Chapter 6, Problem 26SGA

To determine

To find:The angular velocity if $3.6$ revolutions are completed in $2$ minutes and round it to the nearest tenth.

Answer to Problem 26SGA

The angular velocity if $3.6$ revolutions are completed in $2$ minutes rounded to nearest tenth $0.2\text{radians}/\sec$ .

Explanation of Solution

Given information:

The number of revolutions is $3.6$ and the time is $2$ minutes.

Formula used: The angular velocity is the quotient of angular displacement to the time is:

  $\omega =\frac{\theta }{t}$

Calculation:

The angular displacement for $n$ revolutions is given by,

  $\theta =2\pi n\text{radians}$

Substitute $3.6$ for $n$ in the formula for angular displacement.

  $\begin{array}{c}\theta =2\pi n\\ =2\pi \left(3.6\right)\\ =7.2\times 3.14\\ =22.608\text{radians}\end{array}$

Convert the time taken into seconds as:

  $2\times 60\text{seconds}=120\text{seconds}$

Substitute $22.608$ for $\theta$ and $120$ for $t$ in the angular velocity formula.

  $\begin{array}{c}\omega =\frac{\theta }{t}\\ =\frac{22.608}{120}\\ =0.1884\text{radians}/\sec \end{array}$

The answer rounded to nearest tenth is equal to $0.2\text{radians}/\sec$ .

Therefore, the angular velocity if $3.6$ revolutions are completed in $2$ minutes rounded to nearest tenth is $0.2\text{radians}/\sec$ .