Answer:
3.63 s
Explanation:
We can solve the problem by using the equivalent SUVAT equations for the angular motion.
To find the angular acceleration, we can use the following equation:
[tex]\omega_f^2 - \omega_i ^2 =2 \alpha \theta[/tex]
where
[tex]\omega_f = 3.14\cdot 10^4 rad/s[/tex] is the final angular speed
[tex]\omega_i = 1.10 \cdot 10^4 rad/s[/tex] is the initial angular speed
[tex]\theta= 2.00 \cdot 10^4 rad[/tex] is the angular distance covered
[tex]\alpha[/tex] is the angular acceleration
Re-arranging the formula, we can find [tex]\alpha[/tex]:
[tex]\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2[/tex]
Now we want to know the time the bit takes starting from rest to reach a speed of [tex]\omega_f=7.85\cdot 10^4 rad/s[/tex]. So, we can use the following equation:
[tex]\alpha = \frac{\omega_f-\omega_i}{t}[/tex]
where:
[tex]\alpha=2.16\cdot 10^4 rad/s^2[/tex] is the angular acceleration
[tex]\omega_f = 7.85\cdot 10^4 rad/s[/tex] is the final speed
[tex]\omega_i = 0[/tex] is the initial speed
t is the time
Re-arranging the equation, we can find the time:
[tex]t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s[/tex]
For constant angular acceleration we can use
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]
here we will have
[tex]\omega_f = 3.14 \times 10^4 rad/s[/tex]
[tex]\omega_i = 1.10 \times 10^4 rad/s[/tex]
[tex]\theta = 2.00 \times 10^4 rad[/tex]
now from above equation
[tex](3.14 \times 10^4)^2 - (1.10 \times 10^4)^2 = 2(\alpha)(2 \times 10^4)[/tex]
[tex]\alpha = 2.16 \times 10^4 rad/s^2[/tex]
now again by kinematics equation
[tex]\omega_f - \omega_i = \alpha t[/tex]
[tex]7.85 \times 10^4 - 0 = 2.16 \times 10^4 t[/tex]
[tex]t = 3.63 s[/tex]
