Question-216859

Question Number 216859 by ajfour last updated on 23/Feb/25

Commented by ajfour last updated on 23/Feb/25

R a d i u s o f i n n e r d i s c i s R . A s i t r o l l s u p

t h e o u t e r c i r c u l a r t r a c k o f r a d i u s 2 R, f i n d

e q u a t i o n o f t r a j e c t o r y o f a p o i n t P o n t h e

w h e e l u n t i l i t c o m e s i n t o c o n t a c t w i t h

t h e o u t e r t r a c k .

Commented by mr W last updated on 23/Feb/25

f o r R = 2 r t h e l o c u s o f P i s a d i a m e t e r

o f t h e o u t e r c i r c l e .

Commented by mr W last updated on 23/Feb/25

Answered by mr W last updated on 23/Feb/25

Commented by mr W last updated on 23/Feb/25

s a y n = \frac{R}{r}

r φ = R θ

\Rightarrow φ = \frac{R θ}{r}

x_{P} = (R - r) \sin θ + r \sin (φ + α - θ)

\Rightarrow \frac{x_{P}}{r} = (n - 1) \sin θ + \sin [(n - 1) θ + α]

y_{P} = R - (R - r) \cos θ + r \cos (φ + α - θ)

\Rightarrow \frac{y_{P}}{r} = n - (n - 1) \cos θ + \cos [(n - 1) θ + α]

Commented by mr W last updated on 23/Feb/25

Commented by mr W last updated on 23/Feb/25

Commented by mr W last updated on 23/Feb/25

Commented by mr W last updated on 23/Feb/25

Commented by mr W last updated on 23/Feb/25

Commented by ajfour last updated on 23/Feb/25

W o w! T h a n k y o u .

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