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

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Question Number 14677 by tawa tawa last updated on 03/Jun/17
A stone of mass 100g is tied to the end of a string of 50 cm long . The stone is whirled as a conical pendulum so that it rotates in horizontal circle radius 30 cm. Determine the angular speed and the tension in the string.
$$\mathrm{A}\:\mathrm{stone}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{100g}\:\mathrm{is}\:\mathrm{tied}\:\mathrm{to}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{string}\:\mathrm{of}\:\mathrm{50}\:\mathrm{cm}\:\mathrm{long}\:.\:\mathrm{The}\:\mathrm{stone}\:\mathrm{is} \\ $$$$\mathrm{whirled}\:\mathrm{as}\:\mathrm{a}\:\mathrm{conical}\:\mathrm{pendulum}\:\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{rotates}\:\mathrm{in}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{radius} \\ $$$$\mathrm{30}\:\mathrm{cm}.\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{speed}\:\mathrm{and}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}. \\ $$
Answered by ajfour last updated on 03/Jun/17
height of cone generated is h=(√(l^2 −r^2 )) =(√((50)^2 −(30)^2 )) cm = 40cm tan θ=(r/h)=(3/4) ; sin θ =(3/5) Tsin θ=mω^2 r Tsin θ=mg ⇒ tan θ=((ω^2 r)/g) or ω = (√((gtan θ)/r)) =(√((10×3)/(4(0.30)))) = 5rad/s T=((mg)/(sin θ))=((0.1×10)/((3/5))) =(5/3)N=1.67N.
$$\:\:{height}\:{of}\:{cone}\:{generated} \\ $$$${is}\:\boldsymbol{{h}}=\sqrt{\boldsymbol{{l}}^{\mathrm{2}} −\boldsymbol{{r}}^{\mathrm{2}} }\:=\sqrt{\left(\mathrm{50}\right)^{\mathrm{2}} −\left(\mathrm{30}\right)^{\mathrm{2}} }\:{cm} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{40}{cm} \\ $$$$\:\:\:\mathrm{tan}\:\theta=\frac{{r}}{{h}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:;\:\:\:\mathrm{sin}\:\theta\:=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{T}\mathrm{sin}\:\theta={m}\omega^{\mathrm{2}} {r} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{T}\mathrm{sin}\:\theta={mg} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\mathrm{tan}\:\theta=\frac{\omega^{\mathrm{2}} {r}}{{g}} \\ $$$${or}\:\:\:\:\:\:\omega\:=\:\sqrt{\frac{{g}\mathrm{tan}\:\theta}{{r}}}\:\:=\sqrt{\frac{\mathrm{10}×\mathrm{3}}{\mathrm{4}\left(\mathrm{0}.\mathrm{30}\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{5}{rad}/{s} \\ $$$$\:\:\:{T}=\frac{{mg}}{\mathrm{sin}\:\theta}=\frac{\mathrm{0}.\mathrm{1}×\mathrm{10}}{\left(\mathrm{3}/\mathrm{5}\right)}\:=\frac{\mathrm{5}}{\mathrm{3}}{N}=\mathrm{1}.\mathrm{67}{N}. \\ $$
Commented by ajfour last updated on 03/Jun/17
Commented by tawa tawa last updated on 03/Jun/17
I really appreciate your effort sir. God bless you sir.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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