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Question Number 23531 by ajfour last updated on 01/Nov/17

Commented by ajfour last updated on 01/Nov/17

If the system consisting of a  ring of radius R, mass M, mounted  on a horizontal axis with a disc  of radius r, mass m connected   as shown, through a rod of mass m_0    be released as shown; find the  initial  angular accelerations  of rod, disc(about its centre),   and ring.  For a much simpler case  please assume R=3r and      M=m=m_0  .

$${If}\:{the}\:{system}\:{consisting}\:{of}\:{a} \\ $$$${ring}\:{of}\:{radius}\:\boldsymbol{{R}},\:{mass}\:\boldsymbol{{M}},\:{mounted} \\ $$$${on}\:{a}\:{horizontal}\:{axis}\:{with}\:{a}\:{disc} \\ $$$${of}\:{radius}\:\boldsymbol{{r}},\:{mass}\:\boldsymbol{{m}}\:{connected}\: \\ $$$${as}\:{shown},\:{through}\:{a}\:{rod}\:{of}\:{mass}\:{m}_{\mathrm{0}} \: \\ $$$${be}\:{released}\:{as}\:{shown};\:{find}\:{the} \\ $$$${initial}\:\:{angular}\:{accelerations} \\ $$$${of}\:{rod},\:{disc}\left({about}\:{its}\:{centre}\right),\: \\ $$$${and}\:{ring}. \\ $$$${For}\:{a}\:{much}\:{simpler}\:{case} \\ $$$${please}\:{assume}\:\boldsymbol{{R}}=\mathrm{3}\boldsymbol{{r}}\:{and} \\ $$$$\:\:\:\:\boldsymbol{{M}}=\boldsymbol{{m}}=\boldsymbol{{m}}_{\mathrm{0}} \:. \\ $$

Commented by ajfour last updated on 01/Nov/17

does this help you figure the  situation sir, everwhere turning  is allowed, frictionless bearings  are present. The ring is mounted  on a horizontal axis ⇒ it is free  to turn about it in a vertical plane.

$${does}\:{this}\:{help}\:{you}\:{figure}\:{the} \\ $$$${situation}\:{sir},\:{everwhere}\:{turning} \\ $$$${is}\:{allowed},\:{frictionless}\:{bearings} \\ $$$${are}\:{present}.\:{The}\:{ring}\:{is}\:{mounted} \\ $$$${on}\:{a}\:{horizontal}\:{axis}\:\Rightarrow\:{it}\:{is}\:{free} \\ $$$${to}\:{turn}\:{about}\:{it}\:{in}\:{a}\:{vertical}\:{plane}. \\ $$

Commented by mrW1 last updated on 01/Nov/17

thank you for this info. i will try.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{this}\:\mathrm{info}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{try}. \\ $$

