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Problems & Exercises

1.

ω = 0 . 737 rev/s ω = 0 . 737 rev/s

3.

(a) 0.26 rad/s20.26 rad/s2

(b) 27rev27rev

5.

(a) 80 rad/s280 rad/s2

(b) 1.0 rev

7.

(a) 45.7 s

(b) 116 rev

9.

a) 600 rad/s2600 rad/s2

b) 450 rad/ss

c) 21.0 m/ss

10.

(a) 0.338 s

(b) 0.0403 rev

(c) 0.313 s

12.

0.50 kg m 2 0.50 kg m 2

14.

(a) 50.4 Nm50.4 Nm

(b) 17.1 rad/s217.1 rad/s2

(c) 17.0 rad/s217.0 rad/s2

16.

3 . 96 × 10 18 s 3 . 96 × 10 18 s

or 1.26 × 10 11 y 1.26 × 10 11 y

18.

I end = I center + m l 2 2 Thus, I center = I end 1 4 ml 2 = 1 3 ml 2 1 4 ml 2 = 1 12 ml 2 I end = I center + m l 2 2 Thus, I center = I end 1 4 ml 2 = 1 3 ml 2 1 4 ml 2 = 1 12 ml 2

19.

(a) 2.0 ms

(b) The time interval is too short.

(c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of 500 Nm500 Nm is reasonable.

20.

(a) 17,500 rpm

(b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs.

(c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity).

21.

(a) 185 J

(b) 0.0785 rev

(c) W=9.81 NW=9.81 N

23.

(a) 2.57×1029 J2.57×1029 J

(b) KErot=2.65×1033 JKErot=2.65×1033 J

25.
KE rot = 434 J KE rot = 434 J
27.

(a) 128 rad/s128 rad/s

(b) 19.9 m19.9 m

29.

(a) 10.4 rad/s210.4 rad/s2

(b) net W=6.11 J net W=6.11 J

34.

(a) 1.49 kJ

(b) 2.52×104 N2.52×104 N

36.

(a) 2.66×1040kgm2/s2.66×1040kgm2/s

(b) 7.07×1033kgm2/s7.07×1033kgm2/s

The angular momentum of the Earth in its orbit around the Sun is 3.77×1063.77×106 times larger than the angular momentum of the Earth around its axis.

38.

22 . 5 kg m 2 /s 22 . 5 kg m 2 /s

40.

25.3 rpm

43.

(a) 0.156 rad/s0.156 rad/s

(b) 1.17×102 J1.17×102 J

(c) 0.188 kgm/s0.188 kgm/s

45.

(a) 3.13 rad/s

(b) Initial KE = 438 J, final KE = 438 J

47.

(a) 1.70 rad/s

(b) Initial KE = 22.5 J, final KE = 2.04 J

(c) 1.50 kgm/s1.50 kgm/s

48.

(a) 5.64×1033kgm2/s5.64×1033kgm2/s

(b) 1.39×1022Nm1.39×1022Nm

(c) 2.17×1015N2.17×1015N

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