Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 155034 by peter frank last updated on 24/Sep/21

Commented by peter frank last updated on 24/Sep/21

$$\mathrm{i}\mathrm{think}{\mathrm{h}}_{\mathrm{o}}=l$$$$u_o=maxspeed$$

Commented by mr W last updated on 24/Sep/21

$$pleasecheckyourtypos!$$$$whatis{h}_0?,whatis{u}_0?,maximum$$$$speedorminimumspeed?$$

Commented by mr W last updated on 24/Sep/21

$$no.thatistheminimumspeedthe$$$$bulletmusthave.ifthespeedisless$$$$thanit,thebobcannotcompletea$$$$circle.$$

Commented by mr W last updated on 24/Sep/21

$$anditshouldbe{u}_0=\frac{4m}{m_0}\sqrt{5gl},not$$$$u_0=\frac{4m_0}{m}\sqrt{5gl}.$$

Answered by mr W last updated on 24/Sep/21

$$fromQ154851:$$$$v_0^2⩾5gl$$$$m_0{u}_0={mv}_0+m_0\frac{3u_0}{4}$$$$u_0=\frac{4m}{m_0}\times v_0⩾\frac{4m}{m_0}\sqrt{5gl}$$

Answered by peter frank last updated on 25/Sep/21

Commented by peter frank last updated on 25/Sep/21

$$\mathrm{before}\mathrm{collision}$$

Answered by peter frank last updated on 25/Sep/21

Commented by peter frank last updated on 25/Sep/21

$$\mathrm{after}\mathrm{collision}$$

Commented by peter frank last updated on 25/Sep/21

$$\mathrm{since}\mathrm{there}\mathrm{are}\mathrm{collision},$$$$\mathrm{momentum}\mathrm{is}\mathrm{conserved}$$$$\mathrm{Momentum}\mathrm{before}\mathrm{collisio}\left({\mathrm{M}}_{\mathrm{bc}}\right)=$$$$\mathrm{Momentum}\mathrm{after}\mathrm{collision}\left({\mathrm{M}}_{\mathrm{apc}}\right)=$$$${\mathrm{u}}_{\mathrm{o}}{\mathrm{m}}_{\mathrm{o}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}=\frac{3}{4}{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}+{\mathrm{Mv}}_{\mathrm{B}}$$$${\mathrm{u}}_{\mathrm{B}}=0$$$${\mathrm{v}}_{\mathrm{B}}=\frac{{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}}{4\mathrm{M}}...\left(\mathrm{i}\right)$$$$\mathrm{Total}\mathrm{Energy}\mathrm{at}\mathrm{the}\mathrm{bottom}=$$$$\mathrm{Total}\mathrm{Energy}\mathrm{at}\mathrm{the}\mathrm{Top}\left(\mathrm{highest}\right)$$$${\mathrm{E}}_{\mathrm{B}}={\mathrm{E}}_{\mathrm{T}}$$$$\mathrm{K}.{\mathrm{E}}_{\mathrm{B}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{B}}=\mathrm{K}.{\mathrm{E}}_{\mathrm{H}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{H}}$$$$\frac{1}{2}{\mathrm{Mv}}_{\mathrm{o}}^2+\mathrm{mg}\left(0\right)=\frac{1}{2}{\mathrm{Mv}}_{\mathrm{H}}^2+\mathrm{mg}(l+l)$$$${\mathrm{v}}_{\mathrm{B}}^2={\mathrm{v}}_{\mathrm{H}}^2+4\mathrm{g}l....\left(\mathrm{ii}\right)$$$${\left(\frac{{\mathrm{m}}_{\mathrm{o}}{\mathrm{u}}_{\mathrm{o}}}{4\mathrm{M}}\right)}^2={\mathrm{V}}_{\mathrm{H}}^2+4\mathrm{g}l$$$${\mathrm{v}}_{\mathrm{H}}^2=\frac{{\mathrm{m}}_{\mathrm{o}}^2{\mathrm{u}}_{\mathrm{o}}^2}{16\mathrm{M}}-4\mathrm{g}l.....\left(\mathrm{iii}\right)$$$$foranyparticletocompletea$$$$circlesafely$$$$\frac{{mv}_H^2}{\mathrm{r}}>\mathrm{mg}$$$$\frac{v_H^2}{\mathrm{r}}>\mathrm{g}$$$${\mathrm{v}}_{\mathrm{H}}^2>\mathrm{gr}.....\left(\mathrm{iv}\right)$$$$\frac{{\mathrm{m}}_{\mathrm{o}}^2{\mathrm{u}}_{\mathrm{o}}^2}{16\mathrm{M}}-4\mathrm{g}l>\mathrm{gr}[\mathrm{r}=l]$$$$\frac{{\mathrm{m}}_{\mathrm{o}}^2{\mathrm{u}}_{\mathrm{o}}^2}{16\mathrm{M}}>5\mathrm{g}l$$$${\mathrm{m}}_{\mathrm{o}}^2{\mathrm{u}}_{\mathrm{o}}^2>80\mathrm{Mg}l$$$${\mathrm{u}}_{\mathrm{o}}>\frac{\sqrt{80\mathrm{Mg}l}}{{\mathrm{m}}_{\mathrm{o}}}$$$${\mathrm{u}}_{\mathrm{o}}>\frac{4\mathrm{M}\sqrt{5\mathrm{g}l}}{m_o}$$$${\mathrm{u}}_{\mathrm{o}}=\frac{4\mathrm{M}\sqrt{5\mathrm{g}l}}{m_o}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com