Question-90819

Question Number 90819 by ajfour last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

m a s s o f r o d i s M, a n d α = 0, b u t

r o d r o t a t e s w i t h c o n s t a n t a n g u l a r

v e l o i t y ω_{0}, t h e n f i n d r e l a t i v e

s p e e d o f s m a l l m a s s m w i t h

r e s p e c t t o r o d, a s i t r e a c h e s

t h e e x t r e m e e n d (a = l) f r o m t h e

h i n g e d e n d (a = 0) .

Commented by ajfour last updated on 26/Apr/20

t h i s o n e f r o m a r e n o w n e d

p h y s i c s b o o k, h a v e a n s w e r .

Answered by ajfour last updated on 26/Apr/20

Answered by mr W last updated on 26/Apr/20

w h e n s l e e v e i s a t p o s i t i o n x :

a n g u l a r s p e e d o f r o d = ω

s p e e d o f s l e e v e a l o n g r o d = u

a c c . o f s l e e v e a l o n g r o d = a = u \frac{d u}{d x}

I ω_{0} = (I + {m x}^{2}) ω

\Rightarrow ω = \frac{\frac{{M l}^{2}}{3}}{\frac{{M l}^{2}}{3} + {m x}^{2}} ω_{0} = \frac{{M l}^{2}}{{M l}^{2} + 3 {m x}^{2}} ω_{0}

m a - m ω^{2} x = 0

\Rightarrow u \frac{d u}{d x} - ω^{2} x = 0

\Rightarrow u \frac{d u}{d x} - {(\frac{{M l}^{2} ω_{0}}{{M l}^{2} + 3 {m x}^{2}})}^{2} x = 0

\Rightarrow \int_{0}^{u_{m a x}} u d u = \int_{0}^{l} {(\frac{{M l}^{2} ω_{0}}{{M l}^{2} + 3 {m x}^{2}})}^{2} x d x

\Rightarrow \frac{u_{m a c}^{2}}{2} = {(\frac{{M l}^{2} ω_{0}}{3 m})}^{2} \int_{0}^{l} \frac{x d x}{{(\frac{{M l}^{2}}{3 m} + x^{2})}^{2}}

\Rightarrow u_{m a x}^{2} = {(\frac{{M l}^{2} ω_{0}}{3 m})}^{2} \int_{0}^{l} \frac{d (\frac{{M l}^{2}}{3 m} + x^{2})}{{(\frac{{M l}^{2}}{3 m} + x^{2})}^{2}}

\Rightarrow u_{m a x}^{2} = {(\frac{{M l}^{2} ω_{0}}{3 m})}^{2} {[\frac{1}{\frac{{M l}^{2}}{3 m} + x^{2}}]}_{l}^{0}

\Rightarrow u_{m a x}^{2} = {(\frac{M l ω_{0}}{3 m})}^{2} [\frac{1}{\frac{M}{3 m}} - \frac{1}{\frac{M}{3 m} + 1}]

\Rightarrow u_{m a x}^{2} = {(\frac{M l ω_{0}}{3 m})}^{2} [\frac{1}{\frac{M}{3 m} (\frac{M}{3 m} + 1)}]

\Rightarrow u_{m a x} = l ω_{0} \sqrt{\frac{M}{M + 3 m}}

Commented by ajfour last updated on 26/Apr/20

E x c e l l e n t S i r! V e r y b a s i c w a y .

t h a n k s S i r!

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