Chapter 9.1, Problem 6PP

To determine

Time taken by the blade to stop.

Answer to Problem 6PP

The time taken by the blade to come to stop is $t=18\text{s}$ .

Explanation of Solution

Given:

The diameter of the circular saw is $d=0.25\text{m}$ .

The rotation of the circular saw blade is $f_{\text{i}}=5\times {10}^3\text{rpm}$ .

The moment of inertia of the blade is $I=8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ .

The average torque produced by the friction in the axle is $\tau =2.3\times {10}^{-1}\text{N}\cdot \text{m}$ .

Formula used:

The expression for the initial angular velocity $\left({\omega }_{\text{i}}\right)$ of a rotating blade is

  ${\omega }_{\text{i}}=2\pi f_{\text{i}}$

Here, $f_{\text{i}}$ is the rotation of the rotating blade.

The expression for the final angular velocity $\left({\omega }_{\text{f}}\right)$ of the rotating blade is

  $\begin{array}{l}{\omega }_{\text{f}}={\omega }_{\text{i}}+\alpha t\\ {\omega }_{\text{f}}={\omega }_{\text{i}}+\frac{\tau }{I}t\\ t=\frac{\left({\omega }_{\text{f}}-{\omega }_{\text{i}}\right)I}{\tau }\end{array}$

Here, $\tau$ is the torque, $I$ is the moment of inertia, and $t$ is the time taken.

Calculation:

The initial angular velocity $\left({\omega }_{\text{i}}\right)$ of the rotating blade as follows:

  $\begin{array}{c}{\omega }_{\text{i}}=2\pi f_{\text{i}}\\ =2\pi \left(\frac{5\times {10}^3\text{rev}}{\text{min}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\\ =524\text{rad/s}\end{array}$

The final angular velocity of the rotating blade is ${\omega }_{\text{f}}=0$ as the blade comes to stop.

Take the average torque produced by the friction in the axle as $\tau =-2.3\times {10}^{-1}\text{N}\cdot \text{m}$ (The negative sign indicates the direction of the torque)

Consider the time taken by the blade to stop is $t$ seconds.

The time taken by the blade to stop as follows:

  $\begin{array}{l}t=\frac{\left(0-524\text{rad/s}\right)\left(8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\right)}{\left(2.3\times {10}^{-1}\text{N}\cdot \text{m}\right)}\\ t=\frac{\left(0-524\text{rad/s}\right)\left(8\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\right)}{\left(-2.3\times {10}^{-1}\text{N}\cdot \text{m}\times \frac{1\left(\text{kg}\cdot {\text{m/s}}^2\right)}{1\text{N}}\right)}\\ t=\left(\frac{-4.192\text{kg}\cdot {\text{m}}^2/\text{s}}{-2.3\times {10}^{-1}{\text{kgm}}^2{\text{/s}}^2}\right)\\ t=18\text{s}\end{array}$

Conclusion:

Therefore, the time taken by the blade to come to stop is $t=18\text{s}$ .