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 * 10^4 rad/s`

`\omega_i = 1.10 * 10^4 rad/s`

`\theta = 2.00 * 10^4 rad`

now from above equation

`(3.14 * 10^4)^2 - (1.10 * 10^4)^2 = 2(\alpha)(2 * 10^4)`

`\alpha = 2.16 * 10^4 rad/s^2`

now again by kinematics equation

`\omega_f - \omega_i = \alpha t`

`7.85 * 10^4 - 0 = 2.16 * 10^4 t`

`t = 3.63 s`

Answer 2

3.63 s

Step-by-step 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=(\omega_f^2-\omega_i^2)/(2\theta)=((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 = (\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=(\omega_f-\omega_i)/(\alpha)=(7.85\cdot 10^4 rad/s-0)/(2.16\cdot 10^4 rad/s^2)=3.63 s`