# Centrifugal force acting on ring [closed]

Take for example the case of a rod rotating about an axis passing through its centre of mass and perpendicular to it. It has a ring hung from one of its sides. The rotation of the rod causes the ring to move out wards and ultimately fall off the rod.

I asked my teacher what causes the ring to move out wards and he said the centrifugal force acting in the frame of reference of the ring causes motion. What exactly is the ring reference frame? Is it a rotating frame? How would u visualize motion in this frame of reference?

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## closed as off-topic by John Rennie, jinawee, centralcharge, akhmeteli, tpg2114Dec 11 '13 at 19:26

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There is no cenrtifugal force acting on the ring. The only force allowed is in the hoop direction, normal to the rod. If the ring was fixed there would be a force to keep it moving in a circle. – ja72 Dec 10 '13 at 19:55
I got you. Indeed the ring moves outward w.r.t the rod. For simplification consider the rod is frictionless and imagine everything in Lab's frame of reference.. consider the ring is initially hung at some point $P$. Find all the real forces acting on ring by Newton's 3rd law when rod is at rest initially. Now start the rotation. Since rod is frictionless no radial inward force acts on the ring so ring tends to move in a straight line w.r.t lab frame and Point $P$ in circular curve w.r.t Lab. But a tangential force acts on the ring which changes its direction in Lab frame. – user Dec 13 '13 at 12:12
which moves the ring in a spiral. finally the ring reaches the end of the rod and when it falls off its direction of motion remains in the same direction as it is now of $P$ i.e tangential not radially outward. In lab's fram neither centripetal nor centrifugal force acted on the ring. It was the normal reaction $N$ of the rod which was always tangential and caused the ring to move in spiral motion in our lab's frame. Now imagine the whole phenomena frome the frame of Rod you'll find that ring which was was at rest on point $P$ moved apart at a particular velocity so in the .... – user Dec 13 '13 at 12:23
frame of rod the ring accelerated and by Newton's law a force $ma$ acted on it. This is the centrifugal force which let the ring to fall off in rod's frame. what about $N$ in rod's frame i left it for you as an exercise to compare this $N$ with what we call weight. – user Dec 13 '13 at 12:36
One more thing do not bother about the frame of reference of the Ring itself. Obviously Ring will be always at rest in its own frame of reference. – user Dec 13 '13 at 12:38

The centrifugal force on the ring is the pseudo force when in the ring's reference frame, which causes it to move outwards, given by

$$\vec{F} = m\frac{v^2}{r} = mr\omega^2$$

Where m is the mass of the object, v is the tangential velocity of the object, and omega is the angular velocity

To find the time required for the ring to fall off, you need calculus, but that is not within the scope of your question.

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In this case, why is centrifugal force equal to value of centripetal force? – user34304 Dec 10 '13 at 16:00
@user34304 "why is centrifugal force equal to value of centripetal force?": to hold newton's second law F=ma. in the ring's frame the net force on it is 0. we intentionally add a pseudo force to hold newton's second law. – user Dec 10 '13 at 16:07
But sir, I feel there is no centripetal force acting on ring as it is not moving in a circle – user34304 Dec 11 '13 at 3:25
@user34304 bro i explained this before in my answer but am frightened of downvotes because this was not your question. lets leave this . you just look in some physics book which explains a bit about accelerating frame of reference like one i have which explains centrifugal force in a very deliberate way <H.C verma's concept of physics part 1>.centripetal force is a real force and will always act on the body whatever your frame of reference is.now consider two boys sitting in a merry go 'round . they are looking at each other and find them at rest and then one of them throws a ball ....... – user Dec 11 '13 at 13:53
@user34304 .... vertically upward. this ball is only under the action of gravity now no centripetal force is acting on it but the two boys are still rotating and they watch the ball going radially outward. now only centrifugal force is acting on the ball as seen from boys frame of reference because both boys are accelerating w.r.t ball. suppose you are watching both the boys from a bench in the park you will find that a force is acting on them which cause them to rotate but you will also find the ball just falling on ground not rotating with those boys. – user Dec 11 '13 at 14:02