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Question Number 106094 by Dwaipayan Shikari last updated on 02/Aug/20

Commented by Dwaipayan Shikari last updated on 02/Aug/20

Dwaipayan Shikari: The wedge's Surface is friction less (where the ball of mass 'm' will roll) and the friction coefficient between wedge and the ground is also negligible.( An oily surface ��) Calculate the velocity of the wedge when the ball will fly off. Additional information is on the picture

Commented by Dwaipayan Shikari last updated on 02/Aug/20

$$Theballwillrollonthiscirculartrack.$$$$$$$$An\mathrm{old}\mathrm{unanswered}\mathrm{question}$$

Commented by mr W last updated on 02/Aug/20

$$ifthefrictionbetweenwedgeand$$$$groundiszero(i.e.\mu =0),thenthe$$$$questionisveryeasy.$$$$butherewehavethecasethat\mu \ne 0.$$

Commented by mr W last updated on 02/Aug/20

$$toJDamiansir:$$$$thewedgeMhasalsovelocity!$$

Commented by Dwaipayan Shikari last updated on 02/Aug/20

$$\mathrm{U}+\mathrm{KE}=u+ke$$$$mgR+0=0+\frac{1}{2}{mv}^2+\frac{1}{2}{MV}^2$$$$2mgR={mv}^2+{MV}^2\to \left(a\right)$$$$$$$$Momentumofthesystemisconserved$$$$(M+m).0=MV+mv$$$$V=-\frac{mv}{M}$$$$2mgR={mv}^2+M.\frac{m^2{v}^2}{M^2}$$$$2gR=v^2(1+\frac{m}{M})$$$$v=\sqrt{\frac{2gR}{(1+\frac{m}{M})}}$$$$V=-\frac{mv}{M}=-\frac{m}{M}\sqrt{\frac{2gR}{(1+\frac{m}{M})}}$$$$\mathrm{Frictional}\mathrm{case}:$$$$\mathrm{mgR}=\frac{1}{2}{\mathrm{MV}}^2+\frac{1}{2}{\mathrm{mv}}^2+{\mathrm{W}}_{\left(\mathrm{by}\mathrm{the}\mathrm{frictional}\mathrm{force}\right)}$$$$$$$$$$

Commented by Dwaipayan Shikari last updated on 02/Aug/20

$$Thismustbeithinkiffrictionisnegligible$$

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