Unformatted text preview: great as the centripetal acceleration at corner B yields
the relationship between the radial distances: ) ( rA ω 2 = n rB ω 2
1 24
43
123
Centripetal
acceleration
at corner A rA = n rB or (1) Centripetal
acceleration
at corner B The radial distance rB at corner B is the length of the short side of the rectangular plate:
rB = L1. The radial distance rA at corner A is the length of a straight line from the rotation
axis to corner A. This line is the diagonal of the plate, so we obtain rA from the Pythagorean
2
theorem (Equation 1.7): rA = L1 + L2 .
2
2
SOLUTION Making the substitutions rA = L1 + L2 and rB = L1 in Equation (1) gives
2 L2 + L2 = n L1
{
11 3
4
242
rA (2) rB Squaring both sides of Equation (2) and solving for the ratio L1/L2 yields
2
2
n 2 L1 = L1 + L2
2 or ( n2 − 1) L12 = L22 or L1
L2 = 1
n2 − 1 Thus, when n = 2.00, the ratio of the lengths of the sides of the rectangle is Chapter 8 Problems 419 L1
1
1
=
=
= 0.577
L2
3
22 − 1 50. REASONING The centripetal acceleration of the trainee is given by ac = rω 2 (Equation 8.11), where ω is the angular speed (in rad/s) of the centrifuge, and r is the radius
of the circular path the trainee follows. Because this radius is the length of the centrifuge
arm, we will solve Equation 8.11 for r. In the second exercise, the trainee’s total
acceleration gains a tangential component aT = rα (Equation 8.10) due to the angular
acceleration α (in rad/s2) of the centrifuge. The angular speed ω and the length r of the
centrifuge arm are both the same as in the first exercise, so the centripetal component of
acceleration ac = rω 2 is unchanged in the second exercise. The two components of the
trainee’s acceleration are perpendicular and, thus, are related to the trainee’s total
2
2
acceleration a by the Pythagorean theorem: a 2 = ac + aT (Equation 1.7). We will solve the
Pythagorean theorem for the trainee’s tangential acceleration aT, and then use aT = rα to determine the angular acceleration α of the centrifuge. SOLUTION
a. Solving ac = rω 2 (Equation 8.11) yields the length r of the centrifuge arm:
r= ac ω 2 = ( 3.2 9.80 m/s 2 ( 2.5 rad/s ) 2 ) = 5.0 m 2
2
b. Solving a 2 = ac + aT for the tangential component of the trainee’s total acceleration, we 2
obtain aT = a 2 − ac . Then, using aT = rα (Equation 8.10), we find that the angular acceleration of the centrifuge is a
α= T=
r 51. 2
a 2 − ac r = ( ) 2 ( ) 2 ( 4.8 ) 9.80 m/s 2 − ( 3.2 ) 9.80 m/s 2 = 7.0 rad/s 2
5.0 m SSM REASONING
a. The tangential speed vT of the sun as it orbits about the center of the Milky Way is related
to the orbital radius r and angular speed ω by Equation 8.9, vT = rω. Before we use this
relation, however, we must first convert r to meters from lightyears. 420 ROTATIONAL KINEMATICS b. The centripetal force is the net force required to keep an object, such as the sun, moving
on a circular path. According to Newton’s second law of motion, the magnitude Fc of the
centripetal force is equal to the product of the object’s mass m and the magnitude ac of its
centripetal acceleration (see Section 5.3): Fc = mac. The magnitude of the centripetal
acceleration is expressed by Equation 8.11 as ac = rω , where r is the radius of the circular
2 path and ω is the angular speed of the object. SOLUTION
a. The radius of the sun’s orbit about the center of the Milky Way is 9.5 × 1015 m 4
20
r = 2.3 × 10 lightyears = 2.2 × 10 m 1 lightyear ( ) The tangential speed of the sun is vT = rω = ( 2.2 × 1020 m ) (1.1 × 10−15 rad/s ) = 2.4 × 105 m/s (8.9) b. The magnitude of the centripetal force that acts on the sun is Fc = mac = m r ω 2
{
Centripetal
force ( )( )( = 1.99 × 1030 kg 2.2 × 1020 m 1.1 × 10−15 rad /s ) 2 = 5.3 × 1020 N 52. REASONING The tangential acceleration and the centripetal acceleration of a point at a
distance r from the rotation axis are given by Equations 8.10 and 8.11, respectively:
aT = rα and a c = r ω 2 . After the drill has rotated through the angle in question, a c = 2 a T ,
or
r ω 2 = 2 rα This expression can be used to find the angular acceleration α . Once the angular
acceleration is known, Equation 8.8 can be used to find the desired angle.
SOLUTION Solving the expression obtained above for α gives α= ω2
2 Chapter 8 Problems 421 Solving Equation 8.8 for θ (with ω0 = 0 rad/s since the drill starts from rest), and using the
expression above for the angular acceleration α gives ω 2 2 ω2
ω2 = 1.00 rad
θ=
=
=
2α 2(ω 2 / 2) 2 ω 2 Note that since both Equations 8.10 and 8.11 require that the angles be expressed in radians,
the final result for θ is in radians. 53. SSM WWW REASONING AND SOLUTION
acceleration of the motorcycle is
a= v − v0
t = From Equation 2.4, the linear 22.0 m/s − 0 m/s
2
= 2.44 m/s
9.00 s Since the tire rolls without slipping, the linear acceleration equals the tangential acceleration
of a point on the outer edge of the tire: a = aT . Solving Equation 8.13 for α gives α= aT 2.44 m/s 2
=
= 8.71 rad/s2
r
0.280 m 54. REASONING AND SOLUTION The bike would travel with the same speed as a point on
the...
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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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