p11 028

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28.

(a) The tangential acceleration, using Eq. 11-22, is

a

t

= αr =



14.2 rad/s

2



(2.83 cm) = 40.2 cm/s

2

.

(b) In rad/s, the angular velocity is ω = (2760)(2π/60) = 289, so

a

r

= ω

2

r = (289 rad/s)

2

(0.0283 m) = 2.36

× 10

3

m/s

2

.

(c) The angular displacement is, using Eq. 11-14,

θ =

ω

2

2α

=

289

2

2(14.2)

= 2.94

× 10

3

rad .

Then, using Eq. 11-1, the distance traveled is

s = = (0.0283 m)



2.94

× 10

3

rad



= 83.2 m .


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