Chapter 16.1, Problem 16.45P

Cylinder A has an initial angular velocity of 720 rpm clockwise, and cylinders B and C are initially at rest. Disks A and B each weigh 5 lb and have radius $r=4$ in. Disk C weighs 20 lb and has a radius of 8 in. The disks are brought together when C is placed gently onto A and B. Knowing that ${\mu }_k=0.25$ between A and C and no slipping occurs between B and C, determine (a) the angular acceleration of each disk, (b) the final angular velocity of each disk.

To determine

(a)

The angular acceleration of each disk.

Explanation of Solution

Given information:

Angular velocity of cylinder A = 720 rpm

Weight of disks A and B = 5lb

Disks A and B radius = 4in

Cylinders A and B mass,

$m_A=m_B$

$=\frac{W_A}{g}$

Here, g = gravity

$g=32.2ft/s^2$

$m_A=\frac{5lb}{32.2ft/s^2}$

$=0.1552slug$

Cylinders A and B moment of inertia,

$I_A=I_B$

$=\frac{m_A{r}^2}{2}$

Here, $r=4in$

$I_A=\frac{\left(0.1552slug\right)\left(4in.\frac{1ft}{12in.}\right)}{2}$

$=0.0086266slug.f{t}^2$

Cylinder C mass,

$m_C=\frac{W_C}{g}$

Here, $W_C$ = Cylinder C weight

$=\frac{20lb}{32.2ft/s^2}$

$=0.6211slug$

Cylinder C moment of inertia,

$I_C=\frac{m_C{\left(2r\right)}^2}{2}$

$I_C=\frac{\left(0.6211slug\right){\left(2\times 4in.\frac{1ft}{12in.}\right)}^2}{2}$

$=0.1380slug.f{t}^2$

Contact point tangential acceleration between cylinders B and C

${\left(a_t\right)}_{BC}=r{\alpha }_B$

$=2r{\alpha }_C$ ........... (Equation A)

${\alpha }_B=2r{\alpha }_C$

Here ${\alpha }_B$ and ${\alpha }_C$ are angular accelerations of cylinders B and C

Friction force between cylinders A and C

$F_{AC}={\mu }_k{N}_{AC}$

$N_{AC}=\frac{F_{AC}}{{\mu }_k}$ ........................(Equation B)

Here, ${\mu }_k$ = Co-efficient of kinetic friction

Cylinder B free body diagram

Moments about point B,

$F_{BC}r=I_B{\alpha }_B$

$F_{BC}=\frac{I_B{\alpha }_B}{r}$

Here, $F_{BC}$ = friction force between cylinders B and C

$F_{BC}=\frac{\left(0.0086266slug.f{t}^2\right)\left(2{\alpha }_C\right)}{\left(4in.\frac{1ft}{12in.}\right)}$ ...................(Equation C)

$=0.05176{\alpha }_C$

Cylinder C free body diagram

Moment about point C,

$F_{AC}\left(2r\right)-F_{BC}\left(2r\right)=I_C{\alpha }_C$

$\left\{F_{AC}\left(2\right)\left(4in.\frac{1ft}{12in.}\right)-\left(0.05176{\alpha }_C\right)\left(2\right)\left(4in.\frac{1ft}{12in.}\right)\right\}==\left(0.1380slug.f{t}^2\right){\alpha }_C$ .....(Equation D)

$0.6666F_{AC}=0.1380{\alpha }_C+0.0345{\alpha }_C$

$F_{AC}=0.2588{\alpha }_C$

Assume P as the contact point between cylinders A and C

Along line CP, components of forces,

$W_B\sin {30}^{\circ }+F_{BC}-F_{AC}\cos {60}^{\circ }-N_{AC}\sin {60}^{\circ }$

Substitute, $\begin{array}{l}{W}_B=20lb\\ F_{BC}=0.05176{\alpha }_C\\ F_{AC}=0.2588{\alpha }_C\\ N_{AC}=\frac{F_{AC}}{{\mu }_k}\\ {\mu }_k=0.25\end{array}$

$\left(20lb\right)\sin {30}^{\circ }+\left(0.05176{\alpha }_C\right)-\left(0.2588{\alpha }_C\right)\cos {60}^{\circ }-\left(\frac{F_{AC}}{{\mu }_k}\right)\sin {60}^{\circ }=0$

$10-0.0776{\alpha }_C-\left(\frac{0.2588{\alpha }_C}{0.25}\right)\sin {60}^{\circ }=0$

