Question

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.50 × 10 4 1.50×104 rad/s to an angular speed of 3.35 × 10 4 3.35×104 rad/s. In the process, the bit turns through 2.02 × 10 4 2.02×104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 6.90 × 10 4 6.90×104 rad/s, starting from rest?

Answer 1

Answer:

0.32 s

Explanation:

Initial angular speed: $\omega_i = 1.50 \cdot 10^4 rad/s$

Final angular speed: $\omega_f = 3.35\cdot 10^4 rad/s$

Angular rotation: $\theta=2.02\cdot 10^4 rad$

The angular acceleration of the drill can be found by using the equation:

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

Re-arranging it, we find $\alpha$, the angular acceleration:

$\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}=\frac{(3.35\cdot 10^4 rad/s)^2-(1.50\cdot 10^4 rad/s)^2}{2(2.02\cdot 10^4 rad)}=22,209 rad/s^2$

Now we want to know the time t the drill takes to accelerate from

$\omega_i =0$

to

$\omega_f = 6.90\cdot 10^4 rad/s$

This can be found by using the equation

$\omega_f = \omega_i + \alpha t$

where $\alpha$ is the angular acceleration we found previously. Solving for t,

$t=\frac{\omega_f - \omega_i}{\alpha}=\frac{22,209 rad/s^2}{6.90\cdot 10^4 rad/s}=0.32 s$