Physics Rotational Motion
Homework Solutions
p. 182 - 187: #5, 6, 8, 9, 11 - 14, 23 - 26, 39, 49, Extra Challenge: 55
5) Linear speed increases as rotational speed increases.
6) Linear speed increases as distance from the axis increases.
8) Towards the center of the circle.
9) Inward
11) Lack of any net force. Newton’s first law – the Law of Inertia
12) Bug on can(inertia, or centrifugal force). Can on bug (centripetal force).
13) No, no, none.
14) It is not part of an action-reaction pair.
23) Fc = mv2 / r = (2) (3)2 / (2.5) = 7.2 N
24) Fc = mv2 / r = (60) (5)2 / (6) = 250 N
25) Fc = mv2 / r = (2) (10)2 / (1.6) = 125 N
26) Fc = mv2 / r = (70) (3)2 / (2) = 315 N
39) The horizontal component of the normal force. The normal force is the reaction force to the
weight of the car.
49) v = Δx / Δt = (2πr)/30 = 2π(10)/30 = 2.1 m/s
55) a) v = 2πr/ Δt = (6.28)(10)/(0.1) = 628 km/s
b) At twice the distance, the speed doubles, or 1256 km/s
c) v = 2πr/Δt r = 4777 km
pp. 206 - 211: # 1 - 4, 21, 25, 26, 28, 49 – 53
1) Force is a push that can accelerate objects. Torque is the force multiplied by the lever arm
that can produce rotational acceleration.
2) At a right angle to the lever arm.
3) They have the same magnitude.
4) The heavier kid should sit closer to the fulcrum.
21) C, D, A, B
25) τ = F d = (300)(3) = 900 Nm
τ = F d = (600)(1.5) = 900 Nm
There is no net torque.
26) Long handle.
Thick handle.
These choice provide the greater lever arm in their respective situations.
28) Both sit on a see-saw until it balances. Then, W d = W d
49) a) τ = F d = (80)(0.25) = 20 Nm
b) F = τ / d = 20 / 0.1 = 200 N
c) Yes, maximum torque is produced when you push perpendicular to the lever arm.
50)
Fd=Fd
(200)(30) = (500) x
6000 = 500 x
x = 12 cm
So, place it at the 62 cm mark on the meter stick.
51)
Fd+Fd=Fd
(100)(25) + (20)(40) = (500) x
2500 + 800 = 500 x
3300 = 500x
x = 6.6 cm
So, place it at the 56.6 cm mark.
52) Answers may vary.
53) Torque of the rock:
τ = F d = (1)(0.25) = 0.25 Nm
The first quarters of the meter stick from the fulcrum cancel each other out, so forget about them.
The remaining half of a meter stick is centered at 0.5 m from the fulcrum, so:
τ=Fd
0.25 = F(.5)
F = 0.5
If a half a meters stick has a mass of 0.5 kg, then a full meter stick has a mass of 1 kg.
pp. 227 - 231: # 1 - 3, 5, 6, 8, 9, 14 - 16, 19, 21, 22, 42 - 46
pp. 257 - 258: # 22 - 27
1) An object rotating about an axis tends to keep rotating about that axis at the same rate, unless
a net torque acts on it.
2) Yes, when the mass of an object is concentrated farther from the axis, there is greater
rotational inertia.
3) Held closer to the massive end (choke up on the bat). There is less rotational inertia, and
therefore less torque is required to swing it.
5) A short on will swing faster due to having less rotational inertia.
6) Bent legs have less rotational inertia.
8) By pulling their arms inward.
9) A solid disk because it has less rotational inertia per unit of mass, so it will accelerate faster.
14) None. Angular momentum is conserved.
15) Angular momentum remains constant, so her rate of rotation doubles.
16) In absence of a net torque, angular momentum does not change, but rate of rotation can.
19) C, B, A
21) a) A, B, C b) A, B, C
22) C, D, B, A
42) Using I = mr2 = (2m)(0.5r)2 = (2m)(0.25r2) = 0.5 mr2
So, it has half of the original rotational inertia.
43) Using I = mr2 = (2m)(2r)2 = (2m)(4r2) = 8 mr2
So, it has eight times the original rotational inertia.
44) Using I = mr2 = (0.5m)(0.5r)2 = (0.5m)(0.25r2) = 1/8 mr2
So, it has one eighth the original rotational inertia.
45) I = mr2 = (1000)(450)2 = 2 x 108 kg m2
4 x 108 km m2
46) I ω = I ω
(1)(2) = (0.33) ω
ω = 6 rot/s
(for each pod)
(for the two-pod system)
For these problems, use F = G m m / d2
22) 9.8 N
23) 2.4 N
24) 2.1 x 1020 N
25) 3.6 x 1022 N
26) 2.7 x 10-8 N
27) 2.2 x 10-7 N
Obstetrician
8.1 times more