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Question Number 213920 by ajfour last updated on 21/Nov/24

Commented by ajfour last updated on 21/Nov/24

$C a n w e f i n d a t w h a t s p e e d d o e s p o i n t$ $P a p p r o a c h t h e g r o u n d j u s t b e f o r e$ $h i t t i n g t h e g r o u n d i f r e l e a s e d a t s a y$ $t h e l o w e r e d g e a t 45 ° t o h o r i z o n t a l .$ $T h e r a d i u s o f s e m i - d i s c i s r, m a s s m .$
	Answered by mr W last updated on 22/Nov/24

	Commented by mr W last updated on 23/Nov/24

	$e = \frac{4 r}{3 π} = k r w i t h k = \frac{4}{3 π}$ $I_{0} = (\frac{1}{2} - \frac{16}{9 π^{2}}) {m r}^{2} = ξ {m r}^{2} w i t h ξ = \frac{1}{2} - \frac{16}{9 π^{2}}$ $t h e c e n t e r p o i n t G o f d i s c m o v e s$ $o n l y v e r t i c a l l y, b e c a u s e n o f o r c e$ $i s a c t i n g i n h o r i z o n t a l d i r e c t i o n .$ $y_{G} = r \sin θ + e \cos θ = r (\sin θ + k \cos θ)$ $v_{G y} = - \frac{{d y}_{G}}{d t} = r (\cos θ - k \sin θ) ω$ $w i t h ω = - \frac{d θ}{d t}$ $a t t = 0 : θ_{0} = \frac{π}{4}$ $y_{G 0} = r (\sin θ_{0} + k \cos θ_{0})$ $m g (y_{G 0} - r \sin θ - e \cos θ) = \frac{I_{0} ω^{2}}{2} + \frac{{m v}_{G y}^{2}}{2}$ $2 m g r (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)$ $= m ξ r^{2} ω^{2} + {m r}^{2} {(\cos θ - k \sin θ)}^{2} ω^{2}$ $2 g (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)$ $= r [ξ + {(\cos θ - k \sin θ)}^{2}] ω^{2}$ $\Rightarrow ω = \sqrt{\frac{g}{r}} \sqrt{\frac{2 (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}{ξ + {(\cos θ - k \sin θ)}^{2}}}$ $y_{P} = 2 r \sin θ$ $v_{P y} = - \frac{{d y}_{P}}{d t} = - 2 r \cos θ \frac{d θ}{d t} = 2 r \cos θ ω$ $\Rightarrow v_{P y} = 2 \cos θ \sqrt{\frac{2 g r (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}{ξ + {(\cos θ - k \sin θ)}^{2}}}$ $a t θ = 0 :$ $v_{P y} = \sqrt{\frac{[\sqrt{2} - \frac{4 (2 - \sqrt{2})}{3 π}] g r}{\frac{3}{8} - \frac{4}{9 π^{2}}}}$ $\approx 1.87948239 \sqrt{g r}$ $+ + + + + + + + + + + + +$ $\frac{d θ}{d t} = - \sqrt{\frac{2 g (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}{r [ξ + {(\cos θ - k \sin θ)}^{2}]}}$ $d t = - \sqrt{\frac{r [ξ + {(\cos θ - k \sin θ)}^{2}]}{2 g (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}} d θ$ $\int_{0}^{T} d t = \sqrt{\frac{r}{g}} \int_{0}^{\frac{π}{4}} \sqrt{\frac{ξ + {(\cos θ - k \sin θ)}^{2}}{2 (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}} d θ$ $T = \sqrt{\frac{r}{g}} \int_{0}^{\frac{π}{4}} \sqrt{\frac{\frac{1}{2} - \frac{16}{9 π^{2}} + {(\cos θ - \frac{4 \sin θ}{3 π})}^{2}}{\sqrt{2} + \frac{4 \sqrt{2}}{3 π} - 2 \sin θ - \frac{8 \cos θ}{3 π}}} d θ$ $\approx 1.47971689 \sqrt{\frac{r}{g}}$ $+ + + + + + + + + + + + +$ $l e t f (θ) = \sqrt{\frac{2 (\sin θ_{0} + k \cos θ_{0} - \sin θ - k \cos θ)}{ξ + {(\cos θ - k \sin θ)}^{2}}}$ $ω = \sqrt{\frac{g}{r}} f (θ)$ $α = \frac{d ω}{d t} = ω \frac{d ω}{d θ} = \frac{g}{r} f (θ) \frac{d f (θ)}{d θ}$ $N (r \cos θ - e \sin θ) = I_{0} α$ $N r (\cos θ - k \sin θ) = m ξ r^{2} ω \frac{d ω}{d θ}$ $\Rightarrow \frac{N}{m g} = \frac{(\frac{1}{2} - \frac{16}{9 π^{2}}) f (θ) \frac{d f (θ)}{d θ}}{\cos θ - k \sin θ}$
	Commented by ajfour last updated on 23/Nov/24
	$v_(Py) ^2 =(2gr){(((1/( (√2)))(k+1)−k)/((1/2)(1+k^2 )−k^2 +(1/2)(1−k)^2 ))} v_(Py) ^2 =(√2)gr{((1−((√2)−1)k)/(1−k))} ≈ 1.423056 (√(gr))$
	$v_{P y}^{2} = (2 g r) {\frac{\frac{1}{\sqrt{2}} (k + 1) - k}{\frac{1}{2} (1 + k^{2}) - k^{2} + \frac{1}{2} {(1 - k)}^{2}}}$ $v_{P y}^{2} = \sqrt{2} g r {\frac{1 - (\sqrt{2} - 1) k}{1 - k}}$ $\approx 1.423056 \sqrt{g r}$
	Commented by mr W last updated on 23/Nov/24

