Chapter 12, Problem 32PQ

(a)

To determine

The speed of each point in meters per second.

Answer to Problem 32PQ

The speed of the point at the outer end of the blade having length $20.0\text{ m}$ is $77.6\text{ m/s}$ and the speed of the point at the outer end of the blade having length $40.0\text{ m}$ is $155\text{ m/s}$ .

Explanation of Solution

Write the equation connecting linear speed and the angular speed of a point.

$v=r\omega$ (I)

Here, $v$ is the linear speed, $r$ is the distance from the point to the axis of rotation and $\omega$ is the angular speed.

It is given that both the blades have the same angular speed.

Use equation (I) to find the expression for the linear speed of the point at the outer end of the blade having length $20.0\text{ m}$ .

$v_{20}=r_{20}\omega$ (II)

Here, $v_{20}$ is the linear speed of the point at the outer end of the blade having length $20.0\text{ m}$, $r_{20}$ is the distance from the point to the axis of rotation of the blade.

Use equation (I) to find the expression for the linear speed of the point at the outer end of the blade having length $40.0\text{ m}$ .

$v_{40}=r_{40}\omega$ (III)

Here, $v_{40}$ is the linear speed of the point at the outer end of the blade having length $40.0\text{ m}$, $r_{40}$ is the distance from the point to the axis of rotation of the blade.

Conclusion:

It is given that the angular speed is $3.88\text{ rad/s}$ .

Substitute $20.0\text{ m}$ for $r_{20}$ and $3.88\text{ rad/s}$ for $\omega$ in equation (II) to find $v_{20}$ .

  $\begin{array}{c}{v}_{20}=\left(20.0\text{ m}\right)\left(3.88\text{ rad/s}\right)\\ =77.6\text{ m/s}\end{array}$

Substitute $40.0\text{ m}$ for $r_{40}$ and $3.88\text{ rad/s}$ for $\omega$ in equation (III) to find $v_{40}$ .

  $\begin{array}{c}{v}_{40}=\left(40.0\text{ m}\right)\left(3.88\text{ rad/s}\right)\\ =155\text{ m/s}\end{array}$

Therefore, the speed of the point at the outer end of the blade having length $20.0\text{ m}$ is $77.6\text{ m/s}$ and the speed of the point at the outer end of the blade having length $40.0\text{ m}$ is $155\text{ m/s}$ .

(b)

To determine

The angular distance travelled by each point.

Answer to Problem 32PQ

The angular distance travelled by each point is $38.8\text{ rad}$ .

Explanation of Solution

Write the equation for the angular velocity.

  $\omega =\frac{\Delta \theta }{\Delta t}$

Here, $\Delta \theta$ is the angular displacement and $\Delta t$ is the time.

Rewrite the above equation for $\Delta \theta$ .

$\Delta \theta =\omega \Delta t$ (IV)

Equation (IV) shows that angular distance travelled by each point is same since both the blades have same angular speed.

Conclusion:

It is given that the motion is considered for $10.0\text{ s}$ .

Substitute $3.88\text{ rad/s}$ for $\omega$ and $10.0\text{ s}$ for $\Delta t$ in equation (IV) to find $\Delta \theta$ .

  $\begin{array}{c}\Delta \theta =\left(3.88\text{ rad/s}\right)\left(10.0\text{ s}\right)\\ =38.8\text{ rad}\end{array}$

Therefore, the angular distance travelled by each point is $38.8\text{ rad}$ .

(c)

To determine

The translational distance travelled by each point.

Answer to Problem 32PQ

The translational distance travelled by the point at the outer end of the blade having length $20.0\text{ m}$ is $776\text{ m}$ and that by the point at the outer end of the blade having length $40.0\text{ m}$ is $1.55\times {10}^3\text{ m}$ .

Explanation of Solution

The translational distance travelled by each point will be equal to the arc length subtended by the angular displacement of each point.

Write the equation for the arc length.

$s=r\Delta \theta$ (V)

Here, $s$ is the arc length.

Use equation (V) to find the expression for translational distance travelled by the point at the outer end of the blade having length $20.0\text{ m}$ .

$s_{20}=r_{20}\Delta \theta$ (VI)

Here, $s_{20}$ is the translational distance travelled by the point at the outer end of the blade having length $20.0\text{ m}$ .

Use equation (V) to find the expression for translational distance travelled by the point at the outer end of the blade having length $40.0\text{ m}$ .

$s_{40}=r_{40}\Delta \theta$ (VII)

Here, $s_{40}$ is the translational distance travelled by the point at the outer end of the blade having length $40.0\text{ m}$ .

Conclusion:

Substitute $20.0\text{ m}$ for $r_{20}$ and $38.8\text{ rad}$ for $\Delta \theta$ in equation (VI) to find $s_{20}$ .

  $\begin{array}{c}{s}_{20}=\left(20.0\text{ m}\right)\left(38.8\text{ rad}\right)\\ =776\text{ m}\end{array}$

Substitute $40.0\text{ m}$ for $r_{40}$ and $38.8\text{ rad}$ for $\Delta \theta$ in equation (VII) to find $s_{40}$ .

  $\begin{array}{c}{s}_{40}=\left(40.0\text{ m}\right)\left(38.8\text{ rad}\right)\\ =1.55\times {10}^3\text{ m}\end{array}$

Therefore, the translational distance travelled by the point at the outer end of the blade having length $20.0\text{ m}$ is $776\text{ m}$ and that by the point at the outer end of the blade having length $40.0\text{ m}$ is $1.55\times {10}^3\text{ m}$ .

(d)

To determine

The magnitude of centripetal acceleration that would be experienced by an object located at each point.

Answer to Problem 32PQ

The magnitude of centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $20.0\text{ m}$ is $301{\text{ m/s}}^2$ and that experienced by an object located at the point at the outer end of the blade having length $40.0\text{ m}$ is $602{\text{ m/s}}^2$ .

Explanation of Solution

Write the equation for the centripetal acceleration.

$a_c={\omega }^{2}r$ (VIII)

Here, $a_c$ is the centripetal acceleration.

Use equation (VIII) to find the expression for the centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $20.0\text{ m}$.

$a_{c,20}={\omega }^2{r}_{20}$ (IX)

Here, $a_{c,20}$ is the centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $20.0\text{ m}$.

Use equation (VIII) to find the expression for the centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $40.0\text{ m}$.

$a_{c,40}={\omega }^2{r}_{40}$ (X)

Here, $a_{c,40}$ is the centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $40.0\text{ m}$.

Conclusion:

Substitute $3.88\text{ rad/s}$ for $\omega$ and $20.0\text{ m}$ for $r_{20}$ in equation (IX) to find $a_{c,20}$ .

  $\begin{array}{c}{a}_{c,20}={\left(3.88\text{ rad/s}\right)}^2\left(20.0\text{ m}\right)\\ =301{\text{ m/s}}^2\end{array}$

Substitute $3.88\text{ rad/s}$ for $\omega$ and $40.0\text{ m}$ for $r_{40}$ in equation (X) to find $a_{c,40}$ .

  $\begin{array}{c}{a}_{c,40}={\left(3.88\text{ rad/s}\right)}^2\left(40.0\text{ m}\right)\\ =602{\text{ m/s}}^2\end{array}$

Therefore, the magnitude of centripetal acceleration that would be experienced by an object located at the point at the outer end of the blade having length $20.0\text{ m}$ is $301{\text{ m/s}}^2$ and that experienced by an object located at the point at the outer end of the blade having length $40.0\text{ m}$ is $602{\text{ m/s}}^2$ .