HW5 solutions - HomeWork 5 solutions Physics 203 1 Chapter...

HomeWork 5 solutions Physics 203
1. Chapter 5 problem 12. “Centripetal acceleration” is another name for “radial component of acceleration”. It’s better
to use the latter term since it’s much more descriptive of what it means.
REASONING
AND
SOLUTION
a. At the equator a person travels in a circle whose radius equals the radius of the earth,
r = R
e
= 6.38 × 10
6
m, and whose period of rotation is T = 1 day = 86 400 s. We have
v = 2 π R
e
/ T = 464 m/s
The centripetal acceleration is
( )
2
2
–2 2
c
6
464 m/s
3.37 10 m/s
6.38 10 m
v
a
r
= = = ×
×
b. At 30.0° latitude a person travels in a circle of radius,
r = R
e
cos 30.0° = 5.53 × 10
6
m
Thus,
v = 2 π r / T = 402 m/s and a
c
= v
2
/ r = 2.92 × 10
–2
m/s
2
2. Problem 20
REASONING When the penny is rotating with the disk (and not sliding relative to it), it is the static frictional
force that provides the centripetal force required to keep the penny moving on a circular path. The
magnitude
MAX
s
f
of the maximum static frictional force is given by
MAX
s s N
f F µ =
(Equation 4.7), where
F
N
is the magnitude of the normal force and µ
s
is the coefficient of static friction. Solving this relation for
µ
s
gives
MAX
s
s
N
f
F
µ =
(1)
Since the maximum centripetal force that can act on the penny is the maximum static frictional force, we
have
MAX
c s
F f =
. Since F
c
= mv
2
/ r (Equation 5.3), it follows that
MAX 2
s
/ f mv r =
. Substituting this
expression into Equation (1) yields
2
MAX
s
s
N N
mv
f
r
F F
µ =
(2)
The speed of the penny can be determined from the period T