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Unformatted text preview: θ ˙ cos(45 .
(351)
m √
2
˙
+mg R cos(45◦ −˙ 2 + Rxθ cos ◦ R θ 2R second
(345)
The square of the speed is then vm = xθ)+ 2R cos(45θ + θ)2+2 . To cos α . order, we
˙˙
may set cos θ ≈ 1 here. So the Lagrangian is ˙
L = (M/2)x2 + (m/2)(x2 + 2Rxθ + R2 θ2 ) + mgR cos θ.
˙
˙
˙˙ (352) The equations of motion are
¨
(M + m)¨ + mRθ = 0,
x and ¨
x + R θ ≈ −g θ ,
¨ (353) where we have used sin θ ≈ θ. To ﬁnd the normal modes, we can use the determinant
method. Or we can just solve for x in the ﬁrst equation and plug into the second,
¨
which gives
M +m
¨
θ=−
M g
θ =⇒ θ(t) = A cos(ω t + φ),
R where ω = M +m
M g
.
R
(354)
79
The corresponding x is
x(t) = − MR
MR
θ=−
A cos(ω t + φ).
M +m
M +m (355) In this mode, the two masses move in opposite directions.
The other solution to Eq. (354) is θ = 0 identically (this solution would pop out of
the determinant method). Such a solution is usually a trivial solution, but in this
problem both equations of motion give x = 0 =⇒ x(t) = Dt + E . So in this mode,
¨
the wheel moves at a constant rate, with the mass always at th...
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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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