ucm solutions

rogers (nr9358) UCM volle (301) 1

This print-out should have 9 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 pointsCalculate the speed at the edge of a disc

of radius 7.5 cm that rotates at the rate of6 rev/s.

Correct answer: 2.82743 m/s.

Explanation:

Let : R = 7.5 cm and

f = 6 rev/s .

T =1

f=

1

6 rev/s= 0.166667 s .

v =D

T=

2 piR

T

=2 pi (7.5 cm)

0.166667 s 1 m

100 cm

= 2.82743 m/s .

002 10.0 pointsA race car moves along a circular track at anangular speed of 0.313 rad/s.If the cars centripetal acceleration is 17.8

m/s2, what is the distance between the carand the center of the track?

Correct answer: 181.69 m.

Explanation:Basic Concept:

ac = r2

Given:

= 0.313 rad/s

ac = 17.8 m/s2

Solution:

r =ac2

=17.8 m/s2

(0.313 rad/s)2

= 181.69 m

003 (part 1 of 2) 10.0 pointsA dog sits 2.5 m from the center of a merry-go-round.a) If the dog undergoes a 1.9 m/s2 cen-

tripetal acceleration, what is the dogs linearspeed?



ac =vt

2

r

Given:

ac = 1.9 m/s2

r = 2.5 m

Solution:

vt =acr

=(1.9 m/s2)(2.5 m)

= 2.17945 m/s

004 (part 2 of 2) 10.0 pointsb) What is the angular speed of the merry-go-round?

Correct answer: 0.87178 rad/s.


ac = r2

Given:

ac = 1.9 m/s2

r = 2.5 m

Solution:

=

acr

=

1.9 m/s2

2.5 m= 0.87178 rad/s


005 (part 1 of 2) 10.0 pointsThe orbit of a Moon about its planet is ap-proximately circular, with a mean radius of4.44 108 m. It takes 25.8 days for the Moonto complete one revolution about the planet.Find the mean orbital speed of the Moon.


Explanation:Dividing the length C = 2pir of the trajectoryof the Moon by the time

T = 25.8 days = 2.22912 106 s

of one revolution (in seconds!), we obtain thatthe mean orbital speed of the Moon is

v =C

T=

2 pi r

T

=2 pi (4.44 108 m)2.22912 106 s

= 1251.5 m/s .

006 (part 2 of 2) 10.0 pointsFind the Moons centripetal acceleration.

Correct answer: 0.00352757 m/s2.

Explanation:Since the magnitude of the velocity is con-stant, the tangential acceleration of the Moonis zero. The centripetal acceleration is

ac =v2

r

=(1251.5 m/s )2

4.44 108 m= 0.00352757 m/s2 .

007 (part 1 of 2) 10.0 pointsThe radius of the Earth is about 6.37 106 m.a) What is the centripetal acceleration of a

point on the equator?

Correct answer: 0.0336877 m/s2.

Explanation:

Let : requator = 6.37 106 m = 2 pi rad/day

ac =vt

2

r= r 2

= (6.37 106 m) (2 pi rad/day)2

(1 day

24 h

)2(1 h

3600 s

)2= 0.0336877 m/s2

008 (part 2 of 2) 10.0 pointsb) What is the centripetal acceleration of apoint at the North Pole?

Correct answer: 0 m/s2.

Explanation:The Pole is 0 m from the axis of rotation,

soac = (0 m) (2 pi rad/day)

2

= 0 m/day2

= 0 m/s2

009 10.0 pointsWhat angular speed (in revolutions perminute) is needed for a centrifuge to producean acceleration of 707 times the gravitationalacceleration 9.8 m/s2 at a radius of 10.5 cm ?

Correct answer: 2453.01 rev/min.

Explanation:The centripetal acceleration in the cen-

trifuge,ac =

2 r ,

has to be equal to 707 times the free-fall ac-celeration g = 9.8m/s2. Therefore

707 g = 2 r ,

so

=

acr


=

707 g

r

=

6928.6 m/s2

10.5 cm 1m100 cm

= 256.879 rad/s

=60 s/min

2 pi rad/rev 256.879 rad/s

= 2453.01 rev/min .

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