A thin ring of radius is rolling without slipping on the horizontal surface AB with velocity. Then the ring strikes the inclined surface BC and starts rolling up the plane. What fraction of initial energy is lost in the process ?

Question

Infinity Learn · Accepted Answer

Angular momentum of the ring about the point of impact P is conserved.Just after the impactLPf  $$=$$  mVR  $$+$$  $$mR2$$  VR  $$=$$  $$2mVRJust$$ before the impact, LPi  $$=$$  $$mV0$$ Rcosϕ $$+$$  $$mR2$$.  $$V0R$$∴    $$2mVR$$ $$=$$  $$mV0R$$ $$cos$$ϕ $$+$$  $$mV0R$$⇒   $$V=V02$$   $$1+$$ $$cos$$ϕInitial Kinetic energy,Ki $$=$$  $$12$$  $$mV02$$  $$+$$  $$12$$  $$mR2$$  $$V0R2$$  $$=$$  $$mV02Final$$ Kinetic energy,Kf $$=$$  $$12$$  mV $$2$$  $$+$$  $$12$$  $$mR2$$  $$VR2$$  $$=$$  mV $$2$$ $$=$$  $$14$$  $$mV02$$   $$1+$$  $$cos6002=$$  $$916$$   $$mV02$$∴   Fraction of initial energy lost due to $$impact=$$  Ki −  KfKi  $$=$$ $$1$$−  $$916$$  $$=$$  $$716$$

A thin ring of radius is rolling without slipping on the horizontal surface AB with velocity. Then the ring strikes the inclined surface BC and starts rolling up the plane. What fraction of initial energy is lost in the process ?

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