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

Commented by ajfour last updated on 01/Nov/17

$I f t h e s y s t e m c o n s i s t i n g o f a$ $r i n g o f r a d i u s R, m a s s M, m o u n t e d$ $o n a h o r i z o n t a l a x i s w i t h a d i s c$ $o f r a d i u s r, m a s s m c o n n e c t e d$ $a s s h o w n, t h r o u g h a r o d o f m a s s m_{0}$ $b e r e l e a s e d a s s h o w n; f i n d t h e$ $i n i t i a l a n g u l a r a c c e l e r a t i o n s$ $o f r o d, d i s c (a b o u t i t s c e n t r e),$ $a n d r i n g .$ $F o r a m u c h s i m p l e r c a s e$ $p l e a s e a s s u m e R = 3 r a n d$ $M = m = m_{0} .$
Commented by ajfour last updated on 01/Nov/17

$d o e s t h i s h e l p y o u f i g u r e t h e$ $s i t u a t i o n s i r, e v e r w h e r e t u r n i n g$ $i s a l l o w e d, f r i c t i o n l e s s b e a r i n g s$ $a r e p r e s e n t . T h e r i n g i s m o u n t e d$ $o n a h o r i z o n t a l a x i s \Rightarrow i t i s f r e e$ $t o t u r n a b o u t i t i n a v e r t i c a l p l a n e .$
Commented by mrW1 last updated on 01/Nov/17

