Respuesta :
Answer:
Explanation:
Given
initial angular velocity is [tex]\omega _0[/tex]
when motor is turned of it started decelerating with [tex]-\alpha [/tex] angular acceleration
angle turned before coming to halt
[tex]\omega ^2-\omega _0^2=2(\alpha )\theta [/tex]
here final angular velocity is zero
[tex]-\omega _0^2=-2\alpha \theta [/tex]
[tex]\theta =\frac{\omega _0^2}{2\alpha }[/tex]
No of revolutions will be [tex]N=\frac{\theta }{2\pi }[/tex]
[tex]N=\frac{\omega _0^2}{4\pi \alpha }[/tex]
The expression for the number of revolutions that the grinding wheel will rotate through before coming to rest is [tex]n = \frac{\omega_{o}^{2}}{4\pi\cdot \alpha}[/tex].
How to derive an expression for the deceleration of the grinding wheel
In this question we must assume that the grinding wheel decelerates unifomrly in time and a revolution is equivalent to 2π radians. The angular accelaration has the following formula:
[tex]\alpha (t) = -\alpha[/tex] (1)
Where α is the angular acceleration, in radians per square second.
The change in angular position ([tex]\Delta \theta[/tex]) is found by integrating (1) twice:
[tex]\Delta \theta = \omega_{o}\cdot t - \frac{1}{2}\cdot \alpha\cdot t^{2}[/tex] (2)
Where:
- t - Time, in seconds
- [tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
And the number of revolutions done by the grinding wheel is:
[tex]n = \frac{\omega_{o}\cdot t -\frac{1}{2}\cdot \alpha\cdot t^{2} }{2\pi}[/tex] (3)
The time needed for the grinding wheel is found by the equation of the angular velocity:
[tex]t = \frac{\omega_{o}}{\alpha}[/tex] (4)
And we replace the time by (4) in (3):
[tex]n = \frac{\frac{\omega_{o}^{2}}{\alpha} -\frac{1}{2}\cdot \alpha\cdot \left(\frac{\omega_{o}}{\alpha} \right)^{2}}{2\pi}[/tex]
[tex]n = \frac{\omega_{o}^{2}}{2\pi\cdot \alpha} -\frac{\omega_{o}^{2}}{4\pi\cdot \alpha}[/tex]
[tex]n = \frac{\omega_{o}^{2}}{4\pi\cdot \alpha}[/tex]
The expression for the number of revolutions that the grinding wheel will rotate through before coming to rest is [tex]n = \frac{\omega_{o}^{2}}{4\pi\cdot \alpha}[/tex]. [tex]\blacksquare[/tex]
To learn more on rotational motion, we kindly invite to check this verified question: https://brainly.com/question/15120445
