Respuesta :
Answer:
[tex]\theta=107.2917\ rad[/tex]
Explanation:
Given:
- diameter of the CD, [tex]d=12\ cm=0.12\ m[/tex]
- initial rotational speed of the CD, [tex]\omega_i=0\ rad.s^{-1}[/tex]
- rotational acceleration of CD, [tex]\alpha=1\ rad.s^{-2}[/tex]
- maximum speed of rotation after acceleration, [tex]\omega_m=5\ rad.s^{-1}[/tex]
- time of constant angular velocity, [tex]t_c=15\ s[/tex]
- time taken to stop the rotation after the acceleration is stopped, [tex]t_s=12\ s[/tex]
- time of observation from the start, [tex]t=25\ s[/tex]
Now the time taken to get to the maximum angular velocity:
using equation of motion,
[tex]\omega_m=\omega_i+\alpha.t_a[/tex]
[tex]5=0+1\times t_a[/tex]
[tex]t_a=5\ s[/tex]
Now we observe the time to be in the deceleration phase, by:
[tex]\delta t=t-(t_a+t_c)[/tex]
[tex]\delta t=25-(15+5)[/tex]
[tex]\delta t=5\ s[/tex]
- So the phase after 25 seconds from the start is the phase of deceleration of the disk by 5 seconds.
Now the angular acceleration during the phase:
using eq. of motion,
[tex]\omega_m=\omega_f-\alpha_d.t_s[/tex]
where:
[tex]\omega_f=0\ rad.s^{-1}=[/tex] final angular velocity of the disk during the phase
[tex]\alpha_d=[/tex] acceleration during the deceleration phase
[tex]5=0-\alpha_d\times 12[/tex]
[tex]\alpha_d=-\frac{5}{12} \ rad.s^{-1}[/tex]
Now the speed at the instant of 5 seconds of the deceleration phase:
[tex]\omega_5=\omega_m-\alpha.\delta t[/tex]
[tex]\omega_5=5-\frac{5}{12} \times 5[/tex]
[tex]\omega_5=2.9167\ rad.s^{-1}[/tex]
The angular distance travelled by the disk during this 5 seconds:
[tex]\theta_5=\omega_m\times \delta t+\frac{1}{2} \alpha.\delta t^2[/tex]
[tex]\theta_5=5\times 5+0.5\times (-\frac{5}{12} )\times 5^2[/tex]
[tex]\theta_5=19.7917\ rad[/tex] ....................................(1)
Now the angular distance travelled in the phase of acceleration:
[tex]\theta_a=\omega_i.t_a+\frac{1}{2} .\alpha.t_a^2[/tex]
[tex]\theta_a=0+\frac{1}{2} \times 1\times 5^2[/tex]
[tex]\theta_a=12.5\ rad[/tex] .....................................(2)
Now the angular acceleration during the constant angular speed phase:
[tex]\theta_c=\omega_m\times t_c[/tex]
[tex]\theta_c=5\times 15[/tex]
[tex]\theta_c=75\ rad[/tex] ........................(3)
Now the total angular distance travelled by any point on the disk will be same:
so, from (1), (2) and (3)
[tex]\theta=\theta_c+\theta_a+\theta_5[/tex]
[tex]\theta=75+12.5+19.7917[/tex]
[tex]\theta=107.2917\ rad[/tex]
