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Question Number 216324 by Jubr last updated on 04/Feb/25

Commented by Jubr last updated on 04/Feb/25

$$ThatlastsymbolisP$$

Answered by mr W last updated on 04/Feb/25

$$T=5mg$$$$l_0=5a$$$$\mathrm{\Delta }l=\frac{l_0}{8mg}\times 5mg=\frac{5l_0}{8}=\frac{25a}{8}$$$$l=5a+\frac{25a}{8}=\frac{65a}{8}$$$$SP=b=l-2a=\frac{49a}{8}$$$$\sin \theta =\frac{3mg}{T}=\frac{3}{5}$$$$r=b\cos \theta =\frac{49a}{8}\times \frac{4}{5}=\frac{49a}{10}$$$$3m{\omega }^{2}r=T\cos \theta =5mg\times \frac{4}{5}$$$$3{\omega }^2\times \frac{49a}{10}=4g$$$$\Rightarrow \omega =\frac{2}{21}\sqrt{\frac{30g}{a}}$$

Commented by Jubr last updated on 04/Feb/25

$$Thankssir.$$

Commented by mr W last updated on 06/Feb/25

$$Qremainsstationary,thatmeans$$$$it{doesn}^{\prime }tmoveatall.itsvelocity$$$$andangularvelocityarezero!$$$$butitrotatesaboutitsaxiswith$$$$\omega .$$

Commented by Jubr last updated on 06/Feb/25

$$Pleasesir,whatisthe$$$$angularvelocityofQ$$

Commented by Jubr last updated on 10/Feb/25

$$Thankssir.$$

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