College Physics for AP® Courses

# Chapter 10

College Physics for AP® CoursesChapter 10

### Problems & Exercises

1.

$ω = 0 . 737 rev/s ω = 0 . 737 rev/s size 12{ω= {underline {0 "." "737 rev/s"}} } {}$

3.

(a) $−0.26 rad/s2−0.26 rad/s2 size 12{ - 0 "." "26 rad/s" rSup { size 8{2} } } {}$

(b) $27rev27rev size 12{"27""rev"} {}$

5.

(a) $80 rad/s280 rad/s2 size 12{80 rad/s" rSup { size 8{2} } } {}$

(b) 1.0 rev

7.

(a) 45.7 s

(b) 116 rev

9.

a) $600 rad/s2600 rad/s2 size 12{ {underline {6"00 rad/s" rSup { size 8{2} } }} } {}$

c) 21.0 m/s

10.

(a) 0.338 s

(b) 0.0403 rev

(c) 0.313 s

12.

$0.50 kg ⋅ m 2 0.50 kg ⋅ m 2 size 12{5 "." "00""kg" cdot m rSup { size 8{2} } } {}$

14.

(a) $50.4 N⋅m50.4 N⋅m$

(b) $17.1 rad/s217.1 rad/s2 size 12{"17" "." 1"rad/s" rSup { size 8{2} } } {}$

(c) $17.0 rad/s217.0 rad/s2 size 12{"17" "." 0"rad/s" rSup { size 8{2} } } {}$

16.

$3 . 96 × 10 18 s 3 . 96 × 10 18 s size 12{3 "." "96" times "10" rSup { size 8{"18"} } s} {}$

or $1.26 × 10 11 y 1.26 × 10 11 y size 12{1 "." "26" times "10" rSup { size 8{"11"} } y} {}$

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 $500 N⋅m500 N⋅m size 12{"500 N" cdot m} {}$ 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) $W=9.81 NW=9.81 N size 12{W= {underline {9 "." "81 N"}} } {}$

23.

(a) $2.57×1029 J2.57×1029 J size 12{9 "." "736" times "10" rSup { size 8{"37"} } " kg" "." m rSup { size 8{2} } } {}$

(b) $KErot=2.65×1033 JKErot=2.65×1033 J size 12{"KE" rSub { size 8{"tot"} } = {underline {2 "." "65" times "10" rSup { size 8{"33"} } " J"}} } {}$

25.

$KE rot = 434 J KE rot = 434 J size 12{ ital "KE" rSub { size 8{ ital "rot"} } = {"434 J"} } {}$

27.

(a) $128 rad/s128 rad/s size 12{ {underline {"128 rad/s"}} } {}$

(b) $19.9 m19.9 m$

29.

(a) $10.4 rad/s210.4 rad/s2 size 12{α= {underline {"19" "." "5 rad/s" rSup { size 8{2} } }} } {}$

(b) $net W=6.11 J net W=6.11 J size 12{" net W"= {underline {6 "." "31 J"}} } {}$

34.

(a) 1.49 kJ

(b) $2.52×104 N2.52×104 N size 12{I= {underline {9 "." "61" times "10" rSup { size 8{3} } " N"}} } {}$

36.

(a) $2.66×1040kg⋅m2/s2.66×1040kg⋅m2/s size 12{2 "." "66" times "10" rSup { size 8{"40"} } "kg" cdot m rSup { size 8{2} } "/s"} {}$

(b) $7.07×1033kg⋅m2/s7.07×1033kg⋅m2/s size 12{7 "." "07" times "10" rSup { size 8{"33"} } "kg" cdot m rSup { size 8{2} } "/s"} {}$

The angular momentum of the Earth in its orbit around the Sun is $3.77×1063.77×106 size 12{3 "." "76" times "10" rSup { size 8{6} } } {}$ times larger than the angular momentum of the Earth around its axis.

38.

$22 . 5 kg ⋅ m 2 /s 22 . 5 kg ⋅ m 2 /s size 12{"22" "." "5 kg" cdot m rSup { size 8{2} } "/s"} {}$

40.

25.3 rpm

43.

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

(b) $1.17×10−2 J1.17×10−2 J size 12{1 "." "17" times "10" rSup { size 8{ - 2} } " J"} {}$

(c) $0.188 kg⋅m/s0.188 kg⋅m/s size 12{0 "." "188 kg" cdot "m/s"} {}$

45.

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

47.

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

(c) $1.50 kg⋅m/s1.50 kg⋅m/s size 12{1 "." "50 kg" cdot "m/s"} {}$

48.

(a) $5.64×1033kg⋅m2/s5.64×1033kg⋅m2/s size 12{5 "." "65" times "10" rSup { size 8{"33"} } "kg" "." m rSup { size 8{2} } "/s"} {}$

(b) $1.39×1022N⋅m1.39×1022N⋅m size 12{1 "." "39" times "10" rSup { size 8{"22"} } N cdot m} {}$

(c) $2.17×1015N2.17×1015N size 12{2 "." "18" times "10" rSup { size 8{"15"} } N} {}$

### Test Prep for AP® Courses

1.

(b)

3.

(d)

5.

(d)

You are given a thin rod of length 1.0 m and mass 2.0 kg, a small lead weight of 0.50 kg, and a not-so-small lead weight of 1.0 kg. The rod has three holes, one in each end and one through the middle, which may either hold a pivot point or one of the small lead weights.

7.

(a)

9.

(c)

11.

(a)

13.

(a)

15.

(b)

17.

(c)

19.

(b)

21.

(b)

23.

(c)

25.

(d)

27.

A door on hinges is a rotational system. When you push or pull on the door handle, the angular momentum of the system changes. If a weight is hung on the door handle, then pushing on the door with the same force will cause a different increase in angular momentum. If you push or pull near the hinges with the same force, the resulting angular momentum of the system will also be different.

29.

$τ = ΔL Δt τ = ΔL Δt$

$τ⋅Δt = ΔL τ⋅Δt = ΔL$

By substituting,

120 N•m • 1.2 s = 144 N•m•s.

The angular momentum of the globe after 1.2 s is 144 N•m•s.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.