Question

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.34 x 104 rad/s to an angular speed of 4.37 x 104 rad/s. In the process, the bit turns through 3.10 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.43 x 104 rad/s, starting from rest?

Answer 1

Answer:

2.66 s

Explanation:

Given The initial angular speed ($\omega_o$) =  1.34 x 10⁴ rad/s, the final angular speed (ω) = 4.37 x 10⁴ rad/s, the angular acceleration (α), distance (θ) = 3.10 x 10⁴ rad.

Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\\\\Substituting:\\(4.37*10^4)^2=(1.34*10^4)^2+2(3.1*10^4)\alpha\\\\(4.37*10^4)^2-(1.34*10^4)^2=2(3.1*10^4)\alpha\\\\2(3.1*10^4)\alpha=1.73*10^9\\\\\alpha=2.79*10^4\ rad/s^2$

The constant acceleration = 2.79 * 10⁴ rad/s.

Since there is constant acceleration, to reach a final speed of  7.43 x 10⁴ rad/s, we use the formula:

$\omega =\omega_o +\alpha t\\\\\omega_o=0,\omega=7.43*10^4\ rad/s,\alpha=2.79*10^4\ rad/s^2\\\\Substituting:\\\\7.43*10^4=0+2.79*10^4t\\\\7.43*10^4=2.79*10^4t\\\\t=2.66\ s$