trans = (1/2) m*v^2  (*where m=mass, v=velocity)

K rot = (1/2)I*w^2 (*where I =moment of inertia, w=angular velocity)

Furthermore, I = (1/2)mR^2  and  w = V/r

and so K rot = (1/2)((1/2)mR^2) * (V/r)^2


plugging everything back into the equation, we get:

magnetic Force = (K trans + K rot)final

magnetic Force = ((1/2) m*v^2s) + ( (1/2)((1/2)mR^2) * (V/r)^2)


magnetic force can then be rewritten as simply:

I*d*B*L  (*since the cross product when theta = 0, is simply one, we are left with just this)


And so we get:


I*d*B*L = ((1/2) m*v^2s) + ( (1/2)((1/2)mR^2) * (V/r)^2)

and solving for velocity we should get:

answered Feb 23, 2012 by pokemonmaster ~Expert~ (3,854 points)
edited Mar 8, 2012 by pokemonmaster