Chapter 17.2, Problem 17.75P

To determine

(a)

The time required to reduce theangular velocityof cylinder A to $5$ rad/sec.

Cylinder A has the angular velocity of $30$ rad/sec. cylinders are connected by best tension.

Answer to Problem 17.75P

Time required for cylinder A to attain $5$ rad/sec angular velocity is $t=0.56$ sec.

Explanation of Solution

Given:

Mass of cylinder, $m=6$ kg

Radius of cylinder, $r=125$ mm $=0.125$ m

Angular velocity of cylinder $A={\left({\omega }_A\right)}_1=30$ rad/sec

Concept used:

Impulse momentum principle.

Moment of inertia

Calculation:

According to impulse-momentum principle for cylinder B,

Taking moment about B,

$I{\left({\omega }_B\right)}_1–Ptr=I{\left({\omega }_B\right)}_2$ ----------------(1)

We have,

Moment of inertia, $I=\frac{1}{2}m{r}^2=\frac{1w}{2g}{r}^2=\frac{w{r}^2}{2g}$

As per kinematics. $v_{AB}=r{w}_B$

$W_B=\frac{v_{AB}}{r}$

Considering C as aninstantenous center of cylinder A.

$\begin{array}{l}{w}_A=\frac{v_{AB}}{2r}=\frac{1}{2}{w}_B\hfill \\ I{\left(w_B\right)}_1–Ptr=I{\left(w_B\right)}_2\hfill \\ I{\left(w_B\right)}_1–I{\left(w_B\right)}_2=Ptr\hfill \\ Ptr=\frac{1}{2} m{r}^2\left[{\left(w_B\right)}_1–{\left(w_B\right)}_2\right]\hfill \end{array}$

According to the impulse-momentum principle in cylinder A,

Taking moment about C,

$\begin{array}{l}I{\left(w_A\right)}_1+m{\left(v_A\right)}_{1}r+2Ptr–mgtr=I{\left(w_A\right)}_2+m{\left(v_A\right)}_{2}r\hfill \\ I\left[{\left(w_A\right)}_1-{\left(w_A\right)}_2\right]+mr\left[{\left(v_A\right)}_1-{\left(v_A\right)}_2\right]+2Ptr–mgtr=0\hfill \\ \frac{3}{2}m{r}^2\left[\frac{1}{2}{\left(w_B\right)}_1–\frac{1}{2}{\left(w_B\right)}_2\right]+2\left[\frac{1}{2}m{r}^2{\left(w_B\right)}_1-{\left(w_B\right)}_2\right]–mgtr=0\hfill \\ \frac{7}{4}m{r}^2\left[{\left(w_B\right)}_1-{\left(w_B\right)}_2\right]-mgtr=0\hfill \\ t=\frac{7r\left[{\left(w_B\right)}_1-{\left(w_B\right)}_2\right]}{4g}\hfill \\ =\frac{7\times \left(0.125\right)\left[30-5\right]}{4\times 9.81}\hfill \\ t=0.5574sec.\hfill \end{array}$

Conclusion:

Thus we can find the time required to reduce angular velocity from $30$ rad/sec to $5$ rad/sec is $0.557$ sec by applying the impulse-momentum principle on cylinder A and B.

To determine

(b)

The tension in the portion of the belt connecting the two cylinders.

Answer to Problem 17.75P

Belt tension between two cylinders is $16.83$ N.

Explanation of Solution

Given:

The radius of the cylinder, $r=125$ mm $=0.125$ m

Angular velocity of cylinder $A={\left({\omega }_A\right)}_1=30$ rad/sec

Concept used:

Impulse momentum principle

Moment of inertia

Calculation:

As per calculation is done in subpart (a),

We have,

$\begin{array}{l}I{\left({\omega }_B\right)}_1–Ptr=I{\left({\omega }_B\right)}_2 \hfill \\ Ptr={\left({\omega }_B\right)}_1–I{\left({\omega }_B\right)}_2 \hfill \\ =I\left[{\left({\omega }_B\right)}_1–{\left({\omega }_B\right)}_2\right]\hfill \\ =\frac{1}{2}m{r}^2\left[30-5\right]\hfill \\ P=\frac{m{r}^2\left(30-5\right)}{2tr}\hfill \\ P=6\times {0.125}^2\times \frac{\left(30-5\right)}{0.557}\times 0.125\hfill \end{array}$

$P=16.83$ N.

Conclusion:

Thus the tension in belt between cylinder A and B is found to be $16.83$ N by simple calculation.