Recent content by srijag

When a conducting body moves through the field, the charges in the system experience force 'qvb' and hence, opposite charges move to extreme ends and potential difference is established. this continues until qvb= qe(i.e; force due to electric field in the direction opposite to that of qvb.)

http://thecraftycanvas.com/library/finding-forces-acting-upon-objects-on-an-inclined-plane-or-ramp-with-free-body-diagrams/ I hope that helps.

For the emf to be induced in a moving conductor, you don't require varying magnetic field. So, that's the reason i considered them to be two rods. In that case, emf will be produced in both as in the case of two cells connected in the same circuit. I need help with that part of the problem.

Nothing is specified about the magnetic field. So it might also be a constant field which means there won't be any EMF induced.

If you apply Kirchoff's loop rule over here, the total EMF will be zero.

The unit is indeed dimensionless. Radian is dimensionless. If you try to evaluate the dimensions of angular acceleration, it is T^-2. So the units of angular acceleration being s^-2 makes sense.

The ring is performing translational motion and the magnetic field is orthogonal to the motion of the ring. Does it matter if the magnetic flux is changing because I considered the two halves of the ring to be separate rods. That way emf induced in each 'rod' is E= Bl(2R) where R is the radius...

The equation does NOT show that acceleration is zero. It is Rw^2. Since V=Rw, it is equal to v^2/R

a(point of contact) = a(point of contact w.r.t CM) + a(cm) = a(p, tangential) + a(P, radial) + a(cm) i Since the rolling is uniform, a(cm) is zero as well as α(cm) i.e; angular acceleration of centre of mass and since a(P, tangential) = Rα(cm), it becomes...

if you consider the horizontal to be the x-axis, then, the point of contact makes an angle 3π/2. Now a(point of contact)= a(cm)i + a(P,cm) = a(cm)i + a(P, tangential) -i + a(P, radial) = a(cm)i - Rα(cm) + R(w^2) α+ angular acceleration of...

You have to derive that. Also, acceleration of centre of mass is zero only in a uniform pure rolling.

Yes. Pure rolling is same as rotation about point of contact. About the point of contact and the acceleration, mathematically you can derive for any point on the surface. And if you substitute the condition for the pioint of contact, you'll find out that the acceleration of point of contact is...

I was referring to that question. Anyway, the point of contact (in fact, all the points on the surface) have acceleration toward the centre of mass of the object. Be it, uniform pure rolling or accelerated pure rolling, the acceleration will be towards the centre of mass. Also even if there...

The velocity of the point of contact during rolling motion is zero.

Consider a magnetic field perpendicular to a conducting ring moving with a velocity, v.When the ring is moving on the ground in translational motion alone, will emf be induced? I am slightly confused because if you consider the two halves of the ring as two rods, emf will be induced in both of...