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Question Number 29924 by ajfour last updated on 13/Feb/18

Commented by ajfour last updated on 13/Feb/18

$O n a p l a n k a s m a l l b l o c k m i s$ $k e p t . T h e p l a n k i s f r e e t o r o t a t e$ $a b o u t a h o r i z o n t a l a x i s a t i t s e n d .$ $I f r e l e a s e d a s s h o w n, f i n d$ $a n g l e, a n g u l a r v e l o c i t y o f p l a n k$ $w h e n n o r m a l r e a c t i o n b e t w e e n$ $p l a n k a n d b l o c k b e c o m e s z e r o .$ $(A s s u m e f r i c t i o n c o e f f i c i e n t μ) .$
Commented by 33 last updated on 13/Feb/18

$n i c e q u e s t i o n s i r$
Commented by ajfour last updated on 13/Feb/18

Commented by mrW2 last updated on 15/Feb/18

$t h e n o r m a l f o r c e g e t s t o z e r o, w h e n$ $t h e b l o c k r e a c h e s t h e p o s i t i o n x = \frac{2 L}{3} .$ $B u t t h e c o r e s p o n d i n g a n g l e p o s i t i o θ$ $s e e m s n o t t o b e i n r e l a t i o n w i t h x,$ $t h e r e f o r e I h a v e n o s o l u t i o n .$
Commented by ajfour last updated on 15/Feb/18

$l e t u s a s s u m e a o r b i s m u c h l e s s$ $a n d s l i p p i n g b e g i n s w h e n θ = θ_{1}$ $a f t e r t h a t g \sin θ i n c r e a s e s$ $g \cos θ d e c r e a s e s N m u s t b e c o m e$ $z e r o f o r s o m e x t h e r e a f t e r . .$
	Answered by mrW2 last updated on 13/Feb/18

