Respuesta :
Answer:
Part a)
[tex]\omega = 8.17 rad/s[/tex]
Part b)
[tex]N = 7.74 rev[/tex]
Part c)
[tex]\alpha = 0.69 rad/s^2[/tex]
Part d)
[tex]\alpha = 0.48 rad/s^2[/tex]
Part e)
[tex]t = 9.14 s[/tex]
Explanation:
Part a)
Angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(\frac{78}{60})[/tex]
[tex]\omega = 8.17 rad/s[/tex]
Part b)
Since turn table is accelerating uniformly
so we will have
[tex]\theta = \frac{\omega_f + \omega_i}{2} t[/tex]
[tex]\theta = \frac{8.17 + 0}{2}(11.9)[/tex]
[tex]2N\pi = 48.6[/tex]
[tex]N = 7.74 rev[/tex]
Part c)
angular acceleration is given as
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
[tex]\alpha = \frac{8.17 - 0}{11.9}[/tex]
[tex]\alpha = 0.69 rad/s^2[/tex]
Part d)
When its angular speed changes to 120 rpm
then we will have
[tex]\omega_2 = 2\pi (\frac{120}{60})[/tex]
[tex]\omega_2 = 12.56 rad/s[/tex]
number of turns revolved is 15 times
so we have
[tex]\omega_f^2 - \omega_i^2 = 2 \alpha \theta[/tex]
[tex]12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)[/tex]
[tex]\alpha = 0.48 rad/s^2[/tex]
Part e)
now for uniform acceleration we have
[tex]\omega_f - \omega_i = \alpha t[/tex]
[tex]12.56 - 8.17 = 0.48 t[/tex]
[tex]t = 9.14 s[/tex]
This question involves the concept of the equations of motion for angular motion.
(a) Angular speed of the turntable is "8.17 rad/s" for case 1.
(b) The turntable makes "7.8 revs" revolutions in case 1.
(c) The magnitude of the acceleration of the turntable is "0.69 rad/s²" in case 1.
(d) The magnitude of the angular acceleration of the turntable is "1.32 rad/s²" in case 2.
(e) It takes "3.33 s" for the turntable to go from 78 rpm to 120 rpm in case 2.
(a)
We will simply convert the angular frequency to angular speed by the following formula:
[tex]\omega=2\pi f[/tex]
where.
ω = angular speed = ?
f = angular frequency = 78 rev/min
Therefore,
[tex]\omega=(78\ rev/min)(\frac{2\pi\ rad}{1\ rev})(\frac{1\ min}{60\ s})[/tex]
ω = 8.17 rad/s
(c)
We will use the first equation of motion for the angular motion to find the angular acceleration in case 1:
[tex]\omega_f=\omega_i+\alpha_1 t_1[/tex]
where,
[tex]\omega_f[/tex] = final angular speed = 8.17 rad/s
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
[tex]\alpha_1[/tex] = angular acceleration in case 1= ?
t₁ = time taken in case 1 = 11.9 s
Therefore,
[tex]8.17\ rad/s=0\ rad/s+\alpha_1 (11.9\ s)\\\\\alpha_1=\frac{8.17\ rad/s}{11.9\ s}[/tex]
α₁ = 0.69 rad/s²
(b)
We will use the second equation of motion for the angular motion to find the no. of revolutions in case 1:
[tex]\theta_1=\omega_it_1+\frac{1}{2}\alpha_1 t_1^2[/tex]
where,
θ₁ = no. of revolutions =?
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
[tex]\alpha_1[/tex] = angular acceleration in case 1= 0.69 rad/s²
t₁ = time taken in case 1 = 11.9 s
Therefore,
[tex]\theta_1 = (0\ rev/s)(11.9\ s)+\frac{1}{2}(0.69\ rad/s^2)(11.9\ s)^2[/tex]
[tex]\theta_1 = (48.85\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ₁ = 7.8 revs
(d)
We will use the third equation of motion for the angular motion to find the angular acceleration in case 2:
[tex]2\alpha_2 \theta_2 = \omega_f^2-\omega_i^2[/tex]
where,
[tex]\omega_f[/tex] = final angular speed = [tex](120\ rpm)(\frac{2\pi\ rad}{1\ rev})(\frac{1\ min}{60\ s}) = 12.57\ rad/s[/tex]
[tex]\omega_i[/tex] = initial angular speed = 8.17 rad/s
[tex]\alpha_2[/tex] = angular acceleration in case 2 = ?
[tex]\theta_2[/tex] = no. of revolutions in case 2 = [tex](15\ rev)(\frac{2\pi\ rad}{1\ rev}) = 94.25\ rad[/tex]
Therefore,
[tex]2\alpha_2(94.25\ rad) = (12.57\ rad/s)^2-(8.17\ rad/s)^2\\\\\alpha_2=\frac{(12.57\ rad/s)^2-(8.17\ rad/s)^2}{2(94.25\ rad)}[/tex]
α₂ = 1.32 rad/s²
(e)
We will use the first equation of motion for the angular motion to find the time taken in case 2:
[tex]\omega_f=\omega_i+\alpha_2 t_2[/tex]
where,
[tex]\omega_f[/tex] = final angular speed = 12.57 rad/s
[tex]\omega_i[/tex] = initial angular speed = 8.17 rad/s
t₂ = time taken in case 1 = ?
Therefore,
[tex]12.57\ rad/s=8.17\ rad/s+()t_2\\\\t_2=\frac{12.57\ rad/s-8.17\ rad/s}{1.32\ rad/s^2}[/tex]
t₂ = 3.33 s
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
