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

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aristocles aristocles · Answer 1 · 2019-07-14T05:49:58+02:00

Answer:

Part A)

[tex]\alpha = 745 rad/s^2[/tex]

Part B)

[tex]\theta = 4563.1 rad[/tex]

Explanation:

Drill starts from rest so its initial angular speed will be

[tex]\omega_i = 0[/tex]

now after 3.50 s the final angular speed is given as

[tex]f = 2.49 \times 10^4 rev/min[/tex]

[tex]f = {2.49 \times 10^4}{60} = 415 rev/s[/tex]

so final angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega_f = 2607.5 rad/s[/tex]

now we have angular acceleration given as

[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]

[tex]\alpha = \frac{2607.5 - 0}{3.50}[/tex]

[tex]\alpha = 745 rad/s^2[/tex]

Part b)

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

[tex]\theta = \frac{(\omega_f + \omega_i)}{2}\delta t[/tex]

now we have

[tex]\theta = \frac{(2607.5 + 0)}{2}(3.50)[/tex]

[tex]\theta = 4563.1 rad[/tex]