Preparation problems

12-2. What is the linear speed (a) of a point on Earth's equator? (b) At your latitude?

Solution:

(a) We know that linear velocity is given by

vt = wr
where w is the rotational velocity of the object and r is the distance from the axis of rotation. The rotational velocity of Earth can be found by recalling that Earth revolves once every 24 hours. So we know then that it has traveled 2p rad in that time. So its angular velocity must be given by
w
=
2 p rad
T
=
2 p rad
24 h ·3600 s/h
w
=
7.27 ×10-5  rad/s
(1)
Looking up Earth's radius in Appendix E, we get a linear velocity of
vt
=
wrEarth
=
(7.27 ×10-5  rad/s)(6.37 ×106  m)
vt
=
463  m/s

(b) Again, we know that linear velocity is given by

vt = wrCS
where rCS is the distance between Colorado Springs and the axis of Earth's rotation. We know w from Eq. (1). But now we are not at the same distance from the axis of rotation. Drawing a picture, we see that this radius is given by
rCS = rEarth sin(90-qCS)
where qCS is Colorado Springs' latitude in degrees, which happens to be 38.8°. This gives us
vt
=
wrCS
=
wrEarth sin(90-qCS)
=
(7.27 ×10-5  rad/s)(6.37 ×106  m) sin(90-38.8)
vt
=
361  m/s

12-7. A compact disc (CD) player varies the rotation rate of the disc in order to keep the part of the disk from which information is being read moving at a constant linear speed of 1.30 m/s. Compare the rotation rates of a 12.0-cm-diameter CD when information is being read from (a) its outer edge and (b) a point 3.75 cm from the center. Give your answer in rad/s and rpm.

Solution:

(a) We know that linear velocity is given by

vt = wr
where w is the rotational velocity of the object and r is the distance from the axis of rotation. We can therefore solve for w using our known values:
vt
=
wr
w
=
vt
r
w
=
1.3  m/s
0.06  m
w
=
21.67  rad/s = 207  rpm

(b) Again, we know that linear velocity is given by

vt = wr
Using our known values, we again can solve for w, giving us
vt
=
wr
w
=
vt
r
w
=
1.3  m/s
0.0375  m
w
=
34.67  rad/s = 331  rpm

12-18. A car tune-up manual calls for tightening the spark plugs to a torque of 35.0 N·m. To achieve this torque, with what force must you pull on theend of a 24.0-cm-long wrench if you pull (a) at right angles to the wrench shaft and (b) at 110° to the wrench shaft.

Solution:

(a) We can use our simple definition for the magnitude of a torque, or

ê
ê 		®
t
 
 ê
ê 		= ê
ê 		®
r
 
 × ®
F
 
 ê
ê 		= rFsinq
where r is the distance to the application of the force, F is the magnitude of the force, and q is the angle between the two (in this case the angle to the shaft of the wrench). Solving for F, we get
F = t
rsinq
(1)
Putting in our values, we get
F
=
t
rsinq
=
35.0  N ·m
(0.24  m) sin(90°)
F
=
146  N

(b) Referring to Equation (1), we can quickly solve this problem for the force required. Plugging in our known values, we get

F
=
t
rsinq
=
35.0  N ·m
(0.24  m) sin(110°)
F
=
155  N

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Solutions translated from TEX by TTH, version 1.57.