An-object-moves-in-a-circular-path-with-a-constant-speed-in-the-xy-plane-with-the-centre-at-the-origin-When-the-object-is-at-x-2-m-its-velocity-is-4-m-s-j-Then-objects-velocity-at-y-2-m-i

Question Number 16504 by Tinkutara last updated on 23/Jun/17

An object moves in a circular path with

a constant speed in the xy plane with

the centre at the origin . When the

object is at x = - 2 m, its velocity is

- (4 m / s) \overset{\land}{j} . Then objects velocity at

y = 2 m is

(1) 4 m / s \overset{\land}{i}

(2) (- 4 m / s) \overset{\land}{i}

Using this data, find objects acceleration

when it is at y = 2 m

(1) 8 m / s^{2} \overset{\land}{i}

(2) - 8 m / s^{2} \overset{\land}{j}

Answered by ajfour last updated on 23/Jun/17

Commented by ajfour last updated on 23/Jun/17

+ v e y i n t e r c e p t i s (0, 2) i n s t e a d

o f (0, 4) a s i n f i g . a b o v e .

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

Answered by sma3l2996 last updated on 23/Jun/17

\overset{\land}{v} (θ) = 2 w (- s i n θ \hat{i} + c o s θ \overset{\land}{j})

a t x = - 2 m; θ = π

s o : \overset{\land}{v} (π) = 2 w \overset{\land}{j} = - 4 \overset{\land}{j}

w = - 2 r a d / s

\overset{\land}{v} (θ) = 4 (s i n θ \overset{\land}{i} - c o s θ \overset{\land}{j})

a t y = 2 m; θ = \frac{π}{2}

\overset{\land}{v} (\frac{π}{2}) = (4 m / s) \overset{\land}{i}

* *

\overset{\land}{a} = \frac{d \overset{\land}{v}}{d t} = 4 w (c o s θ \overset{\land}{i} + s i n θ \overset{\land}{j}) = - 8 (c o s θ i + s i n θ j)

a t y = 2 m; θ = \frac{π}{2}

s o : \overset{\land}{a} = (- 8 m / s^{2}) \overset{\land}{j}

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!