Answered by mrW1 last updated on 01/Nov/17

let O=center point of ring  let C=center point of disc  θ=rotation of bar (rod) about O  ϕ=rotation of disc about C  φ=rotation of ring about O    φ=θ(1−(r/R))+(r/R)ϕ  α_B =angular acceleration of bar  α_D =angular acceleration of disc  α_R =angular acceleration of ring  ⇒α_R =(1−(r/R))α_B +(r/R)α_D     I_R =MR^2  for ring  I_D =((mr^2 )/2) for disc  I_B =((m_0 (R−r)^2 )/3) for bar (rod)    let F=vertical force between disc and ring  let N=vertical force between bar and disc    I_R α_R =F×R   ...(i)  I_D α_D =−F×r   ...(ii)  ma_D =mα_B (R−r)=mg+N−F   ...(iii)  I_B α_B =m_0 g×((R−r)/2)−N(R−r)   ...(iv)    from (ii)  ⇒ F=−(I_D /r)α_D   from (iii)  ⇒ N=m(R−r)α_B −mg+F  ⇒ N=m(R−r)α_B −mg−(I_D /r)α_D   from (i)  ⇒ I_R α_R =−(R/r)×I_D α_D   ⇒ I_R [(1−(r/R))α_B +(r/R)α_D ]=−(R/r)×I_D α_D   ⇒ (1−(r/R))I_R α_B =−((r/R)×I_R +(R/r)×I_D )α_D   ⇒ α_B =−(r/(R−r))×[1+(R^2 /r^2 )×(I_D /I_R )]α_D    ...(v)    from (iv)  ⇒ I_B α_B =(R−r)[(((2m+m_0 )g)/2)−m(R−r)α_B +(I_D /r) α_D ]  ⇒ I_B α_B =(((2m+m_0 )(R−r)g)/2)−m(R−r)^2 α_B +(R−r)(I_D /r) α_D   ⇒[ I_B +m(R−r)^2 ]α_B =(m+(m_0 /2))(R−r)g+((R/r)−1)I_D  α_D   ⇒{(r/(R−r))×[ I_B +m(R−r)^2 ]×[1+(R^2 /r^2 )×(I_D /I_R )]+((R/r)−1)I_D }α_D =−(m+(m_0 /2))(R−r)g  ⇒ α_D =−(((m+(m_0 /2))(R−r)g)/([ (r/(R−r))×I_B +mr(R−r)](1+(R^2 /r^2 )×(I_D /I_R ))+((R/r)−1)I_D ))  ⇒ α_B =−(r/(R−r))(1+(R^2 /r^2 )×(I_D /I_R ))α_D   ⇒α_R =(1−(r/R))α_B +(r/R)α_D       for R=3r and M=m_0 =m, we get  I_R =MR^2 =9mr^2   I_D =((mr^2 )/2)  I_B =((m_0 (R−r)^2 )/3)=((4mr^2 )/3)  ⇒ α_D =−(3/5)×(g/r)  (↶) for disc  ⇒ α_B =(9/(20))×(g/r)  (↷) for rod  ⇒α_R =(1/(10))×(g/r)  (↷) for ring

