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 30165 by rahul 19 last updated on 17/Feb/18

Commented by ajfour last updated on 19/Feb/18

let us first choose a fixed x-axis.  ω^�  = ωk^�^�     and at time t  θ=ωt  r^� =r(cos θi^� +sin θj^� )    =r(cos ωti^� +sin ωtj^� )=re_r ^�   v^� =(dr/dt)e_r ^� +ωr(−sin ωti^� +cos ωtj^� )    =(dr/dt)e_r ^� +ωre_θ ^�   a^� =(d^2 r/dt^2 )e_r ^� +ω(dr/dt)e_θ ^� +ω(dr/dt)e_θ ^� −ω^2 re_r ^�   as there is complete absence of  force along radial direction, so  radial component of acceleration  is zero.  ⇒  (d^2 r/dt^2 )−ω^2 r =0      let  v_r = (dr/dt)  ⇒  ((v_r dv_r )/dr) = ω^2 r      ∫_0 ^(  v_r ) v_r dr = ω^2 ∫_(R/2) ^(  r) rdr       v_r ^2  = ω^2 (r^2 −(R^2 /4))  v_r = (dr/dt) = ω(√(r^2 −((R/2))^2 ))        ∫_(R/2) ^(  r) (dr/(√(r^2 −((R/2))^2 ))) = ω∫_0 ^(  t) dt  ⇒  ln (r+(√(r^2 −(R^2 /4))))∣_(R/2) ^r = ωt  ⇒  r+(√(r^2 −(R^2 /4))) = (R/2)e^(ωt)     ...(i)        rationalising and taking  reciprocal we also get,        r−(√(r^2 −(R^2 /4))) = (R/2)e^(−ωt)     ...(ii)  hence by adding (i) and (ii)        r = (R/4)(e^(𝛚t) +e^(−𝛚t) )  .  Force (tangential) on block by  disc   = F^�  = m(2ω(dr/dt))e_θ ^�           F^�  =2mωv_r e_θ ^�                =(2mω^2 (√(r^2 −(R^2 /4))) )e_θ ^�   (i)−(ii) gives        (√(r^2 −(R^2 /4))) =(R/4)(e^(ωt) −e^(−ωt) )    hence  F^�  = ((mω^2 R)/2)(e^(ωt) −e^(−ωt) )e_θ ^�   and if x axis is chosen to be  along radial direction rotating  along with disc then i^�  is along e_r ^�   and  j^�  is along e_θ ^�  .  Then F^�  = ((mω^2 R)/2)(e^(ωt) −e^(−ωt) )j^�   disc also supports weight of block  hence we also have a force    N_z =mgk^�   N^�   the reaction force of disc on  block is   N^�  = F^� +mgk^�    N^�  = ((mω^2 R)/2)(e^(ωt) −e^(ωt) )j^� +mgk^�   .

