Question
Asked Nov 7, 2019

A compact disk, which has a diameter of 12 cm, speeds up uniformly from 0 to 4 rev/s in 3 seconds. What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is A.) 2 rev/s and B.) 3 rev/s?

check_circleExpert Solution
Step 1

The radius of the disk is,

r
2
12a
2
0.01m
=6am
lan
0.06m
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r 2 12a 2 0.01m =6am lan 0.06m

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Step 2

The angular acceleration of the compact disk is,

Substitute the values,

о, — о
t
2л гаd
4гevis
-Огad/s
Irev
(3s)
25.12 гad/s- Orad/s
3s
—837гad/s?
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о, — о t 2л гаd 4гevis -Огad/s Irev (3s) 25.12 гad/s- Orad/s 3s —837гad/s?

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Step 3

The tangential acceleration of a point on the outer ri...

a,-(8.37ad/s")(0.06m)
0.502m/s
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a,-(8.37ad/s")(0.06m) 0.502m/s

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Angular Motion

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