$$\mathrm{let}\:\mathrm{O}=\mathrm{center}\:\mathrm{point}\:\mathrm{of}\:\mathrm{ring} \\ $$$$\mathrm{let}\:\mathrm{C}=\mathrm{center}\:\mathrm{point}\:\mathrm{of}\:\mathrm{disc} \\ $$$$\theta=\mathrm{rotation}\:\mathrm{of}\:\mathrm{bar}\:\left(\mathrm{rod}\right)\:\mathrm{about}\:\mathrm{O} \\ $$$$\varphi=\mathrm{rotation}\:\mathrm{of}\:\mathrm{disc}\:\mathrm{about}\:\mathrm{C} \\ $$$$\phi=\mathrm{rotation}\:\mathrm{of}\:\mathrm{ring}\:\mathrm{about}\:\mathrm{O} \\ $$$$ \\ $$$$\phi=\theta\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{R}}\right)+\frac{\mathrm{r}}{\mathrm{R}}\varphi \\ $$$$\alpha_{\mathrm{B}} =\mathrm{angular}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{bar} \\ $$$$\alpha_{\mathrm{D}} =\mathrm{angular}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{disc} \\ $$$$\alpha_{\mathrm{R}} =\mathrm{angular}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{ring} \\ $$$$\Rightarrow\alpha_{\mathrm{R}} =\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{R}}\right)\alpha_{\mathrm{B}} +\frac{\mathrm{r}}{\mathrm{R}}\alpha_{\mathrm{D}} \\ $$$$ \\ $$$$\mathrm{I}_{\mathrm{R}} =\mathrm{MR}^{\mathrm{2}} \:\mathrm{for}\:\mathrm{ring} \\ $$$$\mathrm{I}_{\mathrm{D}} =\frac{\mathrm{mr}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{for}\:\mathrm{disc} \\ $$$$\mathrm{I}_{\mathrm{B}} =\frac{\mathrm{m}_{\mathrm{0}} \left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} }{\mathrm{3}}\:\mathrm{for}\:\mathrm{bar}\:\left(\mathrm{rod}\right) \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{F}=\mathrm{vertical}\:\mathrm{force}\:\mathrm{between}\:\mathrm{disc}\:\mathrm{and}\:\mathrm{ring} \\ $$$$\mathrm{let}\:\mathrm{N}=\mathrm{vertical}\:\mathrm{force}\:\mathrm{between}\:\mathrm{bar}\:\mathrm{and}\:\mathrm{disc} \\ $$$$ \\ $$$$\mathrm{I}_{\mathrm{R}} \alpha_{\mathrm{R}} =\mathrm{F}×\mathrm{R}\:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{I}_{\mathrm{D}} \alpha_{\mathrm{D}} =−\mathrm{F}×\mathrm{r}\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{ma}_{\mathrm{D}} =\mathrm{m}\alpha_{\mathrm{B}} \left(\mathrm{R}−\mathrm{r}\right)=\mathrm{mg}+\mathrm{N}−\mathrm{F}\:\:\:...\left(\mathrm{iii}\right) \\ $$$$\mathrm{I}_{\mathrm{B}} \alpha_{\mathrm{B}} =\mathrm{m}_{\mathrm{0}} \mathrm{g}×\frac{\mathrm{R}−\mathrm{r}}{\mathrm{2}}−\mathrm{N}\left(\mathrm{R}−\mathrm{r}\right)\:\:\:...\left(\mathrm{iv}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\:\mathrm{F}=−\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{r}}\alpha_{\mathrm{D}} \\ $$$$\mathrm{from}\:\left(\mathrm{iii}\right) \\ $$$$\Rightarrow\:\mathrm{N}=\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)\alpha_{\mathrm{B}} −\mathrm{mg}+\mathrm{F} \\ $$$$\Rightarrow\:\mathrm{N}=\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)\alpha_{\mathrm{B}} −\mathrm{mg}−\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{r}}\alpha_{\mathrm{D}} \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\mathrm{I}_{\mathrm{R}} \alpha_{\mathrm{R}} =−\frac{\mathrm{R}}{\mathrm{r}}×\mathrm{I}_{\mathrm{D}} \alpha_{\mathrm{D}} \\ $$$$\Rightarrow\:\mathrm{I}_{\mathrm{R}} \left[\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{R}}\right)\alpha_{\mathrm{B}} +\frac{\mathrm{r}}{\mathrm{R}}\alpha_{\mathrm{D}} \right]=−\frac{\mathrm{R}}{\mathrm{r}}×\mathrm{I}_{\mathrm{D}} \alpha_{\mathrm{D}} \\ $$$$\Rightarrow\:\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{R}}\right)\mathrm{I}_{\mathrm{R}} \alpha_{\mathrm{B}} =−\left(\frac{\mathrm{r}}{\mathrm{R}}×\mathrm{I}_{\mathrm{R}} +\frac{\mathrm{R}}{\mathrm{r}}×\mathrm{I}_{\mathrm{D}} \right)\alpha_{\mathrm{D}} \\ $$$$\Rightarrow\:\alpha_{\mathrm{B}} =−\frac{\mathrm{r}}{\mathrm{R}−\mathrm{r}}×\left[\mathrm{1}+\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{r}^{\mathrm{2}} }×\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{I}_{\mathrm{R}} }\right]\alpha_{\mathrm{D}} \:\:\:...