Respuesta :
Answer:
a)[tex]\omega_1=8.168\,rad.s^{-1}[/tex]
b)[tex]n_1=7.735 \,rev[/tex]
c)[tex]\alpha_1 =0.6864\,rad.s^{-2}[/tex]
d)[tex]\alpha_2=4.1454\,rad.s^{-2}[/tex]
e)[tex]t_2=1.061\,s[/tex]
Explanation:
Given that:
- initial speed of turntable, [tex]N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}[/tex]
- full speed of rotation, [tex]N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}[/tex]
- time taken to reach full speed from rest, [tex]t_1=11.9\,s[/tex]
- final speed after the change, [tex]N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}[/tex]
- no. of revolutions made to reach the new final speed, [tex]n_2=11\,rev[/tex]
(a)
∵ 1 rev = 2π radians
∴ angular speed ω:
[tex]\omega=\frac{2\pi.N}{60}\, rad.s^{-1}[/tex]
where N = angular speed in rpm.
putting the respective values from case 1 we've
[tex]\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}[/tex]
[tex]\omega_1=8.168\,rad.s^{-1}[/tex]
(c)
using the equation of motion:
[tex]\omega_1=\omega_0+\alpha . t_1[/tex]
here α is the angular acceleration
[tex]78=0+\alpha_1\times 11.9[/tex]
[tex]\alpha_1 = \frac{8.168 }{11.9}[/tex]
[tex]\alpha_1 =0.6864\,rad.s^{-2}[/tex]
(b)
using the equation of motion:
[tex]\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1[/tex]
[tex]8.168^2=0^2+2\times 0.6864\times n_1[/tex]
[tex]n_1=48.6003\,rad[/tex]
[tex]n_1=\frac{48.6003}{2\pi}[/tex]
[tex]n_1=7.735\, rev[/tex]
(d)
using equation of motion:
[tex]\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2[/tex]
[tex]12.5664^2=8.168^2+2\alpha_2\times 11[/tex]
[tex]\alpha_2=4.1454\,rad.s^{-2}[/tex]
(e)
using the equation of motion:
[tex]\omega_2=\omega_1+\alpha_2 . t_2[/tex]
[tex]12.5664=8.168+4.1454\times t_2[/tex]
[tex]t_2=1.061\,s[/tex]
