A-particle-will-leave-a-vertical-circle-of-radius-r-when-its-velocity-at-the-lowest-point-of-the-circle-v-L-is-a-2gr-b-5gr-c-3gr-d-6gr-

Question Number 21604 by Tinkutara last updated on 29/Sep/17

A particle will leave a vertical circle of

radius r, when its velocity at the lowest

point of the circle (v_{L}) is

(a) \sqrt{2 g r}

(b) \sqrt{5 g r}

(c) \sqrt{3 g r}

(d) \sqrt{6 g r}

Answered by ajfour last updated on 29/Sep/17

a t t h e t o p i f i t l e a v e s t h e

c i r c l e, t h e n

m g = m \frac{v_{H}^{2}}{r} o r {m v}_{H}^{2} = m g r \dots (i)

f r o m c o n s e r v a t i o n o f e n e r g y :

2 m g r = \frac{1}{2} m (v_{L}^{2} - v_{H}^{2})

\Rightarrow 4 g r = {m v}_{L}^{2} - m g r [s e e (i)]

v_{L} = \sqrt{5 g r} .

w h e b e v e r v_{L} < \sqrt{5 g r} p a r t i c l e

l e a v e s t h e v e r t i c a l c i r c l e .

S o (a) a n d (c) .

Commented by Tinkutara last updated on 29/Sep/17

But answer is given only (c) not (a) .

Commented by ajfour last updated on 29/Sep/17

b u t w i t h v_{L} = \sqrt{2 g r} i t s w i n g s o n l y

i n t h e l o w e r h a l f v e r t i x a l c i r c l e;

n o c h a n c e o f l e a v i n g t h e c i r c l e .

s o j u s t (c) a n d n o t (a) . s o r r y

f o r t h e m i s t a k e .

Commented by Tinkutara last updated on 29/Sep/17

Thank you very much Sir!

Commented by Tinkutara last updated on 29/Sep/17

Can you please guide Q . 21307 ?