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جستجو

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A narrow groove is cut along a diameter in the surface of a horizontal disc with centre $O$. Particles $P$ and $Q$, of masses $0.2{\text{ }}kg$ and $0.3{\text{ }}kg$ respectively, lie in the groove, and the coefficient of friction between each of the particles and the groove is $\mu $. The particles are attached to opposite ends of a light inextensible string of length $1{\text{ }}m$. The disc rotates with angular velocity $\omega {\text{ }}rad{\text{ }}{s^{ - 1}}$ about a vertical axis passing through $O$ and the particles move in horizontal circles (see diagram).

a) Given that $\mu  = 0.36$ and that both $P$ and $Q$ move in the same horizontal circle of radius $0.5{\text{ }}m$, calculate the greatest possible value of $\omega $ and the corresponding tension in the string.

b) Given instead that $\mu  = 0$ and that the tension in the string is $0.48{\text{ }}N$, calculate

(i) the radius of the circle in which $P$ moves and the radius of the circle in which $Q$ moves,

(ii) the speeds of the particles. 

پاسخ تشریحی :
نمایش پاسخ

a) $0.3\,{\omega ^2} \times 0.5 = T + 0.36 \times 0.3g$

$0.2\,{\omega ^2} \times 0.5 = T - 0.36 \times 0.2g$

$0.1\,{\omega ^2} \times 0.5 = 0.36 \times 0.5g$

$\omega  = 6$

$T = 0.3 \times {6^2} \times 0.5 - 0.36 \times 0.3 \times 10$

$T = 4.32$

b)(i) $0.2\,{\omega ^2}r = 0.3\,{\omega ^2}\left( {1 - r} \right)$

$r = 0.6$

${r_P} = 0.6{\text{ }}m$ and ${r_Q} = 0.4{\text{ }}m$

(ii) $0.48 = 0.2v_P^2/0.6$ or $0.48 = 0.3v_Q^2/0.4$

${v_P} = 1.2$

${v_Q} = 0.8$

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