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

0.32 s

Step-by-step 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 = (\omega_f^2 - \omega_i^2)/(2\theta)=((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=(\omega_f - \omega_i)/(\alpha)=(22,209 rad/s^2)/(6.90\cdot 10^4 rad/s)=0.32 s`