Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
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
	$let angle descended until in contact be θ. mgcos θ=((mv^2 )/R) ((mv^2 )/2)=mgR(1−cos θ) ⇒ mgRcos θ=2mgR(1−cos θ) ⇒ θ=cos^(−1) (2/3) v=(√((2gR)/3)) (vcos θ)t=d−Rsin θ ⇒ d=(((√5)R)/3)+((2vt)/3) .....(i) Rcos θ=(vsin θ)t+((gt^2 )/2) ⇒ ((2R)/3)=(((√5)v)/3)t+((gt^2 )/2) ⇒ 3gt^2 +2(√5)vt−4R=0 t=((−2(√5)v+(√(20v^2 +48Rg)))/(6g)) =((−2(√5)(√((2gR)/3))+(√((((40)/3)+48)Rg)))/(6g)) t =(√(R/g))×((((√(46))−(√(10))))/(3(√3))) ....(ii) Now from (i) d=(((√5)R)/3)+((2vt)/3) time for circular arc path_(−) mgsin φ=((mdv)/dt) v=(√(2gR(1−cos φ)))=2(√(gR))sin (φ/2) dv=(√(gR))cos (φ/2)dφ ⇒ ∫_0 ^( T) dt=(√(R/g))∫_0 ^( θ) ((cos (φ/2))/(2sin (φ/2)cos (φ/2)))dφ T = (√(R/g))ln ∣(1/(sin (φ/2)))−(1/(tan (φ/2)))∣_0 ^θ T =(√(R/g)){ln ((1/(sin (θ/2)))−(1/(tan (θ/2)))) −lim_(φ→0) ln (((2sin^2 (φ/4))/(2sin (φ/4)cos (φ/4))))} second term→−∞ T not defined. tan θ=((2tan (θ/2))/(1−tan^2 (θ/2))) let tan (θ/2)=z ⇒ ((√5)/2)=((2z)/(1−z^2 )) ⇒ z^2 +(4/(√5))z−1=0 ⇒ z=((−(4/(√5))+(√(((16)/5)+4)))/2) tan (θ/2)=(1/(√5)) d=(((√5)R)/3)+(2/3)(√((2gR)/3))[(√(R/g))×((((√(46))−(√(10))))/(3(√3)))] ⇒ d=(R/3){(√5)+((2(√2))/9)((√(46))−(√(10)))} =(((4(√(23))+5(√5))/(27)))R . t_(total) =T+(√(R/g)){((((√(46))−(√(10))))/(3(√3)))} T is not defined.$
	$l e t a n g l e d e s c e n d e d u n t i l i n$ $c o n t a c t b e θ .$ $m g \cos θ = \frac{{m v}^{2}}{R}$ $\frac{{m v}^{2}}{2} = m g R (1 - \cos θ)$ $\Rightarrow m g R \cos θ = 2 m g R (1 - \cos θ)$ $\Rightarrow θ = \cos^{- 1} (2 / 3)$ $v = \sqrt{\frac{2 g R}{3}}$ $(v \cos θ) t = d - R \sin θ$ $\Rightarrow d = \frac{\sqrt{5} R}{3} + \frac{2 v t}{3} . . . . . (i)$ $R \cos θ = (v \sin θ) t + \frac{{g t}^{2}}{2}$ $\Rightarrow \frac{2 R}{3} = \frac{\sqrt{5} v}{3} t + \frac{{g t}^{2}}{2}$ $\Rightarrow 3 {g t}^{2} + 2 \sqrt{5} v t - 4 R = 0$ $t = \frac{- 2 \sqrt{5} v + \sqrt{20 v^{2} + 48 R g}}{6 g}$ $= \frac{- 2 \sqrt{5} \sqrt{\frac{2 g R}{3}} + \sqrt{(\frac{40}{3} + 48) R g}}{6 g}$ $t = \sqrt{\frac{R}{g}} \times \frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}} . . . . (i i)$ $N o w f r o m (i)$ $d = \frac{\sqrt{5} R}{3} + \frac{2 v t}{3}$ $\underline{t i m e f o r c i r c u l a r a r c p a t h}$ $m g \sin ϕ = \frac{m d v}{d t}$ $v = \sqrt{2 g R (1 - \cos ϕ)} = 2 \sqrt{g R} \sin \frac{ϕ}{2}$ $d v = \sqrt{g R} \cos \frac{ϕ}{2} d ϕ$ $\Rightarrow \int_{0}^{T} d t = \sqrt{\frac{R}{g}} \int_{0}^{θ} \frac{\cos (ϕ / 2)}{2 \sin \frac{ϕ}{2} \cos \frac{ϕ}{2}} d ϕ$ $T = \sqrt{\frac{R}{g}} \ln ∣ \frac{1}{\sin \frac{ϕ}{2}} - \frac{1}{\tan \frac{ϕ}{2}} ∣_{0}^{θ}$ $T = \sqrt{\frac{R}{g}} {\ln (\frac{1}{\sin \frac{θ}{2}} - \frac{1}{\tan \frac{θ}{2}})$ $- \underset{ϕ \to 0}{\lim l n} (\frac{2 \sin^{2} \frac{ϕ}{4}}{2 \sin \frac{ϕ}{4} \cos \frac{ϕ}{4}})}$ $s e c o n d t e r m \to - \infty$ $T n o t d e f i n e d .$ $\tan θ = \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} l e t \tan \frac{θ}{2} = z$ $\Rightarrow \frac{\sqrt{5}}{2} = \frac{2 z}{1 - z^{2}} \Rightarrow z^{2} + \frac{4}{\sqrt{5}} z - 1 = 0$ $\Rightarrow z = \frac{- \frac{4}{\sqrt{5}} + \sqrt{\frac{16}{5} + 4}}{2}$ $\tan \frac{θ}{2} = \frac{1}{\sqrt{5}}$ $d = \frac{\sqrt{5} R}{3} + \frac{2}{3} \sqrt{\frac{2 g R}{3}} [\sqrt{\frac{R}{g}} \times \frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}}]$ $\Rightarrow$ $d = \frac{R}{3} {\sqrt{5} + \frac{2 \sqrt{2}}{9} (\sqrt{46} - \sqrt{10})}$ $= (\frac{4 \sqrt{23} + 5 \sqrt{5}}{27}) R .$ $t_{t o t a l} = T + \sqrt{\frac{R}{g}} {\frac{(\sqrt{46} - \sqrt{10})}{3 \sqrt{3}}}$ $T i s n o t d e f i n e d .$
	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

	$t h a n k s f o r s o l v i n g a n d c o n f i r m i n g,$ $m r W S i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum