Motion in one direction 2

網路上的路人甲

2013-10-06 10:47 pm

picture: http://www.webassign.net/serpse8/2-p-069.gif

Two thin rods are fastened to
the inside of a circular ring as
shown in Figure P2.69. One
rod of length D is vertical, and
the other of length L makes an
angle u 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 to point in terms of g and D.
(c) Find an expression for the time interval required for
the blue bead to slide from point to point in terms of
g, L, and u. (d) Show that the two time intervals found in
parts (b) and (c) are equal. Hint: What is the angle between
the chords of the circle and ? (e) Do these results
surprise you? Was your intuitive guess in part (a) correct?
This problem was inspired by an article by Thomas B.
Greenslade, Jr., “Galileo’s Paradox,” Phys. Teach. 46, 294
(May 2008).

thx~!

回答 (1)

天同

2013-10-07 1:28 am

✔ 最佳答案

(a) You make your own guess.

(b) Since the red bead falls under gravity, use equation: s = ut + (1/2)at^2
with s = D, u = 0 m/s, a = g (acceleration due to gravity), t= ta
hence, D = (1/2)g(ta)^2
(ta)^2 = 2D/g
ta = square-root[2D/g]

(c) Acceleration of the blue bead = g.sin(u)
use equation: s = ut + (1/2)at^2
with s = L, u = 0 m/s, a = g.sin(u), t = tb
hence, L = (1/2).(g.sin(u)).(tb)^2
(tb)^2 = 2L/[g.sin(u)]
tb = square-root[2L/[g.sin(u)]

(d) From geometry, angle ABC = 90 degrees, and angle BAC = u
hence, in triangle ABC, sin(u) = L/D
i.e. D = L/sin(u)
hence, tb = square-root[2D/g]
which is the same as ta

(e) Use your own comment.

2013-10-07 17:21:56 補充：
Q: 想請問下點肯定angle ABC=90degrees?
A: This is a theorem in geometry. The triangle formed by the diameter and two chords meet at the circumference is a right angle triangle.

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