Answer:
[tex]\omega_f = 3.584\ rad/s[/tex]
Explanation:
given,
turntable rotate to, θ = 5 rad
time, t = 2.8 s
initial angular speed = 0 rad/s
final angular speed = ?
now, using equation of rotational motion
[tex]\theta = \omega_i t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2[/tex]
[tex]\alpha= \dfrac{10}{2.8^2}[/tex]
α = 1.28 rad/s²
now, calculation of angular velocity
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]\omega_f =0 +1.28\times 2.8[/tex]
[tex]\omega_f = 3.584\ rad/s[/tex]
hence, the angular velocity at the end is equal to 3.584 rad/s
The angular velocity at the end of that time 10rad/s
In order to get the angular velocity [tex]\omega[/tex], we will use the equation of motion expressed as [tex]\omega = \omega_0 + \alpha t[/tex]
[tex]\alpha[/tex] is the angular acceleration
t is the time taken
[tex]\omega_0[/tex] is the initial angular velocity
Get the angular acceleration [tex]\alpha[/tex]
[tex]\theta = \omega_0 t+ \frac{1}{2} gt^2[/tex]
[tex]5=0+\frac{1}{2} \alpha (2.8)\\2\times5 = 2.8 \alpha \\10 = 2.8 \alpha\\\alpha = \frac{10}{2.8}\\\alpha = 3.57rad/s^2[/tex]
Get the angular velocity [tex]\omega[/tex]
[tex]\omega = \omega_0 + 3.57(2.8)\\\omega = 0+10\\\omega =10rad/s\\[/tex]
Hence the angular velocity at the end of that time 10rad/s
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