Respuesta :
Answer:
The answer is "[tex]\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}[/tex]"
Explanation:
[tex]\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}[/tex]
The angular velocity of the wheel is [tex]\omega=\sqrt{\frac{Fd}{mkr^2} }[/tex]
Angular Velocity :
Given that the force applied is F, and the displacement of the string is d.
If the radius of the wheel is r and the wheel has an angular displacement of θ, then:
d = rθ
θ = d/r
now, the angular acceleration is given by:
α = F/mkr
The initial angular velocity will be zero and let the final angular velocity is ω.
According to the third equation of motion:
ω² = 0 + 2αθ
ω² = 2(F/mkr)(d/r)
[tex]\omega=\sqrt{\frac{Fd}{mkr^2} }[/tex]
Learn more about angular velocity:
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