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Question Number 137613 by physicstutes last updated on 04/Apr/21

$$\mathrm{Two}\mathrm{particles}\mathrm{of}\mathrm{mass}m\mathrm{and}4m\mathrm{are}\mathrm{connected}\mathrm{by}\mathrm{a}\mathrm{light}\mathrm{inelastic}$$$$\mathrm{string}\mathrm{of}\mathrm{length}3a,\mathrm{which}\mathrm{passes}\mathrm{through}\mathrm{a}\mathrm{small}\mathrm{smooth}\mathrm{fixed}\mathrm{ring}.$$$$\mathrm{The}\mathrm{heavier}\mathrm{particle}\mathrm{hangs}\mathrm{at}\mathrm{rest}\mathrm{at}\mathrm{a}\mathrm{distance}2a\mathrm{beneath}\mathrm{the}\mathrm{ring},$$$$\mathrm{while}\mathrm{the}\mathrm{lighter}\mathrm{particle}\mathrm{describes}\mathrm{a}\mathrm{horizontal}\mathrm{circle}\mathrm{with}\mathrm{constant}$$$$\mathrm{speed}.$$$$\mathrm{Find}$$$$\left(\mathrm{a}\right)\mathrm{The}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{below}\mathrm{the}\mathrm{ring}.$$$$\left(\mathrm{b}\right)\mathrm{The}\mathrm{angular}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{lighter}\mathrm{particle}.$$

Answered by mr W last updated on 04/Apr/21

$$\left(a\right)$$$$\sin \theta =\frac{1}{4}$$$$h=a\sin \theta =\frac{a}{4}$$$$\left(b\right)$$$$ma\cos \theta {\omega }^2=4mg\cos \theta$$$$\Rightarrow \omega =2\sqrt{\frac{g}{a}}$$

Commented by physicstutes last updated on 04/Apr/21

$$\mathrm{sir}\mathrm{why}\mathrm{is}\sin \theta =\frac{1}{4}?$$

Commented by mr W last updated on 04/Apr/21

Commented by physicstutes last updated on 06/Apr/21

$$\mathrm{sorry}\mathrm{to}\mathrm{disturb}\mathrm{sir},\mathrm{but}\mathrm{when}\mathrm{i}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{triangle}\mathrm{containing}$$$$\theta \mathrm{i}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{seem}\mathrm{to}\mathrm{get}\mathrm{how}\sin \theta =\frac{1}{4},\mathrm{is}\sin \theta \mathrm{usually}\mathrm{the}\mathrm{ratio}$$$$\mathrm{of}\mathrm{tensions}\mathrm{or}\mathrm{of}\mathrm{masses},\mathrm{thats}\mathrm{my}\mathrm{crazy}\mathrm{doubt}$$

Commented by mr W last updated on 06/Apr/21

$$inY-directionwehave$$$$4mg\sin \theta -mg=0$$$$\Rightarrow \sin \theta =\frac{1}{4}$$$$inX-directionwehave$$$$4mg\cos \theta -mr{\omega }^2=0$$$$4mg\cos \theta -ma\cos \theta {\omega }^2=0$$$$4g-a{\omega }^2=0$$$$\Rightarrow {\omega }^2=\frac{4g}{a}$$

Commented by physicstutes last updated on 06/Apr/21

$$\mathrm{wow}\mathrm{i}\mathrm{get}\mathrm{nicely}\mathrm{now}.\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

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