One twirls a circular ring (of$$ mass M and radius $$R) near the tip of one’s finger as shown in Figure $$1$$. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω$$0$$ . The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure$$ $$2). The coefficient of friction between the ring and the finger is μ and the acceleration due to gravity is g. Take r≪RThe total kinetic energy of the ring is The minimum value of ω$$0$$ below which the ring will drop down is

Question

One twirls a circular ring (of$$ mass M and radius $$R) near the tip of one’s finger as shown in Figure $$1$$. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω$$0$$ . The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure$$ $$2). The coefficient of friction between the ring and the finger is  μ and the acceleration due to gravity is g. Take r≪RThe total kinetic energy of the ring is The minimum value of ω$$0$$ below which the ring will drop down is

Infinity Learn · Accepted Answer

Centre of mass of ring moves with angular velocity ω$$0$$ in a circle of radius $$R-r$$ centered about the center of the circle of radius r.VCM of Ring $$=(R$$−$$r)$$ω$$0For$$ no slippingKE of $$ring=12MvCM2+12Ic$$ω$$02KE$$ of ring $$=12M(R$$−$$r)2$$ω$$02+12MR2$$ω$$02KE$$ of ring $$=12MR2(1$$−$$rR)2$$ω$$02+12MR2$$ω$$02$$ Since, r<>rThen  μ$$N=Mg$$⇒μM̸(R$$−$$r)ω$$02=Mg$$⇒ω$$0=g$$μ(R$$−$$r)

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