ωπ π π ω by gjjur4356

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									                                     Assignment #8
                                    Physics 102/104
                         Augustana Campus, University of Alberta
                                       Fall 2009
                                       Solutions

Q1:   Textbook question 8-15:

      If there were a great migration of people toward the Earth’s equator, how would this
      affect the length of the day?

A1:   Such a migration would increase the moment of inertia of the Earth. From angular
      momentum conservation ( L = Iω = constant ) this would lead to a decrease in the angular
      velocity of the Earth and therefore the length of the day would increase (since
      T = 2π / ω ).

      So that you don’t cancel your winter vacation to the tropics, let’s try to estimate the size
      of the effect. The moment of inertia of the Earth is approximately (modeling the Earth by
      a solid sphere)
                       I = 5 MR 2 = 2 (6 × 10 24 kg )(6 × 10 6 m ) 2 = 9 × 10 37 kg m 2
                           2
                                    5
      Next, suppose one-billion people, with an average mass of 50 kg, migrate from a latitude
      of 45° (whereby r = R sin 45° = R / 2 ) to the equator. The increase in the Earth’s
      moment of inertia will be
                       ∆I = Nm( R 2 − r 2 ) = (10 9 )(50 kg ) 1 (6 × 10 6 m) 2 = 9 × 10 23 kg m 2
                                                              2

      This is one part in 1014 of I, and ω and T will also change by the same fraction, therefore
                                               T    10 5 s
                                         ∆T = 14 = 14 = 10 −9 s
                                             10      10
      In other words, the day would lengthen by a nanosecond.

Q2:   Textbook problem 8-10:

      What is the linear speed of a point (a) on the equator, (b) on the Arctic Circle (latitude
      66.5° N), and (c) at a latitude of 45.0° N, due to the Earth’s rotation?

A2:   We compute the linear speed from the angular speed via v = ωr . The Earth’s angular
      speed is easily computed from the length of a day:
                            2π      2π  1 day  1 h                −5 -1
                        ω=       =                     = 7.27 × 10 s
                             T 1 day  24 h  3600 s 
      If the radius of the Earth is R and the latitude (measured from the Equator) is φ , then we
      can compute the rotational radius, r, appropriate to any latitude via the triangle in the
      sketch below.
                                                            r


                                          φ                        R




      We quickly see that r = R cos φ . We now have all the ingredients to compute the various
      linear speeds requested by the problem.
      (a) v = (7.27 × 10 −5 s -1 )(6.38 × 10 6 m ) cos(0°) = 464 m/s
      (b) v = (7.27 × 10 −5 s -1 )(6.38 × 10 6 m ) cos(66.5°) = 185 m/s
      (c) v = (7.27 × 10 −5 s -1 )(6.38 × 10 6 m) cos(45.0°) = 328 m/s

Q3:   Textbook problem 8-38:

      The forearm in Figure 8-45 of the text accelerates a 3.6 kg ball at 7.0 m/s 2 by means of
      the triceps muscle. (In the diagram, the elbow is labeled as the axis of rotation, the ball is
      31 cm from the elbow, and the triceps attaches to the other side of the elbow at a distance
      of 2.5 cm from the axis of rotation.) Calculate (a) the torque needed, and (b) the force
      that must be exerted by the triceps muscle. Ignore the mass of the arm.

A3:   (a) We can compute the torque using the rotational analogue of Newton’s Second Law:
                                               τ = Iα
      The moment of inertia of the ball about the elbow (since we are neglecting the mass of
      the arm itself is
                               I = MR 2 = (3.6 kg )(0.31 m) 2 = 0.346 kg m 2
      The angular acceleration, meanwhile, is connected to the tangential acceleration of the
      ball:
                                            a (7.0 m/s 2 )
                                       α= =                    = 22.6 s -2
                                            R     (0.31 m)
      Putting these elements together, we obtain the torque supplied by the triceps:
                               τ = (0.346 kg m 2 )(22.6 s -2 ) = 7.8 Nm

      (b) Since the force acts with maximum leverage, the lever arm is the full 2.5 cm
      separation between the tendon and the elbow, leading to
                                    τ (7.8 Nm)
                                F= =            = 3.1 × 10 2 N
                                    r (0.025 m)
Q4:   Textbook problem 8-64:

      Hurricanes can involve winds in excess of 120 km/h at the outer edge. Make a crude
      estimate of (a) the energy, and (b) the angular momentum, of such a hurricane,
      approximating it as a rigidly rotating uniform cylinder of air (density 1.3 kg/m 3 ) of
      radius 100 km and height 4.0 km.

A4:   We’ll start by estimating the total mass of air associated with a hurricane:
               M = ρV = ρπR 2 h = (1.3 kg/m 3 )π (10 5 m) 2 ( 4000 m) = 1.6 × 1014 kg
      Treating the hurricane (somewhat unrealistically) as a rigid uniform cylinder, we can then
      estimate the moment of inertia:
                        I = 1 MR 2 = 1 (1.6 × 1014 kg )(10 5 m) 2 = 8.2 × 10 23 kg m 2
                            2        2
      Meanwhile, the rotational speed of the hurricane is
                                     v (120 km/h )  1 h                    − 4 -1
                               ω= =                             = 3.3 × 10 s
                                    R      (100 km)  3600 s 

      (a) The kinetic energy associated with this mass of rotating air is
                       E = 1 Iω 2 = 1 (8.2 × 10 23 kg m 2 )(3.3 × 10 −4 s -1 ) 2 = 5 × 1016 J
                            2       2
      (FYI: This works out to ten-million tons of TNT, and this is just the wind energy of the
      hurricane; additional energy is stored as latent heat in the water vapor.)

