Question-144420

Question Number 144420 by ajfour last updated on 25/Jun/21

Commented by ajfour last updated on 25/Jun/21

T w o c y l i n d e r s o n e o n t o p o f

a n o t h e r a r e r e l e a s e d a n d

m o t i o n i n i t i a t e d w i t h a m i l d

d i s t u r b a n c e . I f a l l s u r f a c e s

a r e f r i c t i o n l e s s f i n d t h e

s p e e d s u a n d v o f t h e c y l i n d e r s

j u s t w h e n t h e y l o o s e c o n t a c t

w i t h e a c h o t h e r .

Answered by ajfour last updated on 25/Jun/21

Commented by ajfour last updated on 25/Jun/21

e n e r g y {c o n s e r v a}^{n} : \dots . (i)

m g (2 R) (1 - \cos θ)

= \frac{1}{2} m {v^{2} + {(u_{r} \cos θ - v)}^{2} + u_{r}^{2} \sin^{2} θ}

m o m e n t u m {c o n s e r v a}^{n} :

m v = m (u_{r} \cos θ - v)

\Rightarrow u_{r} \cos θ = 2 v \dots (i i)

m (c e n t r i p e t a l a c c .) = F_{r}

\frac{{m u}_{r}^{2}}{2 R} = m g \cos θ \dots . (i i i)

\Rightarrow 2 g R = \frac{u_{r}^{2}}{\cos θ} \dots . (A)

n o w u s i n g (i i)

\frac{v^{2}}{u_{r}^{2}} = \frac{\cos^{2} θ}{4} \dots . . (B)

u s i n g a l l (A), (B) i n . . (i)

\frac{2 (1 - \cos θ)}{\cos θ} = \frac{\cos^{2} θ}{2} + 1 - \cos^{2} θ

\Rightarrow 4 - 4 \cos θ = - \cos^{3} θ + 2 \cos θ

\Rightarrow \cos^{3} θ - 6 \cos θ + 4 = 0

\Rightarrow \cos θ = \sqrt{3} - 1

n o w v^{2} = \frac{u_{r}^{2} \cos^{2} θ}{4} = \frac{g R \cos^{3} θ}{2}

v^{2} = \frac{g R {(\sqrt{3} - 1)}^{3}}{2} = g R (3 \sqrt{3} - 5)

v = \sqrt{g R (3 \sqrt{3} - 5)}

Answered by mr W last updated on 25/Jun/21

Commented by mr W last updated on 25/Jun/21

x_{B} = x_{A} - 2 R \sin θ

y_{B} = R + 2 R \cos θ

ω = \frac{d θ}{d t}

v = \frac{{d x}_{A}}{d t}

a = \frac{d v}{d t}

u_{B x} = \frac{{d x}_{B}}{d t} = v - 2 R \cos θ ω = - v

\Rightarrow v = R \cos θ ω

u_{B y} = \frac{{d y}_{B}}{d t} = - 2 R \sin θ ω

\frac{1}{2} m (v^{2} + u_{B x}^{2} + u_{B y}^{2}) = m g 2 R (1 - \cos θ)

v^{2} + {(v - 2 R \cos θ ω)}^{2} + {(- 2 R \sin θ ω)}^{2} = 4 g R (1 - \cos θ)

R^{2} \cos^{2} θ ω^{2} + R^{2} \cos^{2} θ ω^{2} + 4 R^{2} \sin^{2} θ ω^{2} = 4 g R (1 - \cos θ)

(1 + \sin^{2} θ) ω^{2} = 2 g (1 - \cos θ)

\Rightarrow ω^{2} = \frac{2 g (1 - \cos θ)}{R (1 + \sin^{2} θ)}

\Rightarrow ω = \sqrt{\frac{2 g (1 - \cos θ)}{R (2 - \cos^{2} θ)}}

2 ω \frac{d ω}{d θ} = \frac{2 g}{R} [\frac{\sin θ}{1 + \sin^{2} θ} - \frac{2 \sin θ \cos θ (1 - \cos θ)}{{(1 + \sin^{2} θ)}^{2}}]

ω \frac{d ω}{d θ} = \frac{g \sin θ (\cos^{2} θ - 2 \cos θ + 2)}{R {(1 + \sin^{2} θ)}^{2}}

a = \frac{d v}{d t} = R ω (- \sin θ ω + \cos θ \frac{d ω}{d θ})

m a = N \sin θ

N = 0 \Rightarrow a = 0

- \sin θ ω + \cos θ \frac{d ω}{d θ} = 0

\tan θ ω^{2} = ω \frac{d ω}{d θ} = \frac{g \sin θ (\cos^{2} θ - 2 \cos θ + 2)}{R {(1 + \sin^{2} θ)}^{2}}

\tan θ \frac{2 g (1 - \cos θ)}{R (1 + \sin^{2} θ)} = \frac{g \sin θ (\cos^{2} θ - 2 \cos θ + 2)}{{(1 + \sin^{2} θ)}^{2}}

\frac{2 (1 - \cos θ)}{\cos θ} = \frac{\cos^{2} θ - 2 \cos θ + 2}{1 + \sin^{2} θ}

\frac{2 - 2 \cos θ}{\cos θ} = \frac{\cos^{2} θ - 2 \cos θ + 2}{2 - \cos^{2} θ}

\cos^{3} θ - 6 \cos θ + 4 = 0

(\cos θ - 2) (\cos^{2} θ + 2 \cos θ - 2) = 0

\Rightarrow \cos θ = \sqrt{3} - 1

\Rightarrow θ = \cos^{- 1} (\sqrt{3} - 1) \approx 42.9 °

v = \sqrt{\frac{2 g R (1 - \cos θ) \cos^{2} θ}{2 - \cos^{2} θ}}

\Rightarrow v = (2 - \sqrt{3}) \sqrt{(\sqrt{3} + 1) g R} \approx 0.443 \sqrt{g R}

Commented by ajfour last updated on 25/Jun/21

Thanks Sir, right answer!