Question-21038

Question Number 21038 by ajfour last updated on 10/Sep/17

Commented by ajfour last updated on 10/Sep/17

F i n d a n g u l a r v e l o c i t y a n d a n g u l a r

a c c e l e r a t i o n o f r o d i f e n d A m o v e s

t o t h e r i g h t w i t h c o n s t a n t v e l o c i t y

v .

Answered by mrW1 last updated on 11/Sep/17

\tan θ = \frac{h}{x}

\frac{1}{\cos^{2} θ} \times \frac{d θ}{dt} = - \frac{h}{x^{2}} \times \frac{dx}{dt} = - {(\frac{h}{x})}^{2} \times \frac{1}{h} \times \frac{dx}{dt}

\frac{1}{\cos^{2} θ} \times \frac{d θ}{dt} = - \tan^{2} θ \times \frac{1}{h} \times \frac{dx}{dt}

\frac{d θ}{dt} = - \sin^{2} θ \times \frac{1}{h} \times \frac{dx}{dt}

\Rightarrow angular velocity ω = \sin^{2} θ \times \frac{v}{h}

\frac{d ω}{dt} = 2 \sin θ \cos θ \frac{d θ}{dt} \times \frac{v}{h} + \sin^{2} θ \times \frac{dv}{dt}

\frac{dv}{dt} = 0

\frac{d θ}{dt} = ω = \sin^{2} θ \times \frac{v}{h}

\Rightarrow angular acceleration \frac{d ω}{dt} = \sin 2 θ \sin^{2} θ {(\frac{v}{h})}^{2}

Commented by ajfour last updated on 11/Sep/17

t h a n k s s i r, i h a d m i s s e d m a r k i n g

h i n q u e s t i o n, t h a n k s f o r

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

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