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Question Number 154851 by peter frank last updated on 22/Sep/21

Commented by peter frank last updated on 22/Sep/21

$$\mathrm{at}{\mathrm{U}}_{\mathrm{o}}^2\mathrm{m}/\mathrm{s}.$$

Commented by peter frank last updated on 22/Sep/21

$$\mathrm{A}\mathrm{particle}\mathrm{of}\mathrm{mass}{\mathrm{M}}_{\mathrm{o}}\mathrm{is}\mathrm{tied}\mathrm{to}$$$$\mathrm{the}\mathrm{string}\mathrm{of}\mathrm{length}\left(\mathrm{L}\right)\mathrm{and}\mathrm{hang}$$$$\mathrm{freely}.\mathrm{if}\mathrm{the}\mathrm{particle}\mathrm{displaced}$$$$\mathrm{hanging}\left(\mathrm{P}\right)\mathrm{at}{\mathrm{U}}_{\mathrm{o}}\mathrm{m}/\mathrm{s}.\mathrm{show}\mathrm{that}$$$$\mathrm{if}\mathrm{particle}\mathrm{complete}\mathrm{a}\mathrm{circle}$$$${\mathrm{U}}_{\mathrm{o}}^2⩾5\mathrm{gL}$$

Commented by mr W last updated on 23/Sep/21

$$\frac{{mu}_1^2}{L}=T_1+mg$$$$with{T}_1=tensioninstringatthe$$$$highestpoint,whichshouldbe⩾0.$$$$\Rightarrow u_1^2⩾gL$$$$\frac{1}{2}{mu}_0^2=\frac{1}{2}{mu}_1^2+mg\left(2L\right)$$$$u_0^2=u_1^2+4gL⩾gL+4gL=5gL$$

Commented by mr W last updated on 23/Sep/21

Commented by peter frank last updated on 23/Sep/21

$$\mathrm{sir}\mathrm{how}\mathrm{you}\mathrm{draw}\mathrm{this}?$$

Commented by mr W last updated on 23/Sep/21

$$withnospecialapp.ijustusedan$$$$appinmysmartphoneforfoto$$$$editing.$$

Commented by peter frank last updated on 23/Sep/21

$$\mathrm{From}\mathrm{tha}\mathrm{figure}\mathrm{above}$$$$\mathrm{Total}\mathrm{energy}\mathrm{at}\mathrm{the}\mathrm{bottom}\left({\mathrm{E}}_{\mathrm{b}}\right)=$$$$\mathrm{Total}\mathrm{energy}\mathrm{at}\mathrm{Top}\left({\mathrm{E}}_{\mathrm{T}}\right)$$$${\mathrm{E}}_{\mathrm{B}}={\mathrm{E}}_{\mathrm{T}}$$$${\mathrm{E}}_{\mathrm{B}}=\mathrm{P}.{\mathrm{E}}_{\mathrm{B}}+\mathrm{K}.{\mathrm{E}}_{\mathrm{B}}$$$${\mathrm{E}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{o}}\mathrm{gh}+\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2[\mathrm{h}=0]$$$${\mathrm{E}}_{\mathrm{B}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2....\left(\mathrm{i}\right)$$$${\mathrm{E}}_{\mathrm{T}}=\mathrm{P}.{\mathrm{E}}_{\mathrm{B}}+\mathrm{K}.{\mathrm{E}}_{\mathrm{B}}$$$${\mathrm{E}}_{\mathrm{T}}={\mathrm{m}}_{\mathrm{o}}{\mathrm{gh}}_{\mathrm{T}}+\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{v}}_{\mathrm{T}}$$$${\mathrm{E}}_{\mathrm{T}}={\mathrm{m}}_{\mathrm{o}}\mathrm{g}\left(2\mathrm{l}\right)+\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{v}}_{\mathrm{T}}$$$${\mathrm{E}}_{\mathrm{T}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}+{2\mathrm{m}}_{\mathrm{o}}\mathrm{gl}...\left(\mathrm{ii}\right)$$$${\mathrm{E}}_{\mathrm{B}}={\mathrm{E}}_{\mathrm{T}}$$$$\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2=\frac{1}{2}{\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2+{2\mathrm{m}}_{\mathrm{o}}\mathrm{gl}$$$${\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2={\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2-{4\mathrm{m}}_{\mathrm{o}}\mathrm{gl}....\left(\mathrm{iii}\right)$$$$\mathrm{from}$$$$\mathrm{T}=\frac{{\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2}{\mathrm{r}}-\mathrm{mg}$$$$\mathrm{for}\mathrm{particle}\mathrm{to}\mathrm{complete}\mathrm{the}\mathrm{circle}$$$$\frac{{\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2}{\mathrm{r}}⩾\mathrm{mg}$$$${\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2⩾\mathrm{mgr}....\left(\mathrm{iv}\right)$$$${\mathrm{m}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{T}}^2={\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2-{4\mathrm{m}}_{\mathrm{o}}\mathrm{gl}...\left(\mathrm{iii}\right)$$$${\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2-{4\mathrm{m}}_{\mathrm{o}}\mathrm{gl}⩾{\mathrm{m}}_{\mathrm{o}}\mathrm{gr}[\mathrm{r}=\mathrm{l}]$$$${\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}^2⩾{5\mathrm{m}}_{\mathrm{o}}\mathrm{g}l$$$${\mathrm{u}}_{\mathrm{o}}^2⩾5\mathrm{g}l$$$$$$

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