Answer:
The angle is 3.95 rad.
Explanation:
The angle can be calculated as follows:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
Where:
[tex]\omega_{f}[/tex]: is the final angular speed
ω₀: is the initial angular speed = 0 (it starts from rest)
α: is the angular acceleration
θ: is the angle=?
The centripetal acceleration is:
[tex]a_{c} = \omega_{f}^{2}*r[/tex]
And the tangential acceleration is:
[tex] a_{T} = \alpha*r [/tex]
Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:
[tex]a_{c} = 7.9a_{T}[/tex]
[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]
Now, the angle is:
[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]
[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]
Therefore, the angle is 3.95 rad.
I hope it helps you!
The angular distance traveled by the electric drill is 3.95 radians.
The given parameters;
The angular distance traveled by the electric drill is calculated as follows;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;
[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]
Substitute the value of angular acceleration into the first equation;
[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]
Thus, the angular distance traveled by the electric drill is 3.95 radians.
Learn more about angular distance here: https://brainly.com/question/12680957
