Chapter 12.7, Problem 50E

(a)

To determine

To find:the angular velocity if the sander disk in radian per second.

Answer to Problem 50E

  $\theta =10\pi$ radian per second.

Explanation of Solution

Given:

The disk of a sander has a radius of $6$ inches and is rotating at a speed of $5$ revolutions per second.

Concept used:

One revolution $=2\pi rad$ .

  $\omega =\frac{2\pi }{t}$ .

Angular displacement $=\theta$ .

Angular velocity $=\omega$ .

Calculation:

Consider the radius of disk is $r=6$ inches.

Angular velocity of disk is:

  $\omega =5\text{ revolutions}$ per second.

But $1$ revolutions $=2\pi rad$ .

Angular displacement:

An object at the edge moves through an angle relative to its starting position known as the angular displacement or angle of rotation.

Each revolution equals $2\pi$ radians.

Thus, the angular displacement is

  $\theta =5\times 2\pi =10\pi$ radians per second.

Therefore, the angular velocity of the sander disk is:

  $\theta =10\pi$ radian per second.

Hence, $\theta =10\pi$ radian per second.

(b)

To determine

To find: the linear velocity of a point on the edge of the sander disk in feet per second.

Answer to Problem 50E

The linear velocity is $15.72$ per second.

Explanation of Solution

Given:

The disk of a sander has a radius of $6$ inches and is rotating at a speed of $5$ revolutions per second.

Concept used:

One revolution $=2\pi rad$ .

  $\omega =\frac{2\pi }{t}$ .

Angular displacement $=\theta$ .

Angular velocity $=\omega$ .

Calculation:

Angular velocity:

If an object moves along a circle during the time $t$ units. Then the angular velocity $\omega$ is given by:

  $\omega =\frac{\theta }{t}$ .

Where $\theta$ is the angular displacement in radians the relation between linear velocity $\left(v\right)$ and angular velocity $\left(\omega \right)$ :

  $v=\omega r$ .

Substituting $r=6$ and $\omega =10\pi$ radians per second in $v=\omega r$ to obtain:

  $\begin{array}{l}v=\omega r.\\ v=\left(6\right).\left(10\pi \right).\\ v=188.57\text{inches per second}\text{.}\\ v=\frac{188.57}{12}=\mathrm{15.7167.}\end{array}$

Hence, the linear velocity is $15.72$ per second.