Chapter 17.1, Problem 17.32P

Two uniform cylinders, each of weight $W=14$ lb and radius $r=5$ in., are connected by a belt as shown. Knowing that at the instant shown the angular velocity of cylinder B is 30 rad/s clockwise, determine (a) the distance through which cylinder A will rise before the angular velocity of cylinder B is reduced to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

  

To determine

(a)

Calculate the distance by which cylinder A is raised before the angular velocity of cylinder B is reduced to $5rad/sec$.

Answer to Problem 17.32P

Cylinder A will be raised by $2.06ft$ before angular velocity of cylinder A is reduced to $5rad/sec$.

Explanation of Solution

Given:

Both the cylinders are connected by the belt as shown in the figure.

Weight of cylinders $=W_c=14lb$

Radius of cylinders $=r=5in$

Angular velocity of the cylinder $B={\left({\omega }_B\right)}_1=30rad/sec$

Concept used:

Work and energy principle.

Calculation:

$\begin{array}{l}{v}_D=v_E=r{\omega }_B\\ $ {({\omega }_B)}_2=5rad/\sec ${\omega }_A=\frac{v_D}{cd}=\frac{r{\omega }_B}{2r}=\frac{{\omega }_B}{2}\\ v_A=r{\omega }_A=\frac{r{\omega }_B}{2}\\ v_D=2v_A\end{array}$

Kinetic energy,

$\begin{array}{l}{T}_1=\frac{m{v}_A{}^2}{2}+\frac{I{\omega }_A{}^2}{2}+\frac{I{\omega }_B{}^2}{2}\\ =\frac{m{r}^2{\omega }_B{}^2}{8}+\frac{m{r}^2{\omega }_B{}^2}{16}+\frac{m{r}^2{\omega }_B{}^2}{4}\\ T_1=\frac{7}{16}m{r}^2{\omega }_B{}^2\end{array}$

At position 1:

${({\omega }_B)}_1=30rad/\sec$

At position 2:

${({\omega }_B)}_2=5rad/\sec$

Work is done,

$\begin{array}{l}W=-W_{c}h\\ =-mgh\end{array}$ -------------------------- [weight of the cylinder is the only force that does work]

h is the rise of the cylinder.

Applying principle of work and energy,

$\begin{array}{l}W+T_1=T_2\\ T_2=T_1+W\\ $ \frac{7}{16}m{r}^2{({\omega }_B)}_2{}^2=\frac{7}{16}m{r}^2{({\omega }_B)}_2{}^2-mgh$h=\frac{7}{16}\times \frac{r^2}{g}[{(}_{{\omega }_B}{}^2-{(}_{{\omega }_B}{}^2]\\ =\frac{7}{16}\times \frac{5^2}{{12}^2\times 32.2}[{30}^2-5^2]\\ h=2.063ft.\end{array}$

Conclusion:

Thus, the cylinder will raise by $2.063ft$ when angular velocity of cylinder B is reduced to $5rad/secfrom30rad/sec.$

To determine

(b)

Calculate the tension in belt where two cylinders are connected.

Answer to Problem 17.32P

Tension in the belt connecting two cylinders is $4lb.$

Explanation of Solution

Given:

Weight of cylinders $=W_c=14lb$

Radius of cylinders $=r=5in$

Angular velocity of cylinder, $B={\left({\omega }_B\right)}_1=30rad/sec$

Concept used:

Work and energy conservation principle

Calculation:

We have,

$v_D=2v_A$

D moves twice the distance than A.

Distance = 2h

$\begin{array}{l}{T}_1=\frac{I{({\omega }_B)}_1{}^2}{2}\\ T_2=\frac{I{({\omega }_B)}_2{}^2}{2}\end{array}$

Let the tension in cord be Q.

Hence, work done

$\begin{array}{l}-Q\times dis\tan ce\\ W=-Q\times 2h\end{array}$

As per the work and energy principle,

$\begin{array}{l}{T}_1+W=T_2\\ \frac{I{({\omega }_B)}_1{}^2}{2}-2Qh=\frac{I{({\omega }_B)}_2{}^2}{2}\\ $ Qh=\frac{1}{4}I[{({\omega }_B)}_1{}^2-{({\omega }_B)}_2{}^2]$$ Q\times \frac{7r^2}{16g}[{({\omega }_B)}_1{}^2-{({\omega }_B)}_2{}^2]=\frac{I}{4}[{({\omega }_B)}_1{}^2-{({\omega }_B)}_2{}^2]$Q=\frac{m{r}^2}{8}\times \frac{16g}{7r^2}\\ Q=\frac{2}{7}mg=\frac{2}{7}{W}_c\\ Q=4lb\end{array}$

Conclusion:

In this way, by using work and energy principle we can calculate the tension in belt adjoining two cylinders.