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1.45 (a) The angular momenta
ωo ωo ωo
r 2r ωo
r L1 = I1ω o and L2 = I2ω o,
but they are in opposite directions. Since there are
no external torques angular momentum is conserved. Question 4: Two disks, with identical masses but The initial angular momentum is different radii (r and 2r) are spinning on frictionless L1 + ( −L2 ) = I1ω o − I2ω o = (I1 − I2 )ω o,
and the final angular momentum is bearings with the same angular speed ω o but in opposite (I1 + I2 )ω . directions. The two disks are brought slowly together
and the resulting frictional force between their surfaces
eventually brings them to a common angular velocity.
(a) What is the magnitude of the final angular speed in
terms of ω o ? (b) What is the change in the rotational
kinetic energy? 1
But I1 = M(2 r )2 = 2Mr 2 and I2 = Mr 2.
∴ 2 − Mr 2ω o = 2 + Mr 2ω , 2
Mr ω o = Mr 2ω
∴ ω = ω o.
5 The total initial rotational kinetic energy:
Ki = I1ω o2 + I2ω o2 = (I1 + I2 )ω o2
The total final rotational kinetic energy:
K f = (I1 + I2 )ω 2.
∴ ∆K K f − Ki ω 2 − ω o2
ω o2 3
but ω = ω o, so (ω 2 − ω o2 ) < 0, i.e., ∆K is always < 0.
K i.e., 64% of the kinetic energy is “lost”.
• What’s happened to it?
Note: this is the analog of the inelastic collision we
studied in chapter 8. Question 5: (a) Assuming the Earth to be a
homogeneous sphere of radius R and mass M, show that
the period of rotation about its axis is
4 πM 2
where L is the angular momentum of the Earth about its
axis. (b) Suppose the radius changes by a small amount
∆R . Show that the fractional change in period is
R (c) By how many kilometers would the Earth have to
expand for the period to change by 0.25 days/year so
that leap years would no longer be necessary? (a) If the angular velocity is ω we have
• angular momentum L = Iω ,
• moment of inertia I = MR 2.
The period of rotation is T =
ω ω = 2 π 2 πI 4 π M 2
L 5L I QED. DISCUSSION PROBLEM [10.2]:
A student sits on a stool that is free to rotate. She is
handed a wheel that is spinning in the horizontal plane as
shown below. (b) Let R ⇒ R + ∆R so T ⇒ T + ∆T , then
4 πM ∆T dT
R = 2 2 R . 5L R ∆R dR
R Assuming she (and the stool) are stationary to begin (c) ∆R ≈ R ∆T...
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