      (b) The angular momentum associated with this mass of rotating air is
                        L = Iω = (8.2 × 10 23 kg m 2 )(3.3 × 10 −4 s -1 ) = 3 × 10 20 kg m/s
      Note that, given the crude nature of these estimates, the final results have been rounded to
      a single significant figure.

Q5:   Textbook problem 8-71:

      A 1.4 kg grindstone in the shape of a uniform cylinder of radius 0.20 m acquires a
      rotational rate of 1800 rev/s from rest over a 6.0 s interval at constant angular
      acceleration. Calculate the torque delivered by the motor.

A5:   This is a simple application of the rotational analogue of Newton’s Second Law:
                        (      )
               τ = Iα = 1 MR 2 
                                 ∆ω  1
                                       = 2 (1.4 kg )(0.20 m)
                                                              2 (1800 rev/s)(2π rad/rev)
                                                                                         = 53 Nm
                                 ∆t 
                         2
                                                                         ( 6 .0 s )
BONUS:       Tides dissipate about 3 TW of power in the Earth’s oceans and to conserve
             energy, this energy is lost from the Earth’s rotation and the Moon’s orbit.
             (a) Compute, to one significant figure, the rotational kinetic energy of the Earth
             and the orbital kinetic energy of the Moon.
             (b) Assuming that the power dissipation splits proportionately according to the
             result of (a), use angular momentum and energy conservation to estimate how
             much the distance to the Moon changes each year. Is this observable?

A:   (a) The rotational kinetic energy of the Earth is about

                                  (          )
                                         2π    4π 2 (6 × 10 24 kg )(6 × 10 6 m) 2
                                                     2
              E ROT = 1 Iω 2 =
                      2
                                  MR 2 
                                 1 2
                                 2 5          =               2             2
                                                                                    = 2 × 10 29 J
                                         T          5(24 h ) (3600 s/h )
     The orbital kinetic energy of the Moon is approximately
                                   2πr    2π 2 (7 × 10 22 kg )(4 × 10 8 m) 2
                                                 2
             E ORB = 1 mv 2 = 1 m
                       2      2          =                                    = 4 × 10 28 J
                                    t      (27 d) 2 (24 h/d) 2 (3600 s/h ) 2
     We see that the Earth’s rotational KE is about 5 times larger than the Moon’s orbital KE.
     We are neglecting the Moon’s rotational motion and the fact that the Earth and Moon
     both orbit their common center of mass.

     (b) Let us compare the angular momentum and kinetic energy of the Moon at one instant
     (denoted by the subscript “1”) with the corresponding quantities one year later (denoted
     by the subscript “2”). Angular momentum cannot be dissipated, therefore
                                                                    vr
                       L1 = L2 ⇒ mv1 r1 = mv 2 r2 ⇒ v 2 = 1 1
                                                                    r2
     (Note that if the Earth’s rotational speed changes, and it does, then it is the total angular
     momentum of the Earth-Moon system which must be conserved, and not necessarily the
     individual pieces. A more detailed calculation would account for this, but we’re just after
     a ballpark estimate of the basic effect so we’ll ignore this.)

     Energy is most certainly not conserved if it is being dissipated by the tides. Taking 1/6 of
     the total dissipation rate to be applied to the Moon’s orbit (in the absence of the more
     difficult integrated Earth-Moon calculation),
                                                 Pt
                                    E1 = E 2 +
                                                 6
                                                    Pt
                                 1
                                 2
                                   mv12 = 1 mv 2 +
                                          2
                                               2
                                                    6
                                                     2
                                      v r       ( Pt / 6)
                                    = 1 1  +
                                       v12
                                       r 
                                       2        ( m / 2)
     Note that we have substituted for v 2 using the result from angular momentum
     conservation. Solving for r2 ,
                                                                   2
                                         ( Pt / 6)       r    
                                 v12 −             = v12  1
                                                         r    
                                                               
                                         (m / 2)          2   
                                                           2
                                             Pt  r1 
                                         1−     = 
                                            6 E1  r2 
                                                  
                                              r1
                                   r2 =
                                          1 − Pt / 6 E1
The key factor here is
           Pt    (3 × 1012 W )(1 y)(365 d/y)(24 h/d)(3600 s/h )
               =                                                = 4 × 10 −10
          6 E1                    6(4 × 10 J)
                                            28

This is the amount by which the argument of the square root differs from one. Perhaps
your calculator will take this square root properly, but if it doesn’t (and even if it does),
we can compute the square root using the binomial approximation:
                           (1 + x) n ≅ 1 + nx for small x
                                               1      x
                                     ⇒            ≅1+
                                             1− x     2
This means that r2 will be larger than r1 by a relative factor of 2 × 10 −10 . With
 r1 = 4 × 10 8 m , this means that r2 = r1 + 8 cm . This calculation suggests that the Moon is
receding from the Earth by several centimeters each year as a result of tidal dissipation.
A more detailed calculation (treating the Earth and Moon as one big system and not
neglecting the gravitational potential energy between them) yields a result of 3.8 cm/y. Is
this observable? Yes, because it has been observed: the Apollo 11 astronauts placed a set
of mirrors on the Moon in 1969 that have allowed us to track the rate at which the Moon
is receding from the Earth. This has all sorts of interesting consequences:
(1) The Moon will eventually (in about a billion years) recede far enough from the Earth
so that total solar eclipses will no longer be possible.
(2) The Earth’s rotation will gradually slow down so that the day becomes much longer
than it currently is. (There is evidence in fossil records that the days on Earth were
shorter than 24 hours in the past.)
(3) The Earth and Moon will be completely “tidally locked”. Right now, the same side of
the Moon always faces the Earth, and eventually the same side of the Earth will always
face the Moon.
(4) The Moon will never escape completely from the Earth, as the tidal dissipation
decreases quite quickly once Earth becomes tidally locked.

								
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