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Unformatted text preview: alf the sin(45 − 2 ), − cos(45 −the ,normal mode (θ, α) ∝ (1, 1)
√
√
◦ 0, we have the normal mode (θ , α) ∝ (1, −1) with
◦
with frequency ω=. And sif B1 = θ) + 2 sin α, − cos(45 + θ) − 2 cos α .
(x, y )2 1 R in(45 +
frequency ω2 . It makes sense that the frequency of the second mode is larger than
the frequency of the ﬁrst. a circle, so its speed is Rθ. Calculating x2 + y2 for the
˙
The left mass moves along
˙2
˙2 right sliding on a the
6.45. Massmass (or using rimlaw of cosines), we have
Let 2 be the2 horizontal coordinate2of M , and 2 θ √ the˙angle of ◦ along the hoop,
x
be 2
m
2 let
˙
˙2
˙
v1 = R2 θ ,
and
v2
˙
measured counterclockwise 2 = Rthe + 2R α of 2 2R θα cos(45 the − α). (344)
from θ bottom + the hoop. Then + θ position and
velocity of m are
The Lagrangian is therefore
(x, y )m = (x + R sin θ, R − R cos θ)
√
2
◦
L = (m/2) (2Ry )˙2 + 2R2 αx + 2 θ 2R2θ˙,αRθ sin θ)+ θ − α)
˙2
˙
=⇒
x, ˙ θ
˙
= ( ˙ + R ˙ cos...
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This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.
 Fall '13
 Brandenberger
 mechanics, Centrifugal Force, Force

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