Chapter 8, Problem 59A

(a)

To determine

The ratio of centripetal accelerations for the fast and slow spin cycles.

Answer to Problem 59A

The ratio of centripetal accelerations for the fast and slow spin cycles is $2.73$ .

Explanation of Solution

Given:

The angular velocity of the slow spin cycle is ${\omega }_{\text{slow}}=328\text{rev/min}$ .

The angular velocity of the fast spin cycle is ${\omega }_{\text{slow}}=542\text{rev/min}$ .

The diameter of the drum is $d=0.43\text{m}$ .

Formula used:

The relation between the centripetal acceleration $\left({\alpha }_{\text{c}}\right)$ and the angular velocity $\left(\omega \right)$ is,

${\alpha }_{\text{c}}=r{\omega }^2$

Calculation:

Consider the radius of the drum is denoted by $r$ .

For slow spin cycle, the centripetal acceleration is,

$\begin{array}{l}{\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}=r{\omega }_{\text{slow}}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}=\left(\frac{d}{2}\right){\left(328\text{rev/min}\right)}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}=\left(\frac{0.43\text{m}}{2}\right){\left(328\frac{\text{rev}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}\times \frac{2\pi }{1\text{rev}}\right)}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}=254{\text{m/s}}^2\end{array}$

For fast spin cycle, the centripetal acceleration is,

$\begin{array}{l}{\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}=r{\omega }_{\text{fast}}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}=\left(\frac{d}{2}\right){\left(542\text{rev/min}\right)}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}=\left(\frac{0.43\text{m}}{2}\right){\left(542\frac{\text{rev}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}\times \frac{2\pi }{1\text{rev}}\right)}^2\\ {\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}=693{\text{m/s}}^2\end{array}$

The ratio of centripetal acceleration for the fast to slow spin cycle is,

$\begin{array}{l}\frac{{\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}}{{\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}}=\frac{693{\text{m/s}}^2}{254{\text{m/s}}^2}\\ \frac{{\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}}{{\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}}=2.73\end{array}$

Conclusion:

Thus, the ratio of centripetal accelerations for the fast and slow spin cycles is $2.73$ .

(b)

To determine

The ratio of linear velocity of an object at the surface of the drum, for the fast and slow spin cycles.

Answer to Problem 59A

The ratio of linear velocity of the object for the fast and slow spin cycles is $1.65$ .

Explanation of Solution

Given:

The angular velocity of the slow spin cycle is ${\omega }_{\text{slow}}=328\text{rev/min}$ .

The angular velocity of the fast spin cycle is ${\omega }_{\text{slow}}=542\text{rev/min}$ .

The diameter of the drum is $d=0.43\text{m}$ .

Formula used:

The relation between the linear velocity $\left(v\right)$ and the angular velocity $\left(\omega \right)$ of an object is,

$v=r\omega$

Calculation:

Consider the radius of the drum is denoted by $r$ .

For slow spin cycle, the linear velocity of an object is

$\begin{array}{l}{v}_{\text{slow}\text{spin}}=r{\omega }_{\text{slow}}\\ v_{\text{slow}\text{spin}}=r\left(328\text{rev/min}\right)\end{array}$

For fast spin cycle, the linear velocity of an object is

$\begin{array}{l}{v}_{\text{fast}\text{spin}}=r{\omega }_{\text{fast}}\\ v_{\text{fast}\text{spin}}=r\left(542\text{rev/min}\right)\end{array}$

The ratio of centripetal accelerations for the fast to slow spin cycles is,

$\begin{array}{l}\frac{v_{\text{fast}\text{spin}}}{v_{\text{slow}\text{spin}}}=\frac{r\left(542\text{rev/min}\right)}{r\left(328\text{rev/min}\right)}\\ \frac{v_{\text{fast}\text{spin}}}{v_{\text{slow}\text{spin}}}=\left(\frac{542}{328}\right)\\ \frac{v_{\text{fast}\text{spin}}}{v_{\text{slow}\text{spin}}}=1.65\end{array}$

Conclusion:

Thus, the ratio of linear velocity of the object for the fast and slow spin cycles is $1.65$ .

(c)

To determine

The maximum centripetal acceleration in term of acceleration due to gravity $g$ .

Answer to Problem 59A

The maximum centripetal acceleration in term of acceleration due to gravity $g$ is $44.79g$ .

Explanation of Solution

Given:

The angular velocity of the slow spin cycle is ${\omega }_{\text{slow}}=328\text{rev/min}$ .

The angular velocity of the fast spin cycle is ${\omega }_{\text{slow}}=542\text{rev/min}$ .

The diameter of the drum is $d=0.43\text{m}$ .

Formula used:

The relation between the centripetal acceleration $\left({\alpha }_{\text{c}}\right)$ and the angular velocity $\left(\omega \right)$ is,

${\alpha }_{\text{c}}=r{\omega }^2$

Calculation:

Frompart (a), the centripetal acceleration for fast spin cycle is ${\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}=693{\text{m/s}}^2$ and the centripetal acceleration for slow spin cycle is ${\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}=254{\text{m/s}}^2$

As the two spin cycles are opposite in direction,

So, the difference in the acceleration $\left(a_C\right)$ of the spin cycles will be,

$\begin{array}{c}{\alpha }_C={\left({\alpha }_{\text{c}}\right)}_{\text{fast}\text{spin}}-{\left({\alpha }_{\text{c}}\right)}_{\text{slow}\text{spin}}\\ =693{\text{m/s}}^2-254{\text{m/s}}^2\\ =439{\text{m/s}}^2\end{array}$

The acceleration due to gravity of the Earth as $g=9.80{\text{m/s}}^2$ .

The maximum centripetal acceleration in term of acceleration due to gravity $g$ is,

$\begin{array}{l}\frac{{\alpha }_C}{g}=\frac{439{\text{m/s}}^2}{9.80{\text{m/s}}^2}\\ \frac{{\alpha }_C}{g}=44.79\\ {\alpha }_C=44.79g\end{array}$

Conclusion:

Thus, the maximum centripetal acceleration in term of acceleration due to gravity $g$ is $44.79g$ .