Answer:
![\theta=107.2917\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D107.2917%5C%20rad)
Explanation:
Given:
- diameter of the CD,
![d=12\ cm=0.12\ m](https://tex.z-dn.net/?f=d%3D12%5C%20cm%3D0.12%5C%20m)
- initial rotational speed of the CD,
![\omega_i=0\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_i%3D0%5C%20rad.s%5E%7B-1%7D)
- rotational acceleration of CD,
![\alpha=1\ rad.s^{-2}](https://tex.z-dn.net/?f=%5Calpha%3D1%5C%20rad.s%5E%7B-2%7D)
- maximum speed of rotation after acceleration,
![\omega_m=5\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_m%3D5%5C%20rad.s%5E%7B-1%7D)
- time of constant angular velocity,
![t_c=15\ s](https://tex.z-dn.net/?f=t_c%3D15%5C%20s)
- time taken to stop the rotation after the acceleration is stopped,
![t_s=12\ s](https://tex.z-dn.net/?f=t_s%3D12%5C%20s)
- time of observation from the start,
![t=25\ s](https://tex.z-dn.net/?f=t%3D25%5C%20s)
<u>Now the time taken to get to the maximum angular velocity:</u>
using equation of motion,
![\omega_m=\omega_i+\alpha.t_a](https://tex.z-dn.net/?f=%5Comega_m%3D%5Comega_i%2B%5Calpha.t_a)
![5=0+1\times t_a](https://tex.z-dn.net/?f=5%3D0%2B1%5Ctimes%20t_a)
![t_a=5\ s](https://tex.z-dn.net/?f=t_a%3D5%5C%20s)
<u>Now we observe the time to be in the deceleration phase, by:</u>
![\delta t=t-(t_a+t_c)](https://tex.z-dn.net/?f=%5Cdelta%20t%3Dt-%28t_a%2Bt_c%29)
![\delta t=25-(15+5)](https://tex.z-dn.net/?f=%5Cdelta%20t%3D25-%2815%2B5%29)
![\delta t=5\ s](https://tex.z-dn.net/?f=%5Cdelta%20t%3D5%5C%20s)
- So the phase after 25 seconds from the start is the phase of deceleration of the disk by 5 seconds.
<u>Now the angular acceleration during the phase:</u>
using eq. of motion,
![\omega_m=\omega_f-\alpha_d.t_s](https://tex.z-dn.net/?f=%5Comega_m%3D%5Comega_f-%5Calpha_d.t_s)
where:
final angular velocity of the disk during the phase
acceleration during the deceleration phase
![5=0-\alpha_d\times 12](https://tex.z-dn.net/?f=5%3D0-%5Calpha_d%5Ctimes%2012)
![\alpha_d=-\frac{5}{12} \ rad.s^{-1}](https://tex.z-dn.net/?f=%5Calpha_d%3D-%5Cfrac%7B5%7D%7B12%7D%20%5C%20rad.s%5E%7B-1%7D)
<u>Now the speed at the instant of 5 seconds of the deceleration phase:</u>
![\omega_5=\omega_m-\alpha.\delta t](https://tex.z-dn.net/?f=%5Comega_5%3D%5Comega_m-%5Calpha.%5Cdelta%20t)
![\omega_5=5-\frac{5}{12} \times 5](https://tex.z-dn.net/?f=%5Comega_5%3D5-%5Cfrac%7B5%7D%7B12%7D%20%5Ctimes%205)
![\omega_5=2.9167\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega_5%3D2.9167%5C%20rad.s%5E%7B-1%7D)
<u>The angular distance travelled by the disk during this 5 seconds:</u>
![\theta_5=\omega_m\times \delta t+\frac{1}{2} \alpha.\delta t^2](https://tex.z-dn.net/?f=%5Ctheta_5%3D%5Comega_m%5Ctimes%20%5Cdelta%20t%2B%5Cfrac%7B1%7D%7B2%7D%20%5Calpha.%5Cdelta%20t%5E2)
![\theta_5=5\times 5+0.5\times (-\frac{5}{12} )\times 5^2](https://tex.z-dn.net/?f=%5Ctheta_5%3D5%5Ctimes%205%2B0.5%5Ctimes%20%28-%5Cfrac%7B5%7D%7B12%7D%20%29%5Ctimes%205%5E2)
....................................(1)
<u>Now the angular distance travelled in the phase of acceleration:</u>
![\theta_a=\omega_i.t_a+\frac{1}{2} .\alpha.t_a^2](https://tex.z-dn.net/?f=%5Ctheta_a%3D%5Comega_i.t_a%2B%5Cfrac%7B1%7D%7B2%7D%20.%5Calpha.t_a%5E2)
![\theta_a=0+\frac{1}{2} \times 1\times 5^2](https://tex.z-dn.net/?f=%5Ctheta_a%3D0%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%201%5Ctimes%205%5E2)
.....................................(2)
Now the angular acceleration during the constant angular speed phase:
![\theta_c=\omega_m\times t_c](https://tex.z-dn.net/?f=%5Ctheta_c%3D%5Comega_m%5Ctimes%20t_c)
![\theta_c=5\times 15](https://tex.z-dn.net/?f=%5Ctheta_c%3D5%5Ctimes%2015)
........................(3)
Now the total angular distance travelled by any point on the disk will be same:
so, from (1), (2) and (3)
![\theta=\theta_c+\theta_a+\theta_5](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctheta_c%2B%5Ctheta_a%2B%5Ctheta_5)
![\theta=75+12.5+19.7917](https://tex.z-dn.net/?f=%5Ctheta%3D75%2B12.5%2B19.7917)
![\theta=107.2917\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D107.2917%5C%20rad)