Angular displacement of the cd is 3.25 rev
Explanation:
The question is incomplete. It is not given what is the maximum angular velocity of the cd.
Here we are going to assume that the maximum angular velocity is:
[tex]\omega = 30 rad/s[/tex]
The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:
[tex]\theta = (\frac{\omega_0 + \omega}{2})t[/tex]
where:
[tex]\theta[/tex] is the angular displacement of the cd during the time interval t
[tex]\omega_0[/tex] is the initial angular velocity of the cd
[tex]\omega[/tex] is the final angular velocity
Here we have:
t = 1.36 s
[tex]\omega_0 = 0[/tex] (assuming the cd starts from rest)
Therefore, the angular displacement of the cd during this time is:
[tex]\theta=(\frac{0+30}{2})(1.36)=20.4 rad[/tex]
And since [tex]1 rev = 2 \pi rad[/tex], we can convert into number of revolutions completed:
[tex]\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev[/tex]
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