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Unformatted text preview: nertia for the threeball system is I = m1r12 + m2 r2 + m3r32 (Equation 9.6),
where m1, m2, and m3 are the masses of the balls and r1, r2, and r3 are the distances from the
axis. In system A, the ball whose mass is m1 does not contribute to the moment of inertia,
because the ball is located on the axis and r1 = 0 m. In system B, the ball whose mass is m3
does not contribute to the moment of inertia, because it is located on the axis and r3 = 0 m.
b. The magnitude of the torque is equal to the magnitude F of the force times the lever
arm l (see Equation 9.1). In system A the lever arm is l = 3.00 m. In B the lever arm is
l = 0 m, since the line of action of the force passes through the axis of rotation.
c. According to Newton’s second law for rotational motion, Equation 9.7, the angular
acceleration α is given by α = ( Στ ) / I , where Στ is the net torque and I is the moment of
inertia. The angular velocity ω is given by Equation 8.4 as ω = ω0 + α t, where ω0 is the
initial angular velocity and t is the time. SOLUTION
a. The moment of inertia for each system is
System A I = m1r12 + m2 r22 + m3r32
= ( 9.00 kg )( 0 m ) + ( 6.00 kg )( 3.00 m ) + ( 7.00 kg )( 5.00 m ) = 229 kg ⋅ m 2
2 System B 2 2 I = m1r12 + m2 r22 + m3r32
= ( 9.00 kg )( 5.00 m ) + ( 6.00 kg )( 4.00 m ) + ( 7.00 kg )( 0 m ) = 321 kg ⋅ m 2
2 2 2 Chapter 9 Problems 467 b. A torque that tends to produce a counterclockwise rotation about the axis is a positive
torque. The torque produced by the force has a magnitude that is equal to the product of the
force magnitude and the lever arm:
System A τ = − F l = − ( 424 N ) ( 3.00 m ) = −1270 N ⋅ m The torque is negative because it produces a clockwise rotation about the axis.
System B τ = F l = ( 424 N ) ( 0 m ) = 0 N ⋅ m c. The final angular velocity ω is related to the initial angular velocity ω0, the angular acceleration α, and the time t by ω = ω0 + α t (Equation 8.4). The angular acceleration is given by Newton’s second law for rotational motion as α = ( Στ ) / I (Equation 9.7), where Στ is the net torque and I is the moment of inertia. Since there is only one torque acting on
each system, it is the net torque, so Στ = τ . Substituting this expression for α into
Equation 8.4 yields τ ω = ω0 + α t = ω0 + t
I In both cases the initial angular velocity is ω0 = 0 rad/s, since the systems start from rest.
The final angular velocities after 5.00 s are: System A −1270 N ⋅ m τ ω = ω0 + t = ( 0 rad/s ) + ( 5.00 s ) = −27.7 rad/s
2
I System B τ ω = ω0 + t = ( 0 rad/s ) + I 229 kg ⋅ m 0 N⋅m ( 5.00 s ) = 0 rad/s
321 kg ⋅ m 2 468 ROTATIONAL DYNAMICS 44. REASONING To determine the angular
acceleration of the pulley and the tension in the
cord attached to the block, we will apply
Newton’s second law to the pulley and the
block separately. Only one external forces acts
on the pulley, as its freebody diagram at the
right shows. This force T is due to the tension
in the cord. The torque that results from this
force is the net torque acting on the pulley and
obeys Newton’s second law for rotational
motion (Equation 9.7). Two external forces act
on the block, as its freebody diagram at the
right indicates. These are (1) the force T due to
the tension in the cord and (2) the weight mg of
the block. The net force that results from these
forces obeys Newton’s second law for
translational motion (Equation 4.2b). R +y
+τ T +x
Freebody diagram for pulley T′ mg
Freebody diagram for block SOLUTION Applying Newton’s second law for rotational motion to the pulley gives
Σ τ = +T R = I α (1) where we have written the torque as the magnitude of the tension force times the lever arm
(the radius) as specified by Equation 9.1. In addition, we have assigned this torque a
positive value, since it causes a counterclockwise rotation of the pulley. To obtain a value
for T, we note that the tension has the same magnitude everywhere in the massless cord, so
that T = T ′ . Thus, by applying Newton’s second law for translation (Equation 4.2b), we
obtain
ΣF = T ′ − mg = −ma or T = T ′ = mg − ma
where we have used a to denote the magnitude of the vertical acceleration of the block and
included the minus sign to account for the fact that the block is accelerating downward.
Substituting this result for T into Equation (1) gives ( mg − ma ) R = Iα (2) To proceed further, we must deal with a. Note that the pulley rolls without slipping against
the cord, so a and α are related according to a = Rα (Equation 8.13). With this
substitution, Equation (2) becomes mg − m ( Rα ) R = Iα Solving for α, we find that or mgR = mR 2α + Iα Chapter 9 Problems ( 469 ) ( 2.0 kg ) 9.80 m/s2 ( 0.040 m )
mgR
α=
=
= 180 rad/s 2
2
2
2
−3
mR + I ( 2.0 kg )( 0.040 m ) + 1.1×10 kg ⋅ m
The value for the tension can be now obtained by substituting this value for α into
Equation (1):
TR = Iα or ( )( ) 1.1×10−3 kg ⋅ m 2 180 rad/s 2
Iα
T=
=
= 5.0 N
R
0.040 m 45. SSM REASONING The angular acceler...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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