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Question Number 28050 by ajfour last updated on 19/Jan/18

Commented by ajfour last updated on 19/Jan/18

$T h e b l u e d i s c, r a d i u s r, m a s s m,$ $r o l l s o n t h e g r o u n d i n a c i r c l e o f$ $r a d i u s R . I t r o l l s a n d i t s a x i s$ $m a i n t a i n s a c o n s t a n t a n g l e w i t h$ $t h e h o r i z o n t a l p l a n e . A s s u m i n g$ $n o e n e r g y l o s s e s t h e r e f o r e, f i n d$ $i t s k i n e t i c e n e r g y .$
Commented by mrW2 last updated on 19/Jan/18

$K E = \frac{1}{2} I ω^{2} + \frac{1}{2} {m v}^{2} = \frac{1}{2} \times \frac{{m r}^{2}}{2} ω^{2} + \frac{1}{2} m {(r ω)}^{2} = \frac{3 {m r}^{2} ω^{2}}{4}$
Commented by ajfour last updated on 19/Jan/18

$F i n d ω a l s o i f t h e d i s c h a s t o k e e p$ $r o l l i n g i n t h e c i r c l e o f r a d i u s R,$ $S i r . T h e d i s c i s t h i n, i t s a x i s i s$ $b e n t b y s o m e a c u t e a n g l e .$ $F u r t h e r w e h a v e t w o ω t y p e s . .$
Commented by mrW2 last updated on 19/Jan/18

$K E = \frac{3 m g r}{4} \sqrt{1 - {(\frac{r}{R})}^{2}}$
Commented by ajfour last updated on 19/Jan/18

$c a n n o t c o n f i r m S i r, t h e q u e s t i o n$ $o c c u r e d t o m e s e e i n g a l i t t l e$ $s i m i l a r o n e i n a b o o k .$ $P l e a s e e x p l a i n .$
Commented by mrW2 last updated on 19/Jan/18

$K E = \frac{3 {m v}^{2}}{4}$ $\frac{{m v}^{2}}{R^{'}} = m g \sin θ = m g \frac{r}{R}$ $R^{'} = \sqrt{R^{2} - r^{2}}$ $\frac{{m v}^{2}}{\sqrt{R^{2} - r^{2}}} = m g \frac{r}{R}$ $\Rightarrow v^{2} = g r \sqrt{1 - {(\frac{r}{R})}^{2}}$ $\Rightarrow K E = \frac{3 m g r}{4} \sqrt{1 - {(\frac{r}{R})}^{2}}$
Commented by mrW2 last updated on 19/Jan/18

Commented by ajfour last updated on 19/Jan/18

$w h y s h o u l d \sin θ = \frac{r}{R}, i {c o u l d n}^{'} t$ $f o l l o w, S i r . D o m a k e c l e a r a l i t t l e$ $m o r e .$
Commented by mrW2 last updated on 19/Jan/18

$I t h i n k t h e f o r c e c o m p o n e n t m g \sin θ$ $s h o u l d p o i n t t o t h e c e n t e r o f c i r c l e,$ $t h a t m e a n s \sin θ = \frac{r}{R}, a s t h e d i a g r a m$ $s h o w s .$
Commented by ajfour last updated on 19/Jan/18

$g o o d c l u e, S i r . T h a n k y o u .$
Commented by ajfour last updated on 20/Jan/18

$b u t i n o n e r e v o l u t i o n, t h e r e i s o n e$ $r o t a t i o n e v e n o f t h e d i s c a b o u t$ $v e r t i c a l a x i s, S i r!$
Commented by mrW2 last updated on 20/Jan/18

$Y e s . I n o t i c e d t h i s t o o . T h e d i s c m a k e s$ $o n e r o t a t i o n a b o u t i t s v e r t i c a l a x i s$ $w h e n i t r o l l s a r o u n d a l o n e t h e c i r c l e .$ $T h e a n g u l a r s p e e d i s ω_{3} = \frac{2 π}{\frac{2 π R}{v}} = \frac{v}{R} = \frac{ω_{1} r}{R},$ $t h e c o r r e s p o n d i n g {K E}_{3} = \frac{1}{2} I_{3} ω_{3}^{2}$ $= \frac{1}{2} \times \frac{{m r}^{2}}{4} \times \frac{v^{2}}{R^{2}} = \frac{{m g r}^{3}}{8 R^{2}} \sqrt{1 - {(\frac{r}{R})}^{2}}$ $\Rightarrow {K E}_{t o t a l} = \frac{m g r}{8} [6 + {(\frac{r}{R})}^{2}] \sqrt{1 - {(\frac{r}{R})}^{2}}$
Commented by ajfour last updated on 20/Jan/18

$I f i t i s c o r r e c t, r e a l l y w o n d u r f u l$ $S i r, i a m a m a z e d!$
Commented by ajfour last updated on 20/Jan/18

$I_{3} = \frac{{m r}^{2}}{4} i t h i n k i c a n t b e v e r y s u r e$ $w i t h o u t y o u r h e l p, S i r . .$
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