Chapter 16.2, Problem 16.105P

A drum of 4-in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 Ib and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 Ib. Knowing that the coefficients of static and kinetic friction are ${\mu }_x=0.25$ and ${\mu }_x=0.20$ respectively, determine (a) whether or not the disk slides, (b) the angular acceleration of the disk and the acceleration of G.

To determine

(a)

The whether or not disk slides.

Answer to Problem 16.105P

The disk slides.

Explanation of Solution

Given information:

The radius of the drum is $4\text{in}$, the radius of the disk is $8\text{in}$, the combined weight of disk and drum is $10\text{lb}$, the combined radius of gyration is $6\text{in}$, the magnitude of force $P$ is $5\text{lb}$, the coefficient of static friction is $0.25$ and the coefficient of kinetic friction is $0.20$.

Below figure represent the free body diagram and the kinetic diagram of the drum.

Figure-(1)

Write the expression of acceleration of the drum.

$a=r\alpha$   ............... (I)

Here, the radius of the disk is $r$ and angular acceleration is $\alpha$.

Write the expression of total force applied in the disk and drum.

$F=ma$   ............... (II)

Here, the combined mass is $m$ and the acceleration is $a$.

Write the expression of generated torque in the disk and drum.

$T=I\alpha$   ............... (III)

Here, the of inertia of drum is $I$.

Write the expression of inertia of the disk and drum.

$I=m{k}^2$ .  ............... (IV)

Here, the combined mass is $m$ and the combined radius of gyration is $k$.

Write the expression of combined mass of the disk and drum.

$m=\frac{W}{g}$   ............... (V)

Here, the combined weight of the disk and drum is $W$ and the acceleration due to gravity is $g$.

Write the expression of moment about point $C$ as shown in Figure-(1).

$\sum M_C=\sum {\left(M_C\right)}_{\text{effective}}$   ............... (VI)

Here, the effective moment is ${\left(M_C\right)}_{\text{effective}}$.

Write the expression of moment about point $C$.

$\sum M_C=P\times r_1$   ............... (VII)

Here, the radius of the disk is $r$ and the force is $P$ and $r_1$ is the radius of drum.

Write the expression of total effective moment about point $C$.

$\sum {\left(M_C\right)}_{\text{effective}}=F\times r+T$   ............... (VIII)

Here, the total force is $F$, the radius of the disk is $r$ and the generated torque in the disk and drum is $T$.

Substitute $P\times r_1$ for $\sum M_C$ and $F\times r+T$ for $\sum {\left(M_C\right)}_{\text{effective}}$ in Equation (VI).

$P\times r_1=F\times r+T$   ............... (IX)

Substitute $ma$ for $F$ and $I\alpha$ for $T$ in Equation (IX).

$P\times r_1=\left(ma\right)\times r+I\alpha$   ............... (X)

Substitute $r\alpha$ for $a$ in Equation (X).

$\begin{array}{l}P\times r_1=\left(mr\alpha \right)\times r+I\alpha \\ P\times r_1=\alpha \left(m{r}^2+I\right)\\ \alpha =\frac{P\times r_1}{\left(m{r}^2+I\right)}\end{array}$   ............... (XI)

Write the expression of force equilibrium equation in $y$ direction.

$N+P=W$   ............... (XII)

Here, the normal force in $y$ direction is $N$.

Write the expression of frictional force for static friction.

$F_f={\mu }_{s}N$   ............... (XIII)

Here, the coefficient of static friction is ${\mu }_s$.

Calculation:

Substitute $10\text{lb}$ for $W$ and $32.2{\text{ft}/\text{s}}^2$ for $g$ in Equation (V).

$\begin{array}{c}m=\frac{\left(10\text{lb}\right)}{\left(32.2{\text{ft}/\text{s}}^2\right)}\\ =\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\end{array}$

Substitute $0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m$ and $6\text{in}$ for $k$ in Equation (IV).

$\begin{array}{c}I=\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times {\left(6\text{in}\right)}^2\\ =\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times {\left\{\left(6\text{in}\right)\times \left(\frac{1\text{ft}}{12\text{in}}\right)\right\}}^2\\ =\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times \left(0.25{\text{ft}}^{\text{2}}\right)\\ =\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)\end{array}$

Substitute $5\text{lb}$ for $P$, $4\text{in}$ for $r_1$, $8\text{in}$ for $r$, $0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m$ and $\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)$ for $I$.

$\begin{array}{c}\alpha =\frac{\left(5\text{lb}\right)\times \left(4\text{in}\right)}{\left\{\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times {\left(8\text{in}\right)}^2+\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)\right\}}\\ =\frac{\left(5\text{lb}\right)\times \left(4\text{in}\right)\times \left(\frac{1\text{ft}}{12\text{in}}\right)}{\left[\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times {\left\{\left(8\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\right\}}^2+\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)\right]}\\ =\frac{1.667\text{lb}\cdot \text{ft}}{0.2153\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}}\\ =7.741\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Substitute $7.741\text{rad}/{\text{s}}^{\text{2}}$ for $\alpha$ and $8\text{in}$ for $r$ in Equation (I).

