# A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.65 times...

## Question:

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of {eq}1.65 \times 10^4 \ rad / s {/eq} to an angular speed of {eq}5.25 \times 10^4 \ rad / s {/eq}. In the process, the bit turns through {eq}2.26 \times 10^4 \ rad {/eq}. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of {eq}8.86 \times 10^4 \ rad / s {/eq}, starting from rest?

## Rotational motion

We have to use the concept and equation of the rotational motion to find the solution to this problem. We have to use the rotational kinematics equation to establish the relation between the angular displacement, time and angular velocity

Given

First angular velocity {eq}\displaystyle w_{1} = 1.65*10^{4} \ rad/s {/eq}

Second angular velocity {eq}\displaystyle w_{2} = 5.25*10^{4} \ rad/s {/eq}

Angular displacement {eq}\displaystyle \theta = 2.26*10^{4} \ rad {/eq}

Now using the kinematic equation

{eq}\displaystyle w_{2}^{2} = w_{1}^{2} + 2\alpha \theta \\ (5.25*10^{4})^{2} = (1.65*10^{4})^{2} + 2\alpha*(2.26*10^{4}) \\ \alpha= 54955.75 \ rad/s^{2} {/eq}

Now initial angular velocity {eq}\displaystyle w_{i} = 0 \ rad/s {/eq}

Maximum angular speed {eq}\displaystyle w_{max} = 8.86*10^{4} \ rad/s {/eq}

Now using the kinematic equation

{eq}\displaystyle w_{max} = w_{i} + \alpha*t \\ 8.86*10^{4} = 0 + (54955.75)t \\ t = 1.612 \ s {/eq}