$${let}\:{us}\:{first}\:{choose}\:{a}\:{fixed}\:{x}-{axis}. \\ $$$$\bar {\omega}\:=\:\omega\hat {{k}}\:\:\:{and}\:{at}\:{time}\:{t}\:\:\theta=\omega{t} \\ $$$$\bar {{r}}={r}\left(\mathrm{cos}\:\theta\hat {{i}}+\mathrm{sin}\:\theta\hat {{j}}\right) \\ $$$$\:\:={r}\left(\mathrm{cos}\:\omega{t}\hat {{i}}+\mathrm{sin}\:\omega{t}\hat {{j}}\right)={r}\hat {{e}}_{{r}} \\ $$$$\bar {{v}}=\frac{{dr}}{{dt}}\hat {{e}}_{{r}} +\omega{r}\left(−\mathrm{sin}\:\omega{t}\hat {{i}}+\mathrm{cos}\:\omega{t}\hat {{j}}\right) \\ $$$$\:\:=\frac{{dr}}{{dt}}\hat {{e}}_{{r}} +\omega{r}\hat {{e}}_{\theta} \\ $$$$\bar {{a}}=\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }\hat {{e}}_{{r}} +\omega\frac{{dr}}{{dt}}\hat {{e}}_{\theta} +\omega\frac{{dr}}{{dt}}\hat {{e}}_{\theta} −\omega^{\mathrm{2}} {r}\hat {{e}}_{{r}} \\ $$$${as}\:{there}\:{is}\:{complete}\:{absence}\:{of} \\ $$$${force}\:{along}\:{radial}\:{direction},\:{so} \\ $$$${radial}\:{component}\:{of}\:{acceleration} \\ $$$${is}\:{zero}. \\ $$$$\Rightarrow\:\:\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }−\omega^{\mathrm{2}} {r}\:=\mathrm{0} \\ $$$$\:\:\:\:{let}\:\:{v}_{{r}} =\:\frac{{dr}}{{dt}} \\ $$$$\Rightarrow\:\:\frac{{v}_{{r}} {dv}_{{r}} }{{dr}}\:=\:\omega^{\mathrm{2}} {r} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\:{v}_{{r}} } {v}_{{r}} {dr}\:=\:\omega^{\mathrm{2}} \int_{{R}/\mathrm{2}} ^{\:\:{r}} {rdr} \\ $$$$\:\:\:\:\:{v}_{{r}} ^{\mathrm{2}} \:=\:\omega^{\mathrm{2}} \left({r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${v}_{{r}} =\:\frac{{dr}}{{dt}}\:=\:\omega\sqrt{{r}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\int_{{R}/\mathrm{2}} ^{\:\:{r}} \frac{{dr}}{\sqrt{{r}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:=\:\omega\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:\left({r}+\sqrt{{r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}}\right)\mid_{{R}/\mathrm{2}} ^{{r}} =\:\omega{t} \\ $$$$\Rightarrow\:\:{r}+\sqrt{{r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{{R}}{\mathrm{2}}{e}^{\omega{t}} \:\:\:\:...\left({i}\right) \\ $$$$\:\:\:\:\:\:{rationalising}\:{and}\:{taking} \\ $$$${reciprocal}\:{we}\:{also}\:{get}, \\ $$$$\:\:\:\:\:\:{r}−\sqrt{{r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{{R}}{\mathrm{2}}{e}^{−\omega{t}} \:\:\:\:...\left({ii}\right) \\ $$$${hence}\:{by}\:{adding}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\boldsymbol{{r}}\:=\:\frac{\boldsymbol{{R}}}{\mathrm{4}}\left(\boldsymbol{{e}}^{\boldsymbol{\omega{t}}} +\boldsymbol{{e}}^{−\boldsymbol{\omega{t}}} \right)\:\:. \\ $$$${Force}\:\left({tangential}\right)\:{on}\:{block}\:{by} \\ $$$${disc}\:\:\:=\:\bar {{F}}\:=\:{m}\left(\mathrm{2}\omega\frac{{dr}}{{dt}}\right)\hat {{e}}_{\theta} \\ $$$$\:\:\:\:\:\:\:\:\bar {{F}}\:=\mathrm{2}{m}\omega{v}_{{r}} \hat {{e}}_{\theta} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}{m}\omega^{\mathrm{2}} \sqrt{{r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}}\:\right)\hat {{e}}_{\theta} \\ $$$$\left({i}\right)−\left({ii}\right)\:{gives} \\ $$$$\:\:\:\:\:\:\sqrt{{r}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}}\:=\frac{{R}}{\mathrm{4}}\left({e}^{\omega{t}} −{e}^{−\omega{t}} \right) \\ $$$$\:\:{hence}\:\:\bar {{F}}\:=\:\frac{{m}\omega^{\mathrm{2}} {R}}{\mathrm{2}}\left({e}^{\omega{t}} −{e}^{−\omega{t}} \right)\hat {{e}}_{\theta} \\ $$$${and}\:{if}\:{x}\:{axis}\:{is}\:{chosen}\:{to}\:{be} \\ $$$${along}\:{radial}\:{direction}\:{rotating} \\ $$$${along}\:{with}\:{disc}\:{then}\:\hat {{i}}\:{is}\:{along}\:\hat {{e}}_{{r}} \\ $$$${and}\:\:\hat {{j}}\:{is}\:{along}\:\hat {{e}}_{\theta} \:. \\ $$$${Then}\:\bar {{F}}\:=\:\frac{{m}\omega^{\mathrm{2}} {R}}{\mathrm{2}}\left({e}^{\omega{t}} −{e}^{−\omega{t}} \right)\hat {{j}} \\ $$$${disc}\:{also}\:{supports}\:{weight}\:{of}\:{block} \\ $$$${hence}\:{we}\:{also}\:{have}\:{a}\:{force} \\ $$$$\:\:{N}_{{z}} ={mg}\hat {{k}} \\ $$$$\bar {\boldsymbol{{N}}}\:\:{the}\:{reaction}\:{force}\:{of}\:{disc}\:{on} \\ $$$${block}\:{is}\:\:\:\bar {{N}}\:=\:\bar {{F}}+{mg}\hat {{k}} \\ $$$$\:\bar {{N}}\:=\:\frac{{m}\omega^{\mathrm{2}} {R}}{\mathrm{2}}\left({e}^{\omega{t}} −{e}^{\omega{t}} \right)\hat {{j}}+{mg}\hat {{k}}\:\:. \\ $$

Commented by ajfour last updated on 19/Feb/18

Commented by rahul 19 last updated on 19/Feb/18

WOW …!

$$\mathscr{W}\mathfrak{O}\mathscr{W}\:\ldots! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com