$10-0.9741{\alpha }_C=0$

${\alpha }_C=\frac{10}{0.9741}$

$=10.2654rad/s^2$

$=10.27rad/s^2$

Substitute ${\alpha }_C=10.27rad/s^2$ in equation C,

$F_{BC}=0.05176{\alpha }_C$

$=0.05176\left(10.2654rad/s^2\right)$

$=0.5313lb$

Substitute ${\alpha }_C=10.27rad/s^2$ in equation D,

$F_{AC}=0.2588{\alpha }_C$

$=0.2588\left(10.2654rad/s^2\right)$

$=2.6567lb$

Substitute $F_{AC}=2.6567lb$ in equation B,

$N_{AC}=\frac{2.6567lb}{0.25}$

$=10.6267lb$

Cylinder A free body diagram

Assume Q as the contact point between cylinder B and C

From above figure, calculate force components along line CQ

$N_{BC}-W_C\cos {30}^{\circ }+N_{AC}\cos {60}^{\circ }-F_{AC}\sin {60}^{\circ }$

$N_{BC}-\left(20lb\right)\cos {30}^{\circ }+\left(10.6267lb\right)\cos {60}^{\circ }-\left(2.6567lb\right)\sin {60}^{\circ }=0$

$N_{BC}=14.3079lb$

From figure above, moment at A,

$F_{AC}r=I_A{\alpha }_A$

$\left(2.6567lb\right)\left(4in.\frac{1ft}{12in.}\right)=\left(0.0086266slug.f{t}^2\right){\alpha }_A$

${\alpha }_A=\frac{0.8855}{0.0086266}$

$=102.7rad/s^2$

Substitute ${\alpha }_C=10.27rad/s^2$ in equation A,

${\alpha }_B=2\left(10.2654rad/s^2\right)$

$=20.5rad/s^2$

Conclusion:

Cylinders A, B and C have angular acceleration of ${\alpha }_A=102.7rad/s^2$, ${\alpha }_B=20.5rad/s^2$ and ${\alpha }_C=10.27rad/s^2$ respectively

To determine

(b)

The angular velocities of cylinders A, B and C.

Explanation of Solution

Given information:

Angular velocity of cylinder A = 720 rpm

Weight of disks A and B = 5lb

Disks A and B radius = 4in

Cylinder A angular velocity,

${\omega }_A={\omega }_0-{\alpha }_{A}t$

Substitute $\begin{array}{l}{\omega }_0=720rpm\\ {\alpha }_A=102.69rad/s^2\end{array}$

${\omega }_A=\left(720rpm\frac{2\Pi rad}{60rpm}\right)-\left(102.69rad/s^2\right)t$ ........(Equation E)

${\omega }_A=75.3982-102.69t$

Cylinder A tangential velocity,

${\left(v_t\right)}_{AC}=r{\omega }_A$

${\left(v_t\right)}_{AC}=\left(4in.\frac{1ft}{12in.}\right)\left(75.3982-102.69t\right)$ ........(Equation F)

${\left(v_t\right)}_{AC}=25.1327-34.23t$

Cylinder C angular velocity,

${\omega }_C={\alpha }_{C}t$

${\omega }_C=\left(10.2654rad/s^2\right)t$

${\omega }_C=10.2654t$ ............(Equation G)

Cylinder C tangential velocity,

${\left(v_t\right)}_{CA}=2r{\omega }_C$

${\left(v_t\right)}_{CA}=2\left(4in.\frac{1ft}{12in.}\right)\left(10.2654t\right)$ .......(Equation H)

${\left(v_t\right)}_{CA}=6.8436t$

Equate equation F and H,

$25.1327-34.23t=6.8436t$

$41.0736t=25.1327$

$t=0.6119s$

Substitute the above value in equation E,

${\omega }_A=75.3982-102.69\left(0.6119s\right)$

$=12.5621rad/s\left(\frac{60rpm}{2\Pi rad/s}\right)$

${\omega }_A=120rpm$

Substitute $t=0.6119s$ in equation G,

${\omega }_C=10.265\left(0.6119s\right)$

$=12.56rad/s\left(\frac{60rpm}{2\Pi rad/s}\right)$

${\omega }_C=60rpm$

Cylinder B angular velocity,

${\omega }_B={\alpha }_{B}t$

${\omega }_B=\left(20.5rad/s^2\right)\left(0.6119s\right)$

$=\left(12.5439rad/s\right)\left(\frac{60rpm}{2\Pi rad/s}\right)$

${\omega }_B=120rpm$

Hence, thecylinders A, B and C have angular velocities of ${\omega }_A=120rpm$, ${\omega }_B=120rpm$ and ${\omega }_C=60rpm$ respectively.