A billiard ball (of$$ radius $$R), initially at rest is given a sharp impulse by a cue. The cue is held horizontally at a distance h above the central line as shown in figure. The ball leaves the cue with a speed Vo. It rolls and slides while moving forward and eventually acquires a final speed of $$9/7$$ $$V0$$. If R $$=$$ $$5.0$$ cm then find the value of h in cm.

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Infinity Learn · Accepted Answer

Let $$v0$$ be the linear velocity and ω$$0$$ be the angular velocity imparted.linear impulse $$=J=mv0Angular$$ impulse about centre O $$=J$$⋅$$h=I$$ω$$0h=2/5mR2$$ω$$0mv0$$⇒$$v0=2R2$$ω$$05h$$  ………(i)As$$ linear velocity increases to $$9/7$$ $$v0$$, friction must be in forward direction and hence it opposes angular motion. For linear motion: friction $$($$μ$$mg) increases the velocity. Hence,μ$$mg=ma$$⇒$$a=$$μgLet t be the time when rolling starts.For rotation: frictional torque $$($$μ$$mgR)$$ opposes rotation and hence decreases ω.α$$=$$−μmgRI ⇒ω$$=$$ω$$0$$−μ$$mgR25mR2t$$⇒ω$$R=$$ω$$0R-52$$μgt ⇒$$v=$$ω$$0R-52$$μgt ⇒μgt $$=$$ $$25$$ω$$0R-25.9v07$$..........ii given $$v=9v07After$$ time t, $$V=97v0=R$$ωAlso $$v=v0+$$ μgt⇒$$9v07=v0$$ $$+$$ μgt .........iiiCombining all the equations, we $$get97v0$$−$$v0=$$ω$$02R5-259v07$$ ⇒ω$$0R=2v0$$ From i, h $$=25Rv0$$×$$2v0$$ $$=$$ $$45R$$ $$=$$ $$45$$×$$5$$ cm $$=4$$ cm  (eliminating$$ μ$$gt)

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