Chapter 15.1, Problem 15.26P

Ring C has an inside radius of 55 mm and an outside radius of 60 mm and is positioned between two wheels A and B, each of 24-mm outside radius. Knowing that wheel A rotates with a constant angular velocity of 300 rpm and that no slipping occurs, determine (a) the angular velocity of ring C and of wheel B, (b) the acceleration of the points on A and B that are in contact with C.

Fig. . P15.26

To determine

(a)

The angular velocity of ring $C$ and of wheel $B$.

Answer to Problem 15.26P

The angular velocity of ring $C$

$=12.57 rad s^{-1}.$

The angular velocity of wheel $B$

$=28.81 rad s^{-1}.$

Explanation of Solution

Given information:

Inside radius of ring $C$

$=55 mm.$

Outside radius of ring $C$

$=60 mm.$

Outside radius of wheels $A$ and $B$

$=24 mm.$

Wheel $A$ rotates with a constant angular velocity. ${\omega }_A=300 rpm.$

No slipping occurs.

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,

$v=r\omega$

The point of ring $C$, which is touching the wheel $A$ has the same linear velocity as wheel $A$.

$v_A=v_C$

$r_A{\omega }_A=r_C{\omega }_C$

Here, the outer radius of the ring $C$ should be considered.

$r_A=24 mm$, $r_C=60 mm$, ${\omega }_A=300 rpm$

${\omega }_C=\frac{r_A}{r_C}{\omega }_A$

${\omega }_C=\frac{24 mm}{60 mm}\times 300 rpm=120 rpm$

${\omega }_C=\frac{2\pi }{60}\times 120rpm=12.57 rad s^{-1}$

The ponit of wheel $B$, which is touching the ring $C$ has the same linear velocity as ring $C$.

$v_C=v_B$

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

Here, the inner radius of the ring $C$ should be considered.

$r_B=24 mm$, $r_C=55 mm$, ${\omega }_C=12.57 rad s^{-1}$

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

${\omega }_B=\frac{55 mm}{24 mm}\times 12.57 rad s^{-1}=28.81 rad s^{-1}$

The angular velocity of ring $C$

$=12.57 rad s^{-1}.$

The angular velocity of wheel $B$

$=28.81 rad s^{-1}.$

Conclusion:

The angular velocity of ring $C$

$=12.57 rad s^{-1}.$

The angular velocity of wheel $B$

$=28.81 rad s^{-1}.$

To determine

(b)

The accelerations of the points on $A$ and $B$ that are in contact with $C$.

Answer to Problem 15.26P

The accelerations of the point on $A$ that are in contact with $C$

${(a_A)}_t=0 m s^{-2}$, ${\left(a_A\right)}_n=23.69 m s^{-2}$

The accelerations of the point on $B$ that are in contact with $C$

$=19.92 m s^{-2}.$

${(a_B)}_t=0 m s^{-2}$, ${\left(a_B\right)}_n=19.92 m s^{-2}$

Explanation of Solution

Given information:

Inside radius of ring $C$

$=55 mm.$

Outside radius of ring $C$

$=60 mm.$

Outside radius of wheels $A$ and $B$

$=24 mm.$

Wheel $A$ rotates with a constant angular velocity. ${\omega }_A=300 rpm.$

No slipping occurs.

Concept used:

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 the point on $A$ that is in contact with $C$.

${(a_A)}_t=r_A{\alpha }_A$

${\left(a_A\right)}_n={\omega }_A^2{r}_A$

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

$r_C=24 mm$, ${\alpha }_A=0$

${(a_A)}_t=0 m s^{-2}$

${\left(a_A\right)}_n={\omega }_A{}^2{r}_A$

${\left(a_A\right)}_n={(31.42 rad s^{-1})}^2\times 0.024 m$

${\left(a_A\right)}_n=23.69 m s^{-2}$

Consider the point on $B$ that is in contact with $C$.

$a_B=r_B\times {\alpha }_B+{\omega }_B\times ({\omega }_B\times r_B)$

${\omega }_B=28.81 rad s^{-1}$, $r_B=24 mm$, ${\alpha }_B=0$

${(a_B)}_t=0 m s^{-2}$

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

${\left(a_B\right)}_n={(28.81 rad s^{-1})}^2\times 0.024 m$

${\left(a_B\right)}_n=19.92 m s^{-2}$

The accelerations of the point on $A$ that are in contact with $C.$

${(a_A)}_t=0 m s^{-2}$, ${\left(a_A\right)}_n=23.69 m s^{-2}$

The accelerations of the point on $B$ that are in contact with $C$

$=19.92 m s^{-2}.$

${(a_B)}_t=0 m s^{-2}$, ${\left(a_B\right)}_n=19.92 m s^{-2}$

Conclusion:

The accelerations of the point on $A$ that are in contact with $C$

${(a_A)}_t=0 m s^{-2}$, ${\left(a_A\right)}_n=23.69 m s^{-2}$

The accelerations of the point on $B$ that are in contact with $C$

$=19.92 m s^{-2}.$

${(a_B)}_t=0 m s^{-2}$, ${\left(a_B\right)}_n=19.92 m s^{-2}$