Answer:
0.32 s
Explanation:
Initial angular speed: [tex]\omega_i = 1.50 \cdot 10^4 rad/s[/tex]
Final angular speed: [tex]\omega_f = 3.35\cdot 10^4 rad/s[/tex]
Angular rotation: [tex]\theta=2.02\cdot 10^4 rad[/tex]
The angular acceleration of the drill can be found by using the equation:
[tex]\omega_f^2 - \omega_i^2 = 2 \alpha \theta[/tex]
Re-arranging it, we find [tex]\alpha[/tex], the angular acceleration:
[tex]\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}=\frac{(3.35\cdot 10^4 rad/s)^2-(1.50\cdot 10^4 rad/s)^2}{2(2.02\cdot 10^4 rad)}=22,209 rad/s^2[/tex]
Now we want to know the time t the drill takes to accelerate from
[tex]\omega_i =0[/tex]
to
[tex]\omega_f = 6.90\cdot 10^4 rad/s[/tex]
This can be found by using the equation
[tex]\omega_f = \omega_i + \alpha t[/tex]
where [tex]\alpha[/tex] is the angular acceleration we found previously. Solving for t,
[tex]t=\frac{\omega_f - \omega_i}{\alpha}=\frac{22,209 rad/s^2}{6.90\cdot 10^4 rad/s}=0.32 s[/tex]
