This preview shows page 1. Sign up to view the full content.
Unformatted text preview: (1), we find that the ratio of the centripetal
acceleration of the tip of the second hand to that of the minute hand is ac, second
ac, minute = 4π 2 r
2
Tsecond 4π 2 r
2
Tminute = 2
Tminute
2
Tsecond ( 3600 s )2
=
( 60 s )2 = 3600 _____________________________________________________________________________________________ 8. 2 REASONING The centripetal acceleration is given by Equation 5.2 as ac = v /r. The value
of the radius r is given, so to determine ac we need information about the speed v. But the speed is related to the period T by v = (2π r)/T, according to Equation 5.1. We can substitute
this expression for the speed into Equation 5.2 and see that 248 DYNAMICS OF UNIFORM CIRCULAR MOTION b 2π r / T
v2
ac =
=
r
r g = 4π
2 T 2 r 2 SOLUTION To use the expression obtained in the reasoning, we need a value for the
period T. The period is the time for one revolution. Since the container is turning at
2.0 revolutions per second, the period is T = (1 s)/(2.0 revolutions) = 0.50 s. Thus, we find
that the centripetal acceleration is bg
bg 2
4 π 2 r 4 π 0.12 m
=
= 19 m / s 2
ac =
2
2
T
0.50 s _____________________________________________________________________________________________ 9. SSM REASONING AND SOLUTION Since the magnitude of the centripetal
2 acceleration is given by Equation 5.2, a C = v / r , we can solve for r and find that
r= v2
( 98.8 m / s ) 2
=
= 332 m
a C 3.00(9.80 m / s 2 ) _____________________________________________________________________________________________ 10. REASONING The magnitude of the centripetal acceleration of any point on the helicopter
blade is given by Equation 5.2, a C = v 2 / r , where r is the radius of the circle on which that
point moves. From Equation 5.1: v = 2 π r / T . Combining these two expressions, we obtain aC = 4π 2 r
T2 All points on the blade move with the same period T.
SOLUTION The ratio of the centripetal acceleration at the end of the blade (point 1) to that
which exists at a point located 3.0 m from the center of the circle (point 2) is
a C1
a C2 = 4 π 2 r1 / T 2
4 π r2 / T
2 2 = r1
r2 = 6.7 m
= 2.2
3.0 m _____________________________________________________________________________________________ 11. REASONING AND SOLUTION
T = 2π r/v. The speed is The sample makes one revolution in time T as given by 2
–2
3
2
v = rac = (5.00 × 10 m)(6.25 × 10 )(9.80 m/s ) so that v = 55.3 m/s Chapter 5 Problems 249 The period is
T = 2π (5.00 × 10 –2 m)/(55.3 m/s) = 5.68 × 10 –3 s = 9.47 × 10 –5 min The number of revolutions per minute = 1/T = 10 600 rev/min .
_____________________________________________________________________________________________ 12. REASONING AND SOLUTION
a. At the equator a person travels in a circle whose radius equals the radius of the earth,
6 r = Re = 6.38 × 10 m, and whose period of rotation is T = 1 day = 86 400 s. We have
v = 2πRe/T = 464 m/s
The centripetal acceleration is v 2 ( 464 m/s )
ac = =
= 3.37 × 10–2 m/s 2
r 6.38 × 106 m
2 b. At 30.0° latitude a person travels in a circle of radius,
6 r = Re cos 30.0° = 5.53 × 10 m
Thus,
v = 2π r/T = 402 m/s and 2 ac = v /r = 2.92 × 10 –2 m/s 2 _____________________________________________________________________________________________ 13. REASONING The magnitude Fc of the centripetal force that acts on the skater is given by
Equation 5.3 as Fc = mv 2 / r , where m and v are the mass and speed of the skater, and r is
the distance of the skater from the pivot. Since all of these variables are known, we can find
the magnitude of the centripetal force.
SOLUTION The magnitude of the centripetal force is
mv 2 ( 80.0 kg )( 6.80 m/s )
Fc =
=
= 606 N
r
6.10 m
______________________________________________________________________________
2 14. REASONING The centripetal acceleration depends on the speed v and the radius r of the
curve, according to ac = v2/r (Equation 5.2). The speeds of the cars are the same, and since
they are negotiating the same curve, the radius is the same. Therefore, the cars have the
same centripetal acceleration. However, the magnitude Fc of the centripetal force depends
on the mass m of the car, as well as the speed and the radius of the curve, according to 250 DYNAMICS OF UNIFORM CIRCULAR MOTION 2 Fc = mv /r (Equation 5.3). Since the speed and the radius are the same for each car, the car
with the greater mass, which is car B, experiences the greater centripetal acceleration.
SOLUTION We find the following values for the magnitudes of the centripetal
accelerations and forces:
Car A ac =
Fc = Car B b 27 m / s
v2
=
r
120 m
mA v 2
r r 6.1 m / s 2 (5.2) 2 120 m b Fc = 2 1100
b kg g27 m / sg =
b = 27 m / s
v2
ac =
=
r
120 m
mB v 2 g=
g= 6700 N (5.3) 2 6.1 m / s 2 1600
b kg g27 m / sg =
b
= (5.2) 2 120 m 9700 N (5.3) _____________________________________________________________________________________________ 15. SSM REASONING AND SOLUTION The magnitude of the centripetal force on the ball is given by Equation 5.3: FC = mv 2 / r . Solving for v,...
View
Full
Document
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

Click to edit the document details