$\begin{array}{c}a=\left(8\text{in}\right)\times \left(7.741\text{rad}/{\text{s}}^{\text{2}}\right)\\ =\left(8\text{in}\right)\times \left(\frac{1\text{ft}}{12\text{in}}\right)\times \left(7.741\text{rad}/{\text{s}}^{\text{2}}\right)\\ =\left(0.667\text{ft}\right)\times \left(7.741\text{rad}/{\text{s}}^{\text{2}}\right)\\ =5.16\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Substitute $\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)$ for $m$ and $5.16\text{ft}/{\text{s}}^{\text{2}}$ for $a$ in Equation (II).

$\begin{array}{c}F=\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)\times \left(5.16\text{ft}/{\text{s}}^{\text{2}}\right)\\ =1.5996\text{lb}\end{array}$

Substitute $5\text{lb}$ for $P$ and $10\text{lb}$ for $W$ in Equation (XII).

$\begin{array}{l}N+5\text{lb}=10\text{lb}\\ N=\left(10\text{lb}\right)-\left(5\text{lb}\right)\\ N=5\text{lb}\end{array}$

Substitute $0.25$ for ${\mu }_s$ and $5\text{lb}$ for $N$ in Equation (XIII).

$\begin{array}{c}{F}_f=\left(0.25\right)\times \left(5\text{lb}\right)\\ =1.25\text{lb}\end{array}$

Here, the friction force is less than the total force, hence the disk slides.

Conclusion:

The disk slides.

To determine

(b)

The angular acceleration of the disk

The acceleration of $G$.

Answer to Problem 16.105P

The angular acceleration of the disk is $12.9{\text{rad}/\text{s}}^2$.

The acceleration of $G$ is $3.225{\text{ft}/\text{s}}^2$.

Explanation of Solution

Write the expression of friction force by considering kinetic friction.

$F_k={\mu }_{k}N$   ............... (XIV)

Here, the coefficient of kinetic friction is ${\mu }_k$.

Write the expression of total moment about point $G$.

$\sum M_G=\sum {\left(M_G\right)}_{\text{effective}}$   ............... (XV)

Here, the total effective moment is $\sum {\left(M_G\right)}_{\text{effective}}$.

Write the expression of moment about $G$.

$\sum M_G=P\times r_1$   ............... (XVI)

Here, the radius of the drum is $r_1$ and force is $P$.

Write the expression of total effective moment.

$\sum {\left(M_G\right)}_{\text{effective}}=T+F_{k}r$   ............... (XVII)

Substitute $T+F_{k}r$ for $\sum {\left(M_G\right)}_{\text{effective}}$ and $P\times r_1$ for $\sum M_G$ in Equation (XV).

$P\times r_1=T+F_{k}r$   ............... (XIX)

Substitute $I\alpha$ for $T$ in Equation (XIX).

$\begin{array}{l}P\times r_1=I\alpha +F_k\times r\\ I\alpha =P\times r_1-F_k\times r\\ \alpha =\frac{P\times r_1-F_k\times r}{I}\end{array}$   ............... (XX)

Write the expression of total force in $x$ direction.

$F_k=F$   ............... (XXI)

Here, the total force is $F$.

Substitute $ma$ for $F$ in Equation (XX).

$F_k=ma$   ............... (XXII)

Calculation:

Substitute $0.20$ for ${\mu }_k$ and $5\text{lb}$ for $N$ in Equation (XIV).

$\begin{array}{c}{F}_k=0.20\times \left(5\text{lb}\right)\\ =1\text{lb}\end{array}$

Substitute $5\text{lb}$ for $P$, $4\text{in}$ for $r_1$, $1\text{lb}$ for $F_k$, $8\text{in}$ for $r$ and $\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)$ for $I$ in Equation (XX).

$\begin{array}{c}\alpha =\frac{\left(5\text{lb}\right)\times \left(4\text{in}\right)-\left(1\text{lb}\right)\times \left(8\text{in}\right)}{\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)}\\ =\frac{\left(5\text{lb}\right)\times \left(4\text{in}\right)\times \left(\frac{1\text{ft}}{12\text{in}}\right)-\left(1\text{lb}\right)\times \left(8\text{in}\right)\times \left(\frac{1\text{ft}}{12\text{in}}\right)}{\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)}\\ =\frac{\left(1.667\text{lb}\cdot \text{ft}\right)-\left(0.667\text{lb}\cdot \text{ft}\right)}{\left(0.0775\text{lb}\cdot \text{ft}\cdot {\text{s}}^{\text{2}}\right)}\\ =12.9{\text{rad}/\text{s}}^2\end{array}$

Substitute $1\text{lb}$ for $F_k$ and $\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)$ for $m$ in Equation (XXII).

$\begin{array}{l}1\text{lb}=\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)a\\ a=\frac{1\text{lb}}{\left(0.31\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\right)}\\ a=3.225{\text{ft}/\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of the disk is $12.9{\text{rad}/\text{s}}^2$.

The acceleration of $G$ is $3.225{\text{ft}/\text{s}}^2$.