$thank you for this info . i will try .$
	Answered by mrW1 last updated on 01/Nov/17
	$let O=center point of ring let C=center point of disc θ=rotation of bar (rod) about O ϕ=rotation of disc about C φ=rotation of ring about O φ=θ(1−(r/R))+(r/R)ϕ α_B =angular acceleration of bar α_D =angular acceleration of disc α_R =angular acceleration of ring ⇒α_R =(1−(r/R))α_B +(r/R)α_D I_R =MR^2 for ring I_D =((mr^2 )/2) for disc I_B =((m_0 (R−r)^2 )/3) for bar (rod) let F=vertical force between disc and ring let N=vertical force between bar and disc I_R α_R =F×R ...(i) I_D α_D =−F×r ...(ii) ma_D =mα_B (R−r)=mg+N−F ...(iii) I_B α_B =m_0 g×((R−r)/2)−N(R−r) ...(iv) from (ii) ⇒ F=−(I_D /r)α_D from (iii) ⇒ N=m(R−r)α_B −mg+F ⇒ N=m(R−r)α_B −mg−(I_D /r)α_D from (i) ⇒ I_R α_R =−(R/r)×I_D α_D ⇒ I_R [(1−(r/R))α_B +(r/R)α_D ]=−(R/r)×I_D α_D ⇒ (1−(r/R))I_R α_B =−((r/R)×I_R +(R/r)×I_D )α_D ⇒ α_B =−(r/(R−r))×[1+(R^2 /r^2 )×(I_D /I_R )]α_D ...(v) from (iv) ⇒ I_B α_B =(R−r)[(((2m+m_0 )g)/2)−m(R−r)α_B +(I_D /r) α_D ] ⇒ I_B α_B =(((2m+m_0 )(R−r)g)/2)−m(R−r)^2 α_B +(R−r)(I_D /r) α_D ⇒[ I_B +m(R−r)^2 ]α_B =(m+(m_0 /2))(R−r)g+((R/r)−1)I_D α_D ⇒{(r/(R−r))×[ I_B +m(R−r)^2 ]×[1+(R^2 /r^2 )×(I_D /I_R )]+((R/r)−1)I_D }α_D =−(m+(m_0 /2))(R−r)g ⇒ α_D =−(((m+(m_0 /2))(R−r)g)/([ (r/(R−r))×I_B +mr(R−r)](1+(R^2 /r^2 )×(I_D /I_R ))+((R/r)−1)I_D )) ⇒ α_B =−(r/(R−r))(1+(R^2 /r^2 )×(I_D /I_R ))α_D ⇒α_R =(1−(r/R))α_B +(r/R)α_D for R=3r and M=m_0 =m, we get I_R =MR^2 =9mr^2 I_D =((mr^2 )/2) I_B =((m_0 (R−r)^2 )/3)=((4mr^2 )/3) ⇒ α_D =−(3/5)×(g/r) (↶) for disc ⇒ α_B =(9/(20))×(g/r) (↷) for rod ⇒α_R =(1/(10))×(g/r) (↷) for ring$
	$let O = center point of ring$ $let C = center point of disc$ $θ = rotation of bar (rod) about O$ $φ = rotation of disc about C$ $ϕ = rotation of ring about O$ $ϕ = θ (1 - \frac{r}{R}) + \frac{r}{R} φ$ $α_{B} = angular acceleration of bar$ $α_{D} = angular acceleration of disc$ $α_{R} = angular acceleration of ring$ $\Rightarrow α_{R} = (1 - \frac{r}{R}) α_{B} + \frac{r}{R} α_{D}$ $I_{R} = {MR}^{2} for ring$ $I_{D} = \frac{{mr}^{2}}{2} for disc$ $I_{B} = \frac{m_{0} {(R - r)}^{2}}{3} for bar (rod)$ $let F = vertical force between disc and ring$ $let N = vertical force between bar and disc$ $I_{R} α_{R} = F \times R . . . (i)$ $I_{D} α_{D} = - F \times r . . . (ii)$ ${ma}_{D} = m α_{B} (R - r) = mg + N - F . . . (iii)$ $I_{B} α_{B} = m_{0} g \times \frac{R - r}{2} - N (R - r) . . . (iv)$ $from (ii)$ $\Rightarrow F = - \frac{I_{D}}{r} α_{D}$ $from (iii)$ $\Rightarrow N = m (R - r) α_{B} - mg + F$ $\Rightarrow N = m (R - r) α_{B} - mg - \frac{I_{D}}{r} α_{D}$ $from (i)$ $\Rightarrow I_{R} α_{R} = - \frac{R}{r} \times I_{D} α_{D}$ $\Rightarrow I_{R} [(1 - \frac{r}{R}) α_{B} + \frac{r}{R} α_{D}] = - \frac{R}{r} \times I_{D} α_{D}$ $\Rightarrow (1 - \frac{r}{R}) I_{R} α_{B} = - (\frac{r}{R} \times I_{R} + \frac{R}{r} \times I_{D}) α_{D}$ $\Rightarrow α_{B} = - \frac{r}{R - r} \times [1 + \frac{R^{2}}{r^{2}} \times \frac{I_{D}}{I_{R}}] α_{D} . . . (v)$ $from (iv)$ $\Rightarrow I_{B} α_{B} = (R - r) [\frac{(2 m + m_{0}) g}{2} - m (R - r) α_{B} + \frac{I_{D}}{r} α_{D}]$ $\Rightarrow I_{B} α_{B} = \frac{(2 m + m_{0}) (R - r) g}{2} - m {(R - r)}^{2} α_{B} + (R - r) \frac{I_{D}}{r} α_{D}$ $\Rightarrow [I_{B} + m {(R - r)}^{2}] α_{B} = (m + \frac{m_{0}}{2}) (R - r) g + (\frac{R}{r} - 1) I_{D} α_{D}$ $\Rightarrow {\frac{r}{R - r} \times [I_{B} + m {(R - r)}^{2}] \times [1 + \frac{R^{2}}{r^{2}} \times \frac{I_{D}}{I_{R}}] + (\frac{R}{r} - 1) I_{D}} α_{D} = - (m + \frac{m_{0}}{2}) (R - r) g$ $\Rightarrow α_{D} = - \frac{(m + \frac{m_{0}}{2}) (R - r) g}{[\frac{r}{R - r} \times I_{B} + mr (R - r)] (1 + \frac{R^{2}}{r^{2}} \times \frac{I_{D}}{I_{R}}) + (\frac{R}{r} - 1) I_{D}}$ $\Rightarrow α_{B} = - \frac{r}{R - r} (1 + \frac{R^{2}}{r^{2}} \times \frac{I_{D}}{I_{R}}) α_{D}$ $\Rightarrow α_{R} = (1 - \frac{r}{R}) α_{B} + \frac{r}{R} α_{D}$ $for R = 3 r and M = m_{0} = m, we get$ $I_{R} = {MR}^{2} = {9 mr}^{2}$ $I_{D} = \frac{{mr}^{2}}{2}$ $I_{B} = \frac{m_{0} {(R - r)}^{2}}{3} = \frac{{4 mr}^{2}}{3}$ $\Rightarrow α_{D} = - \frac{3}{5} \times \frac{g}{r} (↶) for disc$ $\Rightarrow α_{B} = \frac{9}{20} \times \frac{g}{r} (↷) for rod$ $\Rightarrow α_{R} = \frac{1}{10} \times \frac{g}{r} (↷) for ring$
	Commented by mrW1 last updated on 01/Nov/17

	Commented by ajfour last updated on 02/Nov/17

	$y o u r a n s w e r s a r e c o r r e c t$ $S i r .$ $I c h e c k e d y o u r s o l u t i o n,$ $c a r e f u l l y, d i n t f i n d a n y e r r o r,$ $a t t e m p t e d o n m y o w n, g o t t h e$ $s a m e a n s w e r s . α_{D} = - \frac{3 g}{5 r} .$ $T h a n k y o u S i r .$
	Commented by mrW1 last updated on 02/Nov/17

	$Thank you for rechecking!$
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