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Rotation HW Solutions Phys Gladden
Chapter 10
- (a) Consider three of the disks (starting with the one at point O ): ⊕OO. The first one (the one at point O – shown here with the plus sign inside) has rotational inertial (see item (c) in Table 10-2) I = mR^2. The next one (using the parallel-axis theorem) has
I = mR^2 + mh^2
where h = 2 R. The third one has I = mR^2 + m (4 R )^2. If we had considered five of the disks OO⊕OO with the one at O in the middle, then the total rotational inertia is
I = 5( mR^2 ) + 2( m (2 R )^2 + m (4 R )^2 ).
The pattern is now clear and we can write down the total I for the collection of fifteen disks:
I = 15( mR^2 ) + 2( m (2 R )^2 + m (4 R )^2 + m (6 R )^2 + … + m (14 R )^2 ) = mR^2.
The generalization to N disks (where N is assumed to be an odd number) is
I = (2 N^2 + 1) NmR^2.
In terms of the total mass ( m = M /15) and the total length ( R = L /30), we obtain
I = 0.083519 ML^2 ≈ (0.08352)(0.1000 kg)(1.0000 m)^2 = 8.352 ×10−^3 kgm^2.
(b) Comparing to the formula ( e ) in Table 10-2 (which gives roughly I =0.08333 ML^2 ), we find our answer to part (a) is 0.22% lower.
- We use the parallel-axis theorem. According to Table 10-2(i), the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by
I (^) com = M^ a + b 12
c 2 2 h.
A parallel axis through the corner is a distance h^ =^ b a^ /^2 g^2 +b b /^2 g^2 from the center.
Therefore,
186 CHAPTER 10
com^2 (^2 2 )^ (^2 2 )^ (^2 2 )
2 2 4 2
0.172 kg (^) [(0.035 m) +(0.084 m) ] 4.7 10 kg m. 3
I I Mh M^ a b M^ a b M a b −
= = × ⋅
- (a) Using Table 10-2(c) and Eq. 10-34, the rotational kinetic energy is
(^1 2 1 1 2 2 1) (500 kg)(200 rad/s) (1.0 m) (^2 2) 4.9 10 J. 7 2 2 2 4
K = I ω = ^ MR ω = π = ×
(b) We solve P = K / t (where P is the average power) for the operating time t.
t K P
= = ×
×
4 9 10 = 6 2 × 10
J.
8.0 10 W
3 s
which we rewrite as t ≈ 1.0 ×10^2 min.
- Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball. As can be seen from the diagram, the component of the force of gravity that is perpendicular to the rod is mg sin θ. If ℓ^ is the length of the rod, then the torque associated with this force has magnitude
τ= mg ℓ sinθ=(0.75)(9.8)(1.25) sin 30° = 4.6 N m⋅ (^).
For the position shown, the torque is counter-clockwise.
188 CHAPTER 10
- (a) The acceleration vector is obtained by dividing the force vector by the (scalar) mass:
a^ → = F
→ / m = (3.00 m/s^2 )i^^ – (4.00 m/s^2 )j^^ + (2.00 m/s^2 )k^^.
(b) Use of Eq. 11-18 leads directly to
L
→ = (42.0 kg.m^2 /s)i^^ + (24.0 kg.m^2 /s)j^^ + (60.0 kg.m^2 /s)k^^.
(c) Similarly, the torque is
τ = r × F
= (–8.00 N.m)i^^ – (26.0 N.m)j^^ – (40.0 N.m)k^^.
(d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is 7.35 m/s and that of the force is 10.8 N. The dot product of these two vectors is v^ →^.^ F
→ = – 48 (in SI units). Thus, Eq. 3-20 yields
θ = cos−^1 [−48.0/(7.35 × 10.8)] = 127°.
- (a) We choose clockwise as the negative rotational sense and rightwards as the positive translational direction. Thus, since this is the moment when it begins to roll
smoothly, Eq. 11-2 becomes v com = − R ω= −b 011. mgω.
This velocity is positive-valued (rightward) since ω is negative-valued (clockwise) as
shown in Fig. 11-57.
(b) The force of friction exerted on the ball of mass m is − μ k mg (negative since it points
left), and setting this equal to ma com leads to
a com= − μ g = − b0 21. g c 9 8. m s^2 h= −2 1. m s^2
where the minus sign indicates that the center of mass acceleration points left, opposite to its velocity, so that the ball is decelerating.
(c) Measured about the center of mass, the torque exerted on the ball due to the frictional
force is given by τ = − μ mgR. Using Table 10-2(f) for the rotational inertia, the angular
acceleration becomes (using Eq. 10-45)
α = τ^ = −^ μ = −^ μ = −^ = − I
mgR m R
g 2 5 R
2.^.^47
b gb g
b g
rad s^2
where the minus sign indicates that the angular acceleration is clockwise, the same direction as ω (so its angular motion is “speeding up’’).
(d) The center-of-mass of the sliding ball decelerates from v com,0 to v com during time t according to Eq. 2-11: v (^) com = v (^) com,0− μ gt. During this time, the angular speed of the ball
increases (in magnitude) from zero to ω according to Eq. 10-12:
ω = α t = μ gt =
R
v R
com
where we have made use of our part (a) result in the last equality. We have two equations involving v com, so we eliminate that variable and find
t
v g
com,0 (^12) s.
b g.
b gb g
(e) The skid length of the ball is (using Eq. 2-15)
∆ x = v (^) com,0 t − 1 g t = − = m. 2
b^ μ g 2 b.^ gb.^ g b0 21.^ g^ b9 8.^ gb^12.^ g^2 8 6.
(f) The center of mass velocity at the time found in part (d) is
v com = v com,0 − μ gt = 8 5. − b0 21. g b9 8. gb 12. g= 61. m s.
