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```					13.5. Model: Spinning skater, whose arms are outstretched, is a rigid rotating body.
Visualize:

Solve: The speed v  r, where r  140 cm/2  0.70 m. Also, 180 rpm  (180)2 /60 rad/s  6 rad/s. Thus, v 
(0.70 m)(6 rad/s)  13.2 m/s.
Assess: A speed of 13.2 m/s  26 mph for the hands is a little high, but reasonable.

13.8. Model: The turbine is a rigid rotating body.
Solve: The known values are i  3600 rpm  (3600)(2)/60  120 rad/s, ti  0 s, tf  10 min  600 s, f  0
rad/s, and i  0 rad. Using the rotational kinematic equation f  i  (tf – ti), we get 0 rad  (120 rad/s)  (600
s – 0 s). Thus,   0.628 rad/s2. Now,
1
 f   i   i (t f  t i )   (t f  t i )2
2
1
 0 rad  (120 rad/s)(600 s  0 s)  ( 0.628 rad/s2 )(600 s  0 s)2
2
 113,100 rad  18,000 rev
Assess: 18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable.

13.10. Visualize: Please refer to Figure Ex13.10. The coordinates of the three masses mA, mB, and mC are
(0 cm, 0 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively.
Solve: The coordinates of the center of mass are
mA xA  mB xB  mC xC (100 g)(0 cm)  (200 g)(10 cm)  (300 g)(10 cm)
xcm                                                                          8.33 cm
mA  mB  mC                   (100 g  200 g  300 g)
mA yA  mB yB  mC yC (100 g)(0 cm)  (200 g)(10 cm)  (300 g)(0 cm)
ycm                                                                         3.33 cm
mA  mB  mC                   (100 g  200 g  300 g)
13.18. Model: The moment of inertia of any object depends on the axis of rotation.
Visualize:

Solve: (a)     xcm 
m xi i

mA xA  mB xB  mC xC  mD xD
m   i            mA  mB  mC  mD
(100 g)(0 m)  (200 g)(0 m)  (200 g)(0.10 m)  (200 g)(0.10 m)
                                                                  0.0571 m
100 g  200 g  200 g  200 g
mA yA  mB yB  mC yC  mD yD
ycm 
mA  mB  mC  mD
(100 g)(0 m)  (200 g)(0.10 m)  (200 g)(0.10 cm)  (200 g)(0 m)
                                                                     0.0571 m
700 g
(b) The moment of inertia about a diagonal that passes through B and D is
I BD  mArA  mCrC
2      2

where rA  rC  (0.10 m) cos 45  7.07 cm and are the distances from the diagonal. Thus,
IBD  (0.100 kg)rA  (0.200 kg)rC  0.0015 kg m2
2              2

Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.

13.22. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through
the center of mass. Assume that the size of the balls is small compared to 1 m.
Visualize:

We placed the origin of the coordinate system on the 1.0 kg ball.
Solve: The center of mass and the moment of inertia are
(1.0 kg)(0 m)  (2.0 kg)(1.0 m)
xcm                                     0.667 m and ycm  0 m
(1.0 kg  2.0 kg)
Iabout cm   miri2  (1.0 kg)(0.667 m)2  (2.0 kg)(0.333 m)2  0.667 kg m2

We have f  0 rad/s, tf – ti  5.0 s, and i  20 rpm  20(2 rad/60 s)  2  rad/s, so f  i  (tf – ti) becomes
3
 2                           2
0 rad/s      rad/s    (5.0 s)     rad/s2
 3                            15
Having found I and , we can now find the torque  that will bring the balls to a halt in 5.0 s:
2           2             4
  Iabout cm   kg m2      rad/s2       N m  0.279 N m
3           15             45
The magnitude of the torque is 0.279 N m.

13.46. Model: The bar is a rotating rigid body. Assume that the bar is thin.
Visualize: Please refer to Figure Ex13.46.
Solve: The angular velocity   120 rpm  (120)(2)/60 rad/s  4 rad/s. From Table 13.3, the moment of
inertial of a rod about its center is I  12 ML2 . The angular momentum is
1

 1
L  I    (0.50 kg)(2.0 m)2 (4 rad/s)  2.09 kg m2/s
 12 
If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the
page. Consequently,
L  (2.09 kg m2 /s, out of the page)

13.56. Model: The beam is a rigid body of length 3.0 m and the student is a particle.
Visualize:

Solve:   To stay in place, the beam must be in both translational equilibrium (Fnet  0 N) and rotational
equilibrium ( net  0 Nm) . The first condition is

F  y    wbeam  wstudent  F1  F2  0 N
 F1  F2  wbeam  wstudent (100 kg  80 kg)(9.80 m/s2)  1764 N
Taking the torques about the left end of the beam, the second condition is
wbeam (1.5 m) – wstudent (2.0 m)  F2 (3.0 m)  0 N m
 (100 kg)(9.8 m/s2)(1.5 m) – (80 kg)(9.8 m/s2)(2.0 m)  F2 (3.0 m)  0 N m
 F2  1013 N
From F1  F2  1764 N, we get F1  1764 N – 1013 N  751 N.
Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any
point on the body of interest.
13.78. Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from
above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable 
blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum
of the system is conserved.
Visualize: The initial moment of inertia is I1 and the final moment of inertia is I2.
Solve: The initial moment of inertia is I1  Idisk  2 mR2  2 (2.0 kg)(0.10 m)2  0.010 kg m2 and the final
1      1

moment of inertia is
I2  I1  2mR2  0.010 kg m2  2(0.500 kg)  (0.10 m)2  0.010 kg m2  0.010 kg m2  0.020 kg m2
Let 1 and 2 be the initial and final angular velocities. Then
I11 (0.010 kg m2 )(100 rpm)
Lf  Li  2 I2  1I1  2                                 50 rpm
I2       0.020 kg m2

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