A-particle-P-is-moving-on-a-circle-under-the-action-of-only-one-force-acting-always-towards-fixed-point-O-on-the-circumference-Find-ratio-of-d-2-dt-2-and-d-dt-2-

Question Number 22145 by Tinkutara last updated on 11/Oct/17

A particle P is moving on a circle under

the action of only one force acting

always towards fixed point O on the

circumference . Find ratio of \frac{d^{2} ϕ}{{d t}^{2}} and

{(\frac{d ϕ}{d t})}^{2} .

Commented by Tinkutara last updated on 11/Oct/17

Commented by sma3l2996 last updated on 12/Oct/17

w e h a v e \vec{F} = m \vec{a} \Leftrightarrow \vec{a} = - \frac{F}{m} \vec{u} \dots (i)

a n d \vec{O P} = 2 R c o s (ϕ) \vec{u}

s o \vec{v} (P) = \frac{d \vec{O P}}{d t} = 2 R [- \dot{ϕ} s i n (ϕ) \vec{u} + \dot{ϕ} c o s (ϕ) \vec{k}] = 2 R \dot{ϕ} (- s i n ϕ \vec{u} + c o s ϕ \vec{k})

a n d \vec{a} (P) = \frac{d \vec{v}}{d t} = 2 R [\overset{. .}{ϕ} (- s i n ϕ \vec{u} + c o s ϕ \vec{k}) + \dot{ϕ} (- \dot{ϕ} c o s ϕ \vec{u} - \dot{ϕ} s i n ϕ \vec{k} - \dot{ϕ} s i n ϕ \vec{k} - \dot{ϕ} c o s ϕ \vec{u})]

s o : \vec{a} = 2 R [(- \overset{. .}{ϕ} s i n ϕ - 2 \dot{ϕ}^{2} c o s ϕ) \vec{u} + (\overset{. .}{ϕ} c o s ϕ - 2 \dot{ϕ}^{2} s i n ϕ) \vec{k}] \dots (i i)

w e h a v e (i) = (i i)

s o

2 R (- \overset{. .}{ϕ} s i n ϕ - 2 \dot{ϕ}^{2} c o s ϕ) = \frac{- F}{m} a n d \overset{. .}{ϕ} c o s ϕ - 2 \dot{ϕ}^{2} s i n ϕ = 0

s o

\overset{. .}{ϕ} c o s ϕ = 2 \dot{ϕ}^{2} s i n ϕ \Leftrightarrow \overset{. .}{ϕ} = 2 \dot{ϕ}^{2} t a n (ϕ) \dots (3 i)

\overset{. .}{ϕ} s i n ϕ + 2 \dot{ϕ}^{2} c o s ϕ = \frac{F}{2 R m} \dots (4 i)

(3 i) \to (4 i) \Rightarrow 2 \dot{ϕ}^{2} t a n (ϕ) s i n (ϕ) + 2 \dot{ϕ}^{2} c o s ϕ = \frac{F}{2 R m}

2 \dot{ϕ}^{2} c o s ϕ (\frac{t a n (ϕ) s i n ϕ}{c o s ϕ} + 1) = \frac{F}{2 R m}

2 \dot{ϕ}^{2} c o s ϕ ({t a n}^{2} ϕ + 1) = \frac{F}{2 R m}

{(\frac{d ϕ}{d t})}^{2} = \dot{ϕ}^{2} = \frac{F}{4 R m} c o s ϕ

a n d \overset{. .}{ϕ} = 2 \dot{ϕ}^{2} t a n ϕ = 2 (\frac{F}{4 R m} c o s ϕ) t a n ϕ

s o : \frac{d^{2} ϕ}{{d t}^{2}} = \overset{. .}{ϕ} = \frac{F}{2 R m} s i n ϕ

Commented by sma3l2996 last updated on 12/Oct/17

Answered by ajfour last updated on 12/Oct/17

A l t e r n a t e s o l u t i o n :

l e t t h e ∠ a t C b e θ .

θ = 2 ϕ \Rightarrow ω = \frac{d θ}{d t} = 2 \frac{d ϕ}{d t} . . (i)

{(\frac{d ϕ}{d t})}^{2} = \frac{ω^{2}}{4} \dots (a)

α = \frac{d ω}{d t} = 2 \frac{d^{2} ϕ}{{d t}^{2}} \Rightarrow \frac{d^{2} ϕ}{{d t}^{2}} = \frac{α}{2} \dots (b)

(b) \div (a) w i l l g i v e

\frac{d^{2} ϕ / {d t}^{2}}{{(d ϕ / d t)}^{2}} = \frac{α / 2}{ω^{2} / 4} = \frac{2 α}{ω^{2}}

= \frac{2 m (α r)}{m (ω^{2} r)} = \frac{2 {m a}_{T}}{{m a}_{r}}

= \frac{2 F \sin ϕ}{F \cos ϕ} = 2 t a n ϕ .

Commented by Tinkutara last updated on 12/Oct/17

Thank you very much Sir!