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I for axis 2
I for axis 1
Using the expressions from Table 9.1 for the moments of inertia, this result becomes α for axis 2 I for axis 1
=
=
α for axis 1 I for axis 2 1
M B L2
B
3
1
M A L2
A
12 Noting that M A = 2M B and LA = 2 LB , we find that = 4M B L2
B
M A L2
A Chapter 9 Problems 463 α for axis 2 4M B L2
4 M B L2
1
B
B
=
=
=
2
2
2
α for axis 1 M A LA 2M ( 2 L )
B
B
α for axis 2 1
4.6 rad/s 2
= α for axis 1 =
= 2.3 rad/s 2
2
2 39. SSM REASONING The figure below shows eight particles, each one located at a different
corner of an imaginary cube. As shown, if we consider an axis that lies along one edge of
the cube, two of the particles lie on the axis, and for these particles r = 0. The next four
particles closest to the axis have r = l , where l is the length of one edge of the cube. The
remaining two particles have r = d , where d is the length of the diagonal along any one of
the faces. From the Pythagorean theorem, d = l 2 + l 2 = l 2 . l Axis d l d l l According to Equation 9.6, the moment of inertia of a system of particles is given by
I = ∑ mr 2 . SOLUTION Direct application of Equation 9.6 gives
I = ∑ mr 2 = 4(ml 2 ) + 2(md 2 ) = 4(ml 2 ) + 2(2ml 2 ) = 8ml 2
or I = 8(0.12 kg)(0.25 m) 2 = 0.060 kg ⋅ m2 40. REASONING The time t it takes to completely unwind the hose from the reel is related to
the reel’s angular displacement θ and its angular acceleration α by θ = ω0t + 1 α t 2
2
(Equation 8.7). The reel is initially at rest, so ω0 = 0 rad/s. Substituting this into θ = ω0t + 1 α t 2 , and solving for the elapsed time t, we obtain
2 464 ROTATIONAL DYNAMICS θ = ( 0 rad/s ) t + 1 α t 2 = 1 α t 2
2
2 or t= 2θ (1) α The hose unwinds from the reel without slipping, so the total arc length s traversed by a
point on the reel’s rim is equal to the total length L of the hose. The reel’s angular
displacement θ, therefore, is related to the length of the hose and the radius R of the reel by
s = L = Rθ (Equation 8.1). Thus, the reel’s angular displacement is given by θ = L/R, so
that Equation (1) becomes L
2 R
t= (2) α The angular acceleration α of the reel depends upon the net torque and the reel’s moment
inertia I via α = Στ I (Equation 9.7).
SOLUTION Assuming that the hose unwinds in the counterclockwise direction, the torque
exerted on the reel by the tension in the hose is positive. Using Equation 9.1, we have
τ = T R , where T is the magnitude of the tension in the hose and R is the radius of the reel.
This torque is opposed by the frictional torque τf , so the net torque on the reel is
Στ = TR − τ f . Thus, the angular acceleration of the reel is α= Στ TR − τ f
=
I
I (9.7) Substituting Equation 9.7 into Equation (2) now yields the elapsed time: t= L
2 R =
TR − τ f
I 2 LI
=
R (TR − τ f ) ( 2 (15.0 m ) 0.44 kg ⋅ m 2 ) ( 0.160 m ) ( 25.0 N )( 0.160 m ) − 3.40 N ⋅ m = 12 s 41. SSM REASONING The drawing shows the two identical sheets and the axis of rotation
for each.
Axis of Rotation L1 = 0.40 m
L2 = 0.20 m Chapter 9 Problems 465 The time t it takes for each sheet to reach its final angular velocity depends on the angular
acceleration α of the sheet. This relation is given by Equation 8.4 as t = (ω − ω0 ) / α , where ω and ω0 are the final and initial angular velocities, respectively. We know that ω0 = 0 rad/s
in each case and that the final angular velocities are the same. The angular acceleration can
be determined by using Newton’s second law for rotational motion, Equation 9.7, as
α = τ / I , where τ is the torque applied to a sheet and I is its moment of inertia.
SOLUTION Substituting the relation α = τ / I into t = (ω − ω0 ) / α gives t= (ω − ω0 ) = (ω − ω0 ) = I (ω − ω0 )
α τ I τ The time it takes for each sheet to reach its final angular velocity is: tLeft = I Left (ω − ω0 ) τ and tRight = I Right (ω − ω0 ) τ The moments of inertia I of the left and right sheets about the axes of rotation are given by
the following relations, where M is the mass of each sheet (see Table 9.1 and the drawings
2
above): I Left = 1 ML1 and I Right = 1 ML2 . Note that the variables M, ω, ω0, and τ are the
2
3
3
same for both sheets. Dividing the timeexpression for the right sheet by that for the left
sheet gives
I Right (ω − ω0 )
tRight
I Right 1 ML2 L2
2
τ
2
=
=
=3
=2
1 ML2
tLeft
I Left
I Left (ω − ω0 )
L1
1
3
τ
Solving this expression for tRight yields
tRight 2 L2 2 = 8.0 s ( 0.20 m ) = 2.0 s
= tLeft 2 (
)
L ( 0.40 m )2 1 The final angular speed of the arm is ω = vT/r, where
r = 0.28 m. The angular acceleration needed to produce this angular speed is α = (ω − ω0)/t. 42. REASONING AND SOLUTION The net torque required is Στ = Iα. This torque is due solely to the force M, so that
Στ = ML. Thus, 466 ROTATIONAL DYNAMICS ω − ω0 I t
∑τ M=
=
L
L Setting ω0 = 0 rad/s and ω = vT/r, the force becomes
v I T 2 r t = I vT = 0.065 kg ⋅ m ( 5.0 m/s ) = 460 N
M=
L
Lrt ( 0.025 m )( 0.28 m )( 0.10 s ) ( ) 43. REASONING
2
a. The moment of i...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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