Question-73297

Question Number 73297 by mr W last updated on 10/Nov/19

Commented by mr W last updated on 10/Nov/19

a b l o c k o f m a s s m i s r e l e a s e d a t t h e

t o p o f a h e m i s p h e r e .

f i n d t h e t i m e f o r t h e b l o c k t o h i t t h e

g r o u n d a n d t h e p o s i t i o n o f t h e h i t .

t h e h e m i s p h e r e i s f i x e d o n t h e g r o u n d

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

Answered by ajfour last updated on 10/Nov/19

l e t a n g l e d e s c e n d e d u n t i l i n

c o n t a c t b e θ .

m g \cos θ = \frac{{m v}^{2}}{R}

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

\Rightarrow m g R \cos θ = 2 m g R (1 - \cos θ)

\Rightarrow θ = \cos^{- 1} (2 / 3)

v = \sqrt{\frac{2 g R}{3}}

(v \cos θ) t = d - R \sin θ

\Rightarrow d = \frac{\sqrt{5} R}{3} + \frac{2 v t}{3} \dots . . (i)

R \cos θ = (v \sin θ) t + \frac{{g t}^{2}}{2}

\Rightarrow \frac{2 R}{3} = \frac{\sqrt{5} v}{3} t + \frac{{g t}^{2}}{2}

\Rightarrow 3 {g t}^{2} + 2 \sqrt{5} v t - 4 R = 0

t = \frac{- 2 \sqrt{5} v + \sqrt{20 v^{2} + 48 R g}}{6 g}

= \frac{- 2 \sqrt{5} \sqrt{\frac{2 g R}{3}} + \sqrt{(\frac{40}{3} + 48) R g}}{6 g}

t = \sqrt{\frac{R}{g}} \times \frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}} \dots . (i i)

N o w f r o m (i)

d = \frac{\sqrt{5} R}{3} + \frac{2 v t}{3}

\underline{t i m e f o r c i r c u l a r a r c p a t h}

m g \sin ϕ = \frac{m d v}{d t}

v = \sqrt{2 g R (1 - \cos ϕ)} = 2 \sqrt{g R} \sin \frac{ϕ}{2}

d v = \sqrt{g R} \cos \frac{ϕ}{2} d ϕ

\Rightarrow \int_{0}^{T} d t = \sqrt{\frac{R}{g}} \int_{0}^{θ} \frac{\cos (ϕ / 2)}{2 \sin \frac{ϕ}{2} \cos \frac{ϕ}{2}} d ϕ

T = \sqrt{\frac{R}{g}} \ln ∣ \frac{1}{\sin \frac{ϕ}{2}} - \frac{1}{\tan \frac{ϕ}{2}} ∣_{0}^{θ}

T = \sqrt{\frac{R}{g}} {\ln (\frac{1}{\sin \frac{θ}{2}} - \frac{1}{\tan \frac{θ}{2}})

- \underset{ϕ \to 0}{\lim l n} (\frac{2 \sin^{2} \frac{ϕ}{4}}{2 \sin \frac{ϕ}{4} \cos \frac{ϕ}{4}})}

s e c o n d t e r m \to - \infty

T n o t d e f i n e d .

\tan θ = \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} l e t \tan \frac{θ}{2} = z

\Rightarrow \frac{\sqrt{5}}{2} = \frac{2 z}{1 - z^{2}} \Rightarrow z^{2} + \frac{4}{\sqrt{5}} z - 1 = 0

\Rightarrow z = \frac{- \frac{4}{\sqrt{5}} + \sqrt{\frac{16}{5} + 4}}{2}

\tan \frac{θ}{2} = \frac{1}{\sqrt{5}}

d = \frac{\sqrt{5} R}{3} + \frac{2}{3} \sqrt{\frac{2 g R}{3}} [\sqrt{\frac{R}{g}} \times \frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}}]

\Rightarrow

d = \frac{R}{3} {\sqrt{5} + \frac{2 \sqrt{2}}{9} (\sqrt{46} - \sqrt{10})}

= (\frac{4 \sqrt{23} + 5 \sqrt{5}}{27}) R .

t_{t o t a l} = T + \sqrt{\frac{R}{g}} {\frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}}}

T i s n o t d e f i n e d .

Commented by mr W last updated on 10/Nov/19

t h a n k s s i r!

t i m e a l o n g h e m i s p h e r e i s n o t t o

d e t e r m i n e, {i t}^{'} s + \infty .

Answered by mr W last updated on 10/Nov/19

Commented by mr W last updated on 10/Nov/19

a t θ :

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

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

v = R \frac{d θ}{d t}

\frac{d θ}{d t} = 2 \sqrt{\frac{g}{R}} \sin \frac{θ}{2}

\int_{0}^{θ} \frac{d (\frac{θ}{2})}{\sin \frac{θ}{2}} = \sqrt{\frac{g}{R}} \int_{0}^{t_{1}} d t

- \int_{0}^{θ} \frac{d (\cos \frac{θ}{2})}{1 - \cos^{2} \frac{θ}{2}} = t_{1} \sqrt{\frac{g}{R}}

\frac{1}{2} {[\ln \frac{1 - \cos \frac{θ}{2}}{1 + \cos \frac{θ}{2}}]}_{0}^{θ} = t_{1} \sqrt{\frac{g}{R}}

l e t F (θ) = \ln \frac{1 - \cos \frac{θ}{2}}{1 + \cos \frac{θ}{2}}

\Rightarrow t_{1} = \frac{1}{2} \sqrt{\frac{R}{g}} [F (θ) - \lim_{θ \to 0} F (θ)]

s i n c e \lim_{θ \to 0} F (θ) = - \infty, t_{1} c a n n o t b e

d e t e r m i n e d, t_{1} \to + \infty .

a t θ = φ :

N = 0

m g \cos φ = m \frac{v^{2}}{R} = 2 m g (1 - \cos φ)

\cos φ = 2 (1 - \cos φ)

\Rightarrow \cos φ = \frac{2}{3} = 2 \cos^{2} \frac{φ}{2} - 1

\Rightarrow \cos \frac{φ}{2} = \sqrt{\frac{5}{6}}, \sin \frac{φ}{2} = \frac{1}{\sqrt{6}}

\Rightarrow \sin φ = \frac{\sqrt{5}}{3}

v = \frac{2}{\sqrt{6}} \sqrt{g R}

R \cos φ = v \sin φ t_{2} + \frac{1}{2} {g t}_{2}^{2}

0 = - \frac{4}{3} + \frac{2 \sqrt{30}}{9} (\sqrt{\frac{g}{R}} t_{2}) + {(\sqrt{\frac{g}{R}} t_{2})}^{2}

\Rightarrow \sqrt{\frac{g}{R}} t_{2} = \frac{\sqrt{138} - \sqrt{30}}{9}

\Rightarrow t_{2} = \frac{\sqrt{138} - \sqrt{30}}{9} \sqrt{\frac{R}{g}}

d = R \sin φ + v \cos φ t_{2}

d = R \frac{\sqrt{5}}{3} + \frac{2}{\sqrt{6}} \times \frac{2}{3} \times \frac{(\sqrt{138} - \sqrt{30}) R}{9}

d = \frac{(4 \sqrt{23} + 5 \sqrt{5}) R}{27} \approx 1.124 R

Commented by ajfour last updated on 11/Nov/19

t h a n k s f o r s o l v i n g a n d c o n f i r m i n g,

m r W S i r .

Facebook Tweet Pin