Question

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.38 x 104 rad/s to an angular speed of 5.05 x 104 rad/s. In the process, the bit turns through 1.53 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.28 x 104 rad/s, starting from rest?

Answer 1

Let us first find the angular acceleration. Two equations for angular displacement and angular velocity are:

$\phi=\nu_{0}t+\epsilon t^{2}/2$

$\nu_{f}=\nu_{0}+\epsilon t$

where $\nu_{0}$ is initial angular speed and $\epsilon$ is angular acceleration From where we can find that

$\phi=\nu_{0}t+(\nu_{f}-\nu_{0})t/2=t/2(\nu_{0}+\nu_{f})$

And we can find time of acceleration

$t=\frac{\phi}{1/2(\nu_{0}+\nu_{f})}$

then we can find acceleration:

$\epsilon=\frac{\nu_{f}-\nu_{0}}{t}=\frac{\nu_{f}-\nu_{0}}{\frac{\phi}{1/2(\nu_{0}+\nu_{f})}}$

And, finally, the time, needed to reach $\nu_{2}=7.28\cdot 10^{4}$ rad/s speed from rest

$t=\frac{\nu_{2}}{\epsilon}=\frac{\nu_{2}\phi}{(\nu_{f}-\nu_{0})(1/2(\nu_{0}+\nu_{f}))}=$

$=\frac{7.28\cdot 10^{4}\cdot 1.53\cdot 10^{4}}{(5.05\cdot 10^{4}-1.38\cdot 10^{4})(1/2\cdot 1.38\cdot 10^{4}+1/2\cdot 5.05\cdot 10^{4})}\approx 0.944\,s$