A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails (Figure 3)

Question

A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails (Figure 3) that are current of I = 48.0 A (in the direction shown) and rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T 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? A rod of mass m and radius R rests on two parallel rails (Figure 3) that are a distance d apart and 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?

Explanation / Answer

3) Magnetic force is given by Fb = idB = 48*0.12*0.24

= 1.3824N

Let the friction be F. For pure rolling, a = alpha r

(Fb - F)/ m = (Fr/(0.5mr^2))*r

1.3824 - F = 2F

F = 1.3824/3 = 0.4608 N

a = (1.3824 - 0.4608)/0.720 = 1.28 m/s

By third equation of motion,

v = sqrt (2aL)

= sqrt (2*1.28*0.45)

= 1.073 m/s answer

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