Chapter 2: Q84P (page 58)

Two thin rods are fastened to the inside of a circular ring as shown in Figure P2.84. One rod of length D is vertical, and the other of length L makes an angle $θ$ with the horizontal. The two rods and the ring lie in a vertical plane. Two small beads are free to slide without friction along the rods. (a) If the two beads are released from rest simultaneously from the positions shown, use your intuition and guess which bead reaches the bottom first. (b) Find an expression for the time interval required for the red bead to fall from point A to point C in terms of g and D. (c) Find an expression for the time interval required for the blue bead to slide from point B to point C in terms of g, L, and $θ$ . (d) Show that the two time intervals found in parts (b) and (c) are equal. What is the angle between the chords of the circle A B and B C? (e) Do these results surprise you? Was your intuitive guess in part (a) correct?

Short Answer

Expert verified

$(a) T h e b l u e b e a d r e a c h e s a t t h e b o t t o m f i r s t . (b) T h e t i m e t a k e n b y r e d b e a d t o f a l l f r o m p o i n t A t o p o i n t C i s \sqrt{\frac{2 D}{g}} . (c) T h e t i m e t a k e n b y r e d b e a d t o f a l l f r o m p o i n t B t o p o i n t C i s \sqrt{\frac{2 L}{g \sin θ}} . (d) T h e r e d b e a d a n d b l u e b e a d r e a c h e s a t b o t t o m s i m u l \tan e o u s . (e) T h e i n t u i t i v e g u e s s i n p a r t (a) i s i n c o r r e c t .$

Step by step solution

Identification of given data

$T h e l e n g t h o f v e r t i c a l r o d i s D . T h e l e n g t h o f s m a l l r o d i s L . T h e a n g l e o f t h e s m a l l r o d i s θ .$

(a) Determination of bead that reach at bottom first

$T h e r e d b e a d f a l l s w i t h g r a v i t a t i o n a l a c c e l e r a t i o n g a n d b l u e b e a d f a l l s w i t h a c c e l e r a t i o n g \sin θ . T h e a c c e l e r a t i o n f o r r e d b e a d i s m o r e t h a n t h e b l u e b e a d b u t d i s \tan c e f o r r e d b e a d i s a l s o m o r e t h a n b l u e b e a d . T h e t i m e t a k e n b y b l u e d b e a d t o r e a c h a t b o t t o m v a r i e s w i t h t h e a n g l e o f s m a l l r o d . T h e b l u e b e a d r e a c h e s a t t h e b o t t o m f i r s t b e c a u s e d i s \tan c e o f b l u e b e a d f r o m b o t t o m i s l e s s t h a n t h e d i s \tan c e f o r r e d b e a d, b u t d i f f e r e n c e i n t i m e f o r b o t h b e a d s w i l l b e v e r y s m a l l . T h e r e f o r e, t h e b l u e b e a d r e a c h e s t h e a t b o t t o m f i r s t .$

(b) Determination time required by red bead to fall from point A to C

$T h e t i m e t a k e n b y t h e r e d b e a d t o f a l l f r o m p o i n t A t o p o i n t C i s g i v e n a s : D = u t_{r} + \frac{1}{2} g {t_{r}}^{2} D = \frac{1}{2} g {t_{r}}^{2} H e r e, u i s t h e i n i t i a l s p e e d o f r e d b e a d a n d i t s v a l u e i s z e r o . S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . D = (0) t_{r} + \frac{1}{2} g {t_{r}}^{2} t_{r} = \sqrt{\frac{2 D}{g}} . . . . . . (1) T h e r e f o r e, t h e t i m e t a k e n b y r e d b e a d t o f a l l f r o m p o i n t A t o p o i n t C i s \sqrt{\frac{2 D}{g}} .$

(c) Determine time required by blue bead to fall from point B to C

$T h e a c c e l e r a t i o n o f b l u e b e a d i s g i v e n a s : a_{b} = g \sin θ T h e t i m e t a k e n b y r e d b e a d t o f a l l f r o m p o i n t A t o p o i n t C i s g i v e n a s : L = v t_{b} + \frac{1}{2} a_{b} {t_{b}}^{2} H e r e, v i s t h e i n i t i a l s p e e d o f b l u e b e a d a n d i t s v a l u e i s z e r o . S u b s t i t u t e a l l t h e v a l u e s i n t h e a b o v e e q u a t i o n . L = (0) t_{b} + \frac{1}{2} (g \sin θ) {t_{b}}^{2} L = \frac{1}{2} g \sin θ {t_{b}}^{2} t_{b} = \sqrt{\frac{2 L}{g \sin θ}} . . . . . (2) T h e r e f o r e, t h e t i m e t a k e n b y r e d b e a d t o f a l l f r o m p o i n t B t o p o i n t C i s \sqrt{\frac{2 L}{g \sin θ}} .$

(d) Show proof for the same time interval for both beads at point C

$T h e a n g l e o f t h e s m a l l r o d f r o m h o r i z o n t a l i s g i v e n a s : \sin θ = \frac{L}{D} S u b s t i t u t e a l l t h e v a l u e s i n t h e e q u a t i o n (2) . t_{b} = \sqrt{\frac{2 L}{g (\frac{L}{D})}} t_{b} = \sqrt{\frac{2 D}{g}} . . . . . . (3) T h e e q u a t i o n (1) a n d e q u a t i o n (3) s h o w s t h a t t i m e f o r b o t h b e a d s t o r e a c h a t t h e b o t t o m i s t h e s a m e . T h e r e f o r e, t h e r e d b e a d a n d b l u e b e a d r e a c h e s a t t h e b o t t o m s i m u l \tan e o u s l y .$

(e) Analysis for result in part (a)

The intuitive guess for the time interval for both beads is not correct for large angles. If the angle is small, then the time intervals for the beads remain the same.