Question

A dentist's drill starts from rest. After 3.50 s of constant angular acceleration, it turns at a rate of 2.49 104 rev/min. (a) Find the drill's angular acceleration. Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. rad/s2 (along the axis of rotation) (b) Determine the angle through which the drill rotates during this period. rad

Answer 1

Part A)

`\alpha = 745 rad/s^2`

Part B)

`\theta = 4563.1 rad`

Step-by-step explanation:

Drill starts from rest so its initial angular speed will be

`\omega_i = 0`

now after 3.50 s the final angular speed is given as

`f = 2.49 * 10^4 rev/min`

`f = {2.49 * 10^4}{60} = 415 rev/s`

so final angular speed is given as

`\omega = 2\pi f`

`\omega_f = 2607.5 rad/s`

now we have angular acceleration given as

`\alpha = (\omega_f - \omega_i)/(\Delta t)`

`\alpha = (2607.5 - 0)/(3.50)`

`\alpha = 745 rad/s^2`

Part b)

The angle through which it is rotated is given by the formula

`\theta = ((\omega_f + \omega_i))/(2)\delta t`

now we have

`\theta = ((2607.5 + 0))/(2)(3.50)`

`\theta = 4563.1 rad`