	$d i d y o u t a k e t h e l o w e r p o i n t a s a$ $h i n g e ? i t h o u g h t i t w e r e a r o l l e r .$
	Commented by ajfour last updated on 23/Nov/24

	$y e a h r o l l e r S i r, i h a v e u s e d$ $K . E . = \frac{1}{2} I_{c m} ω^{2} + \frac{1}{2} {m v}_{c m}^{2}$
	Commented by ajfour last updated on 23/Nov/24

	Commented by ajfour last updated on 23/Nov/24

	$E x c e l l e n t s i r, t h a n k y o u v v e r y m u c h .$
	Commented by mr W last updated on 23/Nov/24

	Commented by mr W last updated on 23/Nov/24

	$p l e a s e r e c h e c k!$ $a t t = 0 :$ $θ = 45 °$ $P . E . = m g R \sin (θ + β) = m g (r \sin θ + h \cos θ)$ $a t t = T :$ $P . E . = m g h$ $v_{G} = ω r = ω R \cos β$ $K . E . = \frac{I_{0} ω^{2}}{2} + \frac{{m v}_{G}^{2}}{2} = \frac{m ω^{2} R^{2}}{2} (\frac{I_{0}}{{m R}^{2}} + \cos^{2} β)$ $m g R \sin (θ + β) - m g h = \frac{m ω^{2} R^{2}}{2} (\frac{I_{0}}{{m R}^{2}} + \cos^{2} β)$ $2 g [R \sin (θ + β) - h] = ω^{2} (\frac{I_{0}}{m} + r^{2})$ $2 g [r \sin θ + h \cos θ - h] = ω^{2} (\frac{1}{2} - \frac{16}{9 π^{2}} + 1) r^{2}$ $2 g (\sin θ + k \cos θ - k) = ω^{2} (\frac{1}{2} - \frac{16}{9 π^{2}} + 1) r$ $\Rightarrow ω^{2} = \frac{2 g (\sin θ + k \cos θ - k)}{r (\frac{3}{2} - \frac{16}{9 π^{2}})} = \frac{g (\sqrt{2} + \sqrt{2} k - 2 k)}{r (\frac{3}{2} - \frac{16}{9 π^{2}})}$ $v_{P} = 2 r ω = 2 \sqrt{\frac{(\sqrt{2} + \sqrt{2} k - 2 k) g r}{\frac{3}{2} - \frac{16}{9 π^{2}}}}$
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