\left(\mathrm{v}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{iv}\right) \\ $$$$\Rightarrow\:\mathrm{I}_{\mathrm{B}} \alpha_{\mathrm{B}} =\left(\mathrm{R}−\mathrm{r}\right)\left[\frac{\left(\mathrm{2m}+\mathrm{m}_{\mathrm{0}} \right)\mathrm{g}}{\mathrm{2}}−\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)\alpha_{\mathrm{B}} +\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{r}}\:\alpha_{\mathrm{D}} \right] \\ $$$$\Rightarrow\:\mathrm{I}_{\mathrm{B}} \alpha_{\mathrm{B}} =\frac{\left(\mathrm{2m}+\mathrm{m}_{\mathrm{0}} \right)\left(\mathrm{R}−\mathrm{r}\right)\mathrm{g}}{\mathrm{2}}−\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} \alpha_{\mathrm{B}} +\left(\mathrm{R}−\mathrm{r}\right)\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{r}}\:\alpha_{\mathrm{D}} \\ $$$$\Rightarrow\left[\:\mathrm{I}_{\mathrm{B}} +\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} \right]\alpha_{\mathrm{B}} =\left(\mathrm{m}+\frac{\mathrm{m}_{\mathrm{0}} }{\mathrm{2}}\right)\left(\mathrm{R}−\mathrm{r}\right)\mathrm{g}+\left(\frac{\mathrm{R}}{\mathrm{r}}−\mathrm{1}\right)\mathrm{I}_{\mathrm{D}} \:\alpha_{\mathrm{D}} \\ $$$$\Rightarrow\left\{\frac{\mathrm{r}}{\mathrm{R}−\mathrm{r}}×\left[\:\mathrm{I}_{\mathrm{B}} +\mathrm{m}\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} \right]×\left[\mathrm{1}+\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{r}^{\mathrm{2}} }×\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{I}_{\mathrm{R}} }\right]+\left(\frac{\mathrm{R}}{\mathrm{r}}−\mathrm{1}\right)\mathrm{I}_{\mathrm{D}} \right\}\alpha_{\mathrm{D}} =−\left(\mathrm{m}+\frac{\mathrm{m}_{\mathrm{0}} }{\mathrm{2}}\right)\left(\mathrm{R}−\mathrm{r}\right)\mathrm{g} \\ $$$$\Rightarrow\:\alpha_{\mathrm{D}} =−\frac{\left(\mathrm{m}+\frac{\mathrm{m}_{\mathrm{0}} }{\mathrm{2}}\right)\left(\mathrm{R}−\mathrm{r}\right)\mathrm{g}}{\left[\:\frac{\mathrm{r}}{\mathrm{R}−\mathrm{r}}×\mathrm{I}_{\mathrm{B}} +\mathrm{mr}\left(\mathrm{R}−\mathrm{r}\right)\right]\left(\mathrm{1}+\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{r}^{\mathrm{2}} }×\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{I}_{\mathrm{R}} }\right)+\left(\frac{\mathrm{R}}{\mathrm{r}}−\mathrm{1}\right)\mathrm{I}_{\mathrm{D}} } \\ $$$$\Rightarrow\:\alpha_{\mathrm{B}} =−\frac{\mathrm{r}}{\mathrm{R}−\mathrm{r}}\left(\mathrm{1}+\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{r}^{\mathrm{2}} }×\frac{\mathrm{I}_{\mathrm{D}} }{\mathrm{I}_{\mathrm{R}} }\right)\alpha_{\mathrm{D}} \\ $$$$\Rightarrow\alpha_{\mathrm{R}} =\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{R}}\right)\alpha_{\mathrm{B}} +\frac{\mathrm{r}}{\mathrm{R}}\alpha_{\mathrm{D}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{R}=\mathrm{3r}\:\mathrm{and}\:\mathrm{M}=\mathrm{m}_{\mathrm{0}} =\mathrm{m},\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{I}_{\mathrm{R}} =\mathrm{MR}^{\mathrm{2}} =\mathrm{9mr}^{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{D}} =\frac{\mathrm{mr}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{B}} =\frac{\mathrm{m}_{\mathrm{0}} \left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{4mr}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\alpha_{\mathrm{D}} =−\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{g}}{\mathrm{r}}\:\:\left(\curvearrowleft\right)\:\mathrm{for}\:\mathrm{disc} \\ $$$$\Rightarrow\:\alpha_{\mathrm{B}} =\frac{\mathrm{9}}{\mathrm{20}}×\frac{\mathrm{g}}{\mathrm{r}}\:\:\left(\curvearrowright\right)\:\mathrm{for}\:\mathrm{rod} \\ $$$$\Rightarrow\alpha_{\mathrm{R}} =\frac{\mathrm{1}}{\mathrm{10}}×\frac{\mathrm{g}}{\mathrm{r}}\:\:\left(\curvearrowright\right)\:\mathrm{for}\:\mathrm{ring} \\ $$

Commented by mrW1 last updated on 01/Nov/17

Commented by ajfour last updated on 02/Nov/17

 your answers are correct   Sir.   I checked your solution,  carefully, dint find any error,  attempted on my own, got the  same answers. 𝛂_D =−((3g)/(5r)) .  Thank you Sir.

$$\:\boldsymbol{{your}}\:\boldsymbol{{answers}}\:\boldsymbol{{are}}\:\boldsymbol{{correct}}\: \\ $$$$\boldsymbol{{Sir}}. \\ $$$$\:{I}\:{checked}\:{your}\:{solution}, \\ $$$${carefully},\:{dint}\:{find}\:{any}\:{error}, \\ $$$${attempted}\:{on}\:{my}\:{own},\:{got}\:{the} \\ $$$${same}\:{answers}.\:\boldsymbol{\alpha}_{{D}} =−\frac{\mathrm{3}{g}}{\mathrm{5}{r}}\:. \\ $$$${Thank}\:{you}\:{Sir}. \\ $$

Commented by mrW1 last updated on 02/Nov/17

Thank you for rechecking!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{rechecking}! \\ $$

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