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Question Number 14677 by tawa tawa last updated on 03/Jun/17

$$\mathrm{A}\mathrm{stone}\mathrm{of}\mathrm{mass}100\mathrm{g}\mathrm{is}\mathrm{tied}\mathrm{to}\mathrm{the}\mathrm{end}\mathrm{of}\mathrm{a}\mathrm{string}\mathrm{of}50\mathrm{cm}\mathrm{long}.\mathrm{The}\mathrm{stone}\mathrm{is}$$$$\mathrm{whirled}\mathrm{as}\mathrm{a}\mathrm{conical}\mathrm{pendulum}\mathrm{so}\mathrm{that}\mathrm{it}\mathrm{rotates}\mathrm{in}\mathrm{horizontal}\mathrm{circle}\mathrm{radius}$$$$30\mathrm{cm}.\mathrm{Determine}\mathrm{the}\mathrm{angular}\mathrm{speed}\mathrm{and}\mathrm{the}\mathrm{tension}\mathrm{in}\mathrm{the}\mathrm{string}.$$

Answered by ajfour last updated on 03/Jun/17

$$heightofconegenerated$$$$is\mathit{h}=\sqrt{{\mathit{l}}^2-{\mathit{r}}^2}=\sqrt{{\left(50\right)}^2-{\left(30\right)}^2}cm$$$$=40cm$$$$\tan \theta =\frac{r}{h}=\frac{3}{4};\sin \theta =\frac{3}{5}$$$$T\sin \theta =m{\omega }^{2}r$$$$T\sin \theta =mg$$$$\Rightarrow \tan \theta =\frac{{\omega }^{2}r}{g}$$$$or\omega =\sqrt{\frac{g\tan \theta }{r}}=\sqrt{\frac{10\times 3}{4(0.30)}}$$$$=5rad/s$$$$T=\frac{mg}{\sin \theta }=\frac{0.1\times 10}{\left(3/5\right)}=\frac{5}{3}N=1.67N.$$

Commented by ajfour last updated on 03/Jun/17

Commented by tawa tawa last updated on 03/Jun/17

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{effort}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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