Question

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14 104 rad/s. in the process, the bit turns through 2.00 104 rad. assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.85 104 rad/s, starting from rest?

Answer 1

For constant angular acceleration we can use

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

here we will have

$\omega_f = 3.14 \times 10^4 rad/s$

$\omega_i = 1.10 \times 10^4 rad/s$

$\theta = 2.00 \times 10^4 rad$

now from above equation

$(3.14 \times 10^4)^2 - (1.10 \times 10^4)^2 = 2(\alpha)(2 \times 10^4)$

$\alpha = 2.16 \times 10^4 rad/s^2$

now again by kinematics equation

$\omega_f - \omega_i = \alpha t$

$7.85 \times 10^4 - 0 = 2.16 \times 10^4 t$

$t = 3.63 s$

Answer 2

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:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$:

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$. So, we can use the following equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where:

$\alpha=2.16\cdot 10^4 rad/s^2$ is the angular acceleration

$\omega_f = 7.85\cdot 10^4 rad/s$ is the final speed

$\omega_i = 0$ is the initial speed

t is the time

Re-arranging the equation, we can find the time:

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s$