Chapter 15.1, Problem 15.24P

gear reduction system consists of three gears A, B, and C. Knowing that gear A rotates clockwise with a constant angular velocity rpm, determine (a) the angular velocities of gears B and C, (b) the accelerations of the points on gears B and C which are in contact.

Fig. P15.24

To determine

(a)

The angular velocities of gears $B$ and $C$.

Answer to Problem 15.24P

The angular velocity of gear $B$

$=31.42 rad s^{-1}.$

The angular velocity of gear $C$

$=10.47 rad s^{-2}.$

Explanation of Solution

Given information:

Gear $A$ rotates clockwise with a constant angular velocity ${\omega }_A=600 rpm.$

Concept used:

The velocity $(v)$ of the point $P$ of a body rotating about a fixed axis,

$v=r\omega$

$r-$ Radius

$\omega -$ Angular velocity

Calculation:

The velocity of the point $P$ of a body rotating about a fixed axis,

$v=r\omega$

Since, $A$ and $B$ gears are touching at point $P_1$, they have same linear velocity.

Thus,

$v_A=v_B$

$r_A{\omega }_A=r_B{\omega }_B$

${\omega }_B=\frac{r_A{\omega }_A}{r_B}$

$r_A=2 in.$, $r_B=4 in.$, ${\omega }_A=600 rpm$

${\omega }_B=\frac{2 in.\times 600 rpm}{4 in.}=300 rpm$

${\omega }_B=\frac{2\pi }{60}\times 300 rpm=31.42 rad s^{-1}$

$B$ and $C$ gears are touching at point $P_2$, they have same linear velocity.

$v_A=v_B$

$r_B{\omega }_B=r_C{\omega }_C$

${\omega }_C=\frac{r_B{\omega }_B}{r_C}$

$r_B=4 in.$, $r_C=6 in.$, ${\omega }_B=31.42 rad s^{-1}$

${\omega }_C=\frac{2 in.\times 31.42 rad s^{-1}}{6 in.}=10.47 rad s^{-2}$

The angular velocity of gear $B$

$=31.42 rad s^{-1}.$

The angular velocity of gear $C$

$=10.47 rad s^{-2}.$

Conclusion:

The angular velocity of gear $B$

$=31.42 rad s^{-1}.$

The angular velocity of gear $C$

$=10.47 rad s^{-2}.$

To determine

(b)

The accelerations of the points on gears $B$ and $C$ which are in contact.

Answer to Problem 15.24P

The acceleration of the point on gear $B$ which is in contact,

${\left(a_{P_1}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_1}\right)}_n=329 ft s^{-2}$

The acceleration of the point on gear $C$ which is in contact,

${\left(a_{P_2}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_2}\right)}_n=54.29 ft s^{-2}$

Explanation of Solution

Given information:

Gear $A$ rotates clockwise with a constant angular velocity ${\omega }_A=600 rpm.$

Concept used:

The velocity $(v)$ of the point $P$ of a body rotating about a fixed axis,

$v=r\omega$

$r-$ Radius

$\omega -$ Angular velocity

The acceleration $(a)$ of the point $P$ of a body rotating about a fixed axis,

$a_t=r\alpha$

$a_n={\omega }^{2}r$

$r-$ Radius

$\omega -$ Angular velocity

Calculation:

The acceleration of the point $P$ of a body rotating about a fixed axis,

$a_t=r\alpha$

$a_n={\omega }^{2}r$

Consider point $P_1$, where gears $A$ and $B$ are in contact.

${\left(a_{P_1}\right)}_t=r_B{\alpha }_B$

${\left(a_{P_1}\right)}_n={\omega }_B{}^2{r}_B$

${\omega }_B=31.42 rad s^{-1}$, $r_B=4 in.$, ${\alpha }_B=0$

${\left(a_{P_1}\right)}_t=0 ft s^{-2}$

${\left(a_{P_1}\right)}_n={(31.42 rad s^{-1})}^2\times \frac{4 }{12}ft$

${\left(a_{P_1}\right)}_n=329 ft s^{-2}$

Consider point $P_2$, where gears $B$ and $C$ are in contact.

${\left(a_{P_2}\right)}_t=r_C{\alpha }_C$

${\left(a_{P_2}\right)}_n={\omega }_C{}^2{r}_C$

${\omega }_C=10.47 rad s^{-2}$, $r_C=6 in.$, ${\alpha }_C=0$

${\left(a_{P_2}\right)}_t=0 ft s^{-2}$

${\left(a_{P_2}\right)}_n={(10.47 rad s^{-2})}^2\times \frac{6 }{12}ft$

${\left(a_{P_2}\right)}_n=54.29 ft s^{-2}$

The acceleration of the point on gear $B$ which is in contact

${\left(a_{P_1}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_1}\right)}_n=329 ft s^{-2}$

The acceleration of the point on gear $C$ which is in contact

${\left(a_{P_2}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_2}\right)}_n=54.29 ft s^{-2}$

Conclusion:

The acceleration of the point on gear $B$ which is in contact

${\left(a_{P_1}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_1}\right)}_n=329 ft s^{-2}$

The acceleration of the point on gear $C$ which is in contact

${\left(a_{P_2}\right)}_t=0 ft s^{-2}$, ${\left(a_{P_2}\right)}_n=54.29 ft s^{-2}$