	Commented by mrW2 last updated on 14/Feb/18
	$phase 1: no slip between block and plank x=b=constant (b replaces a in question) ω=(dθ/dt) α=(dω/dt) v_θ =xω=bω v_r =(dx/dt)=0 mg cos θ−N=mbα ⇒N=m(g cos θ−bα) f−mg sin θ=mbω^2 ⇒f=m(g sin θ+bω^2 ) ((ML^2 )/3)α=Nb+Mg((Lcos θ)/2) ((ML^2 )/3)α=mb(g cos θ−bα)+Mg((Lcos θ)/2) (((ML^2 +3mb^2 )/3))α=(((2mb+ML)gcos θ)/2) ⇒α=((3(2mb+ML)g)/(2(ML^2 +3mb^2 )))cos θ ⇒ω(dω/dθ)=((3(2mb+ML)g)/(2(ML^2 +3mb^2 )))cos θ ⇒∫_0 ^ω ωdω=((3(2mb+ML)g)/(2(ML^2 +3mb^2 )))∫_0 ^θ cos θ dθ ⇒(ω^2 /2)=((3(2mb+ML)g)/(2(ML^2 +3mb^2 ))) sin θ ⇒ω^2 =((3(2mb+ML)g)/((ML^2 +3mb^2 ))) sin θ ⇒f=mg[1+((3b(2mb+ML))/((ML^2 +3mb^2 )))]sin θ ⇒f=[((ML^2 +9mb^2 +3MLb)/(ML^2 +3mb^2 ))]mg sin θ ⇒N=mg[1−((3b(2mb+ML))/(2(ML^2 +3mb^2 )))]cos θ ⇒N=[((ML(2L−3b))/(2(ML^2 +3mb^2 )))]mg cos θ (f/N)=((ML^2 +9mb^2 +3MLb)/(ML^2 +3mb^2 ))×((2(ML^2 +3mb^2 ))/(ML(2L−3b)))×tan θ ⇒(f/N)=((2(ML^2 +9mb^2 +3MLb))/(ML(2L−3b)))×tan θ=μ tan θ_1 =((μML(2L−3b))/(2(ML^2 +9mb^2 +3MLb))) θ_1 =tan^(−1) {((μML(2L−3b))/(2(ML^2 +9mb^2 +3MLb)))} with η=(m/M) and λ=(b/L) ⇒θ_1 =tan^(−1) {((μ(2−3λ))/(2(1+3λ+9ηλ^2 )))} i.e. at θ=θ_1 block m begins to slip. E.g. η=(m/M)=0.5, λ=(b/L)=0.25, μ=1 ⇒θ_1 =tan^(−1) {((1×(2−3×0.25))/(2(1+3×0.25+9×0.5×0.25^2 )))}=17.1° at θ=θ_1 : N_1 =[((2−3λ)/(2(1+3ηλ^2 )))]mg cos θ_1 ω_1 =(√(((3(1+2ηλ)g)/(L(1+3ηλ^2 ))) sin θ_1 )) in phase 1 the reaction force N is always more than zero if λ<(2/3), i.e. if b<(2/3)L.$
	$p h a s e 1 : n o s l i p b e t w e e n b l o c k a n d p l a n k$ $x = b = c o n s t a n t (b r e p l a c e s a i n q u e s t i o n)$ $ω = \frac{d θ}{d t}$ $α = \frac{d ω}{d t}$ $v_{θ} = x ω = b ω$ $v_{r} = \frac{d x}{d t} = 0$ $m g \cos θ - N = m b α$ $\Rightarrow N = m (g \cos θ - b α)$ $f - m g \sin θ = m b ω^{2}$ $\Rightarrow f = m (g \sin θ + b ω^{2})$ $\frac{{M L}^{2}}{3} α = N b + M g \frac{L \cos θ}{2}$ $\frac{{M L}^{2}}{3} α = m b (g \cos θ - b α) + M g \frac{L \cos θ}{2}$ $(\frac{{M L}^{2} + 3 {m b}^{2}}{3}) α = \frac{(2 m b + M L) g \cos θ}{2}$ $\Rightarrow α = \frac{3 (2 m b + M L) g}{2 ({M L}^{2} + 3 {m b}^{2})} \cos θ$ $\Rightarrow ω \frac{d ω}{d θ} = \frac{3 (2 m b + M L) g}{2 ({M L}^{2} + 3 {m b}^{2})} \cos θ$ $\Rightarrow \int_{0}^{ω} ω d ω = \frac{3 (2 m b + M L) g}{2 ({M L}^{2} + 3 {m b}^{2})} \int_{0}^{θ} \cos θ d θ$ $\Rightarrow \frac{ω^{2}}{2} = \frac{3 (2 m b + M L) g}{2 ({M L}^{2} + 3 {m b}^{2})} \sin θ$ $\Rightarrow ω^{2} = \frac{3 (2 m b + M L) g}{({M L}^{2} + 3 {m b}^{2})} \sin θ$ $\Rightarrow f = m g [1 + \frac{3 b (2 m b + M L)}{({M L}^{2} + 3 {m b}^{2})}] \sin θ$ $\Rightarrow f = [\frac{{M L}^{2} + 9 {m b}^{2} + 3 M L b}{{M L}^{2} + 3 {m b}^{2}}] m g \sin θ$ $\Rightarrow N = m g [1 - \frac{3 b (2 m b + M L)}{2 ({M L}^{2} + 3 {m b}^{2})}] \cos θ$ $\Rightarrow N = [\frac{M L (2 L - 3 b)}{2 ({M L}^{2} + 3 {m b}^{2})}] m g \cos θ$ $\frac{f}{N} = \frac{{M L}^{2} + 9 {m b}^{2} + 3 M L b}{{M L}^{2} + 3 {m b}^{2}} \times \frac{2 ({M L}^{2} + 3 {m b}^{2})}{M L (2 L - 3 b)} \times \tan θ$ $\Rightarrow \frac{f}{N} = \frac{2 ({M L}^{2} + 9 {m b}^{2} + 3 M L b)}{M L (2 L - 3 b)} \times \tan θ = μ$ $\tan θ_{1} = \frac{μ M L (2 L - 3 b)}{2 ({M L}^{2} + 9 {m b}^{2} + 3 M L b)}$ $θ_{1} = \tan^{- 1} {\frac{μ M L (2 L - 3 b)}{2 ({M L}^{2} + 9 {m b}^{2} + 3 M L b)}}$ $w i t h η = \frac{m}{M} a n d λ = \frac{b}{L}$ $\Rightarrow θ_{1} = \tan^{- 1} {\frac{μ (2 - 3 λ)}{2 (1 + 3 λ + 9 η λ^{2})}}$ $i . e . a t θ = θ_{1} b l o c k m b e g i n s t o s l i p .$ $E . g . η = \frac{m}{M} = 0.5, λ = \frac{b}{L} = 0.25, μ = 1$ $\Rightarrow θ_{1} = \tan^{- 1} {\frac{1 \times (2 - 3 \times 0.25)}{2 (1 + 3 \times 0.25 + 9 \times 0.5 \times 0 . 25^{2})}} = 17.1 °$ $a t θ = θ_{1} :$ $N_{1} = [\frac{2 - 3 λ}{2 (1 + 3 η λ^{2})}] m g \cos θ_{1}$ $ω_{1} = \sqrt{\frac{3 (1 + 2 η λ) g}{L (1 + 3 η λ^{2})} \sin θ_{1}}$ $i n p h a s e 1 t h e r e a c t i o n f o r c e N i s$ $a l w a y s m o r e t h a n z e r o i f λ < \frac{2}{3},$ $i . e . i f b < \frac{2}{3} L .$
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