Which expression can be used to calculate centripetal acceleration? 4p2r/T2
4p2r/T
2pr/T
(2pr)2/T2

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 2a.

Strategy

Because v and r are given, the first expression in

a

c

=

v

2

r

;

a

c

=

r

ω

2

a

c

=

v

2

r

;

a

c

=

r

ω

2

is the most convenient to use.

Solution

Entering the given values of v = 25.0 m/s and r=500 m into the first expression for ac gives

a

c

=

v

2

r

=

(

25.0

m/s

)

2

500

m

=

1.25

m/s

2

a

c

=

v

2

r

=

(

25.0

m/s

)

2

500

m

=

1.25

m/s

2

.

Discussion

To compare this with the acceleration due to gravity (g = 9.80 m/s2), we take the ratio of

a

c

g

=

(

1.25

m/s

2

)

(

9.80

m/s

2

)

=

0.128

a

c

g

=

(

1.25

m/s

2

)

(

9.80

m/s

2

)

=

0.128

. Thus, ac=0.128 g and is noticeable especially if you were not wearing a seat belt.

Explanation:

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explanation:

the closer the planet is to the sun, the larger the gravity