Chapter 17, Problem 17.135RP

A uniform disk, initially at rest and of constant thickness, is placed in contact with the belt shown, which moves at a constant speed v = 80 ft/s. Knowing that the coefficient of kinetic friction between the disk and the belt is 0.15, determine (a) the number of revolutions executed by the disk before it reaches a constant angular velocity, (b) the time required for the disk to reach that constant angular velocity.

Fig. P17.135

To determine

Find the number of revolutions executed by the disk before it reaches a constant angular velocity.

Answer to Problem 17.135RP

The number of revolutions executed by the disk is $\underset{\_}{106.7\text{rev}}$.

Explanation of Solution

Given information:

The speed of the belt is $v=80\text{ft/s}$.

The coefficient of kinetic friction between the disk and the belt is ${\mu }_k=0.15$.

The radius of the uniform disk is $r=5\text{in}\text{.}$.

Calculation:

Show the free-body diagram of the disk as in Figure 1.

Find the frictional force $\left(F_f\right)$ using the relation.

$F_f={\mu }_{k}N$

Here, the normal force is N.

Substitute 0.15 for ${\mu }_k$.

$F_f=0.15N$ (1)

Resolve the vertical component of forces to find the normal force.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N\cos 25°-F_f\sin 25°-mg=0\end{array}$

Here, the mass of the disk is m and the acceleration due to gravity is g.

Consider the acceleration due to gravity as $g=32.2{\text{ft/s}}^2$.

Substitute 0.15N for $F_f$.

$\begin{array}{l}N\cos 25°-0.15N\sin 25°-mg=0\\ 0.90631N-0.06339N-mg=0\\ N=1.18635mg\end{array}$

Substitute $1.18635mg$ for N in Equation (1).

$\begin{array}{c}{F}_f=0.15\left(1.18635mg\right)\\ =0.17795mg\end{array}$

Find the mass moment of inertia $\left(\overline{I}\right)$ using the equation.

$\overline{I}=\frac{1}{2}m{r}^2$

Here, the radius of the disk is r.

Find the angular velocity $\left({\omega }_2\right)$ using the relation.

${\omega }_2=\frac{v}{r}$

Find the work done $\left(U_{1\to 2}\right)$ applied in the process using the relation.

$U_{1\to 2}=F_{f}r\theta$

Here, the number of revolutions is $\theta$.

Substitute $0.17795mg$ for $F_f$.

$U_{1\to 2}=0.17795mgr\theta$

The total kinetic energy at initial position is zero.

$T_1=0$

Find the total kinetic energy after reaching the constant angular velocity using the relation.

$T_2=\frac{1}{2}\overline{I}{\omega }_2^2$

Substitute $\frac{1}{2}m{r}^2$ for $\overline{I}$ and $\frac{v}{r}$ for ${\omega }_2$

$\begin{array}{c}{T}_2=\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{v}{r}\right)}^2\\ =\frac{m{v}^2}{4}\end{array}$

Write the equation of conservation of energy and work as follows;

$T_1+U_{1\to 2}=T_2$

Substitute 0 for $T_1$, $0.17795mgr\theta$ for $U_{1\to 2}$, and $\frac{m{v}^2}{4}$ for $T_2$.

$0+0.17795mgr\theta =\frac{m{v}^2}{4}$

Substitute 5 in. for r, $32.2{\text{ft/s}}^2$ for g, and $80\text{ft/s}$ for v.

$\begin{array}{c}0.17795\times 32.2\times 5\text{in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\times \theta =\frac{{80}^2}{4}\\ \theta =670.15824\text{rad}\times \frac{1\text{rev}}{2\pi \text{rad}}\\ =106.7\text{rev}\end{array}$

Therefore, the number of revolutions executed by the disk is $\underset{\_}{106.7\text{rev}}$.

To determine

Find the time required for the disk to reach the constant angular velocity.

Answer to Problem 17.135RP

The time required for the disk is $\underset{\_}{6.98\text{s}}$.

Explanation of Solution

Given information:

The speed of the belt is $v=80\text{ft/s}$.

The coefficient of kinetic friction between the disk and the belt is ${\mu }_k=0.15$.

The radius of the uniform disk is $r=5\text{in}\text{.}$.

Calculation:

Show the free-body diagram of the principle of impulse-momentum as in Figure 2.

Take moment about point A as follows;

$\begin{array}{l}{\displaystyle \sum M_A=\overline{I}{\omega }_2}\\ 0+F_{f}tr=\overline{I}{\omega }_2\end{array}$

Substitute $0.17795mg$ for $F_f$, $\frac{1}{2}m{r}^2$ for $\overline{I}$, and $\frac{v}{r}$ for ${\omega }_2$.

$\begin{array}{l}0.17795mgtr=\frac{1}{2}m{r}^2\frac{v}{r}\\ 0.17795gt=\frac{v}{2}\end{array}$

Substitute $32.2{\text{ft/s}}^2$ for g and $80\text{ft/s}$ for v.

$\begin{array}{l}0.17795\times 32.2\times t=\frac{80}{2}\\ t=6.98\text{s}\end{array}$

Therefore, the time required for the disk is $\underset{\_}{6.98\text{s}}$.