Chapter 15.5, Problem 15.173P

Pin P slides in a circular slot cut in the plate shown at a constant relative speed $u=90$ mm/s. Knowing that at the instant shown the plate rotates clockwise about A at the constant rate $\omega =3$ rad/s, determine the acceleration of the pin if it is located at (a) point A, (b) point B, (c) point C.

Fig. P15.173

To determine

(a)

Acceleration of pin at point A.

Answer to Problem 15.173P

The acceleration of pin P at point A is $0.621m/s^2$ upwards.

Explanation of Solution

Given information:

Constant speed $u$ of pin P relative to the plate is $90mm/s$.

Constant angular velocity $\omega$ of plate is $3rad/s$ clockwise.

The Coriolis acceleration is a combination of

$a_P=a_{P^1}+a_{P/F}+a_C$

Where

$a_P$ - Absolute acceleration of particle P.

$a_{P^1}$ - Acceleration of point P¹ of moving frame F coinciding with P.

$a_{P/F}$ - Acceleration of P relative to moving frame F.

$a_C$ -The Coriolis acceleration.

The Coriolis acceleration is defined as,

$a_C=2\Omega \times v_{P/F}$

Calculation:

Point A is the origin.

Therefore, position vector $r_A$ of point A

$r_A=0$

The relative velocity $v_{A/F}$ of point A with respect to frame

$v_{A/F}=u=0.09m/s\leftarrow$

The acceleration $a_{A^1}$ of coinciding point $A^1$ with point A

$a_{A^1}=0$

The relative acceleration $a_{A/F}$ of point A with respect to the frame

$a_{A/F}=\frac{v_{A/F}{}^2}{\rho }=\frac{{\left(0.09m/s\right)}^2}{0.1}=0.081m/s^2\uparrow$

The Coriolis acceleration $a_C$

$\begin{array}{l}{a}_C=2\omega u\\ =2\left(3rad/s\right)\left(0.09m/s\right)\\ =0.54m/s^2\uparrow \end{array}$

Therefore, acceleration $a_A$ of pin ‘P’ at point A

$\begin{array}{l}{a}_A=a_{A^1}+a_{A/F}+a_C\\ =0+\left[0.081m/s^2\uparrow \right]+\left[0.54m/s^2\uparrow \right]\\ =0.621m/s^2\uparrow \end{array}$

Conclusion:

The acceleration of pin P at point A is $0.621m/s^2$ upwards.

To determine

(b)

Acceleration of pin at point B

Answer to Problem 15.173P

The acceleration of pin P at point B is $1.767m/s^2$ at an angle $30.61°$ downwards.

Explanation of Solution

Given information:

Constant speed $u$ of pin P relative to the plate is $90mm/s$.

Constant angular velocity $\omega$ of plate is $3rad/s$ clockwise.

The Coriolis acceleration is a combination of

$a_P=a_{P^1}+a_{P/F}+a_C$

Where

$a_P$ - Absolute acceleration of particle P.

$a_{P^1}$ - Acceleration of point P¹ of moving frame F coinciding with P

$a_{P/F}$ - Acceleration of P relative to moving frame F

$a_C$ -The Coriolis acceleration

The Coriolis acceleration is defined as

$a_C=2\Omega \times v_{P/F}$

Calculation:

The position vector $r_B$ of point B

$r_B=0.1\sqrt{2}m\nwarrow 45°$

The relative velocity $v_{B/F}$ of point B with respect to frame

$v_{B/F}=u=0.09m/s\uparrow$

The acceleration $a_{B^1}$ of coinciding point $B^1$ with point B

$\begin{array}{l}{a}_{B^1}=-{\omega }^2{r}_B\\ =-{\left(3rad/s\right)}^{2}0.1\sqrt{2}m\nwarrow 45°\\ =0.9\sqrt{2}\searrow 45°\end{array}$

The relative acceleration $a_{B/F}$ of point B with respect to the frame

$a_{B/F}=\frac{v_{B/F}{}^2}{\rho }=\frac{{\left(0.09m/s\right)}^2}{0.1}=0.081m/s^2\to$

The Coriolis acceleration $a_C$

$\begin{array}{l}{a}_C=2\omega u\\ =2\left(3rad/s\right)\left(0.09m/s\right)\\ =0.54m/s^2\to \end{array}$

The acceleration $a_B$ of pin P at point B

$\begin{array}{l}{a}_B=a_{B^1}+a_{B/F}+a_C\\ =\left[0.9m/s^2\to \right]+\left[0.9m/s^2\downarrow \right]+\left[0.081m/s^2\to \right]+\left[0.54m/s^2\to \right]\\ =\left[1.521m/s^2\to \right]+\left[0.9m/s^2\downarrow \right]\end{array}$

Find the magnitude and angle $\beta$

$\begin{array}{l}{a}_B=\sqrt{{1.521}^2+{0.9}^2}=1.767m/s^2\\ \beta ={\tan }^{-1}\left(\frac{0.9}{1.521}\right)=30.61°\end{array}$

Conclusion:

The acceleration of pin P at point B is $1.767m/s^2$ at an angle $30.61°$ downwards.

To determine

(c)

Acceleration of pin at point C

Answer to Problem 15.173P

The acceleration of pin P at point C is $2.421m/s^2$ downwards.

Explanation of Solution

Given information:

Constant speed $u$ of pin P relative to the plate is $90mm/s$.

Constant angular velocity $\omega$ of plate is $3rad/s$ clockwise.

The Coriolis acceleration is a combination of

$a_P=a_{P^1}+a_{P/F}+a_C$

Where

$a_P$ - Absolute acceleration of particle P.

$a_{P^1}$ - Acceleration of point P¹ of moving frame F coinciding with P

$a_{P/F}$ - Acceleration of P relative to moving frame F

$a_C$ -The Coriolis acceleration

The Coriolis acceleration is defined as

$a_C=2\Omega \times v_{P/F}$

Calculation:

The position vector $r_C$ of point C

$r_C=0.2m\uparrow$

The relative velocity $v_{C/F}$ of point C with respect to frame

$v_{C/F}=u=0.09m/s\to$

The acceleration $a_{C^1}$ of coinciding point $C^1$ with point C

$\begin{array}{l}{a}_{C^1}=-{\omega }^2{r}_C\\ =-{\left(3rad/s\right)}^2\left(0.2m\right)\\ =1.8m/s^2\downarrow \end{array}$

The relative acceleration $a_{C/F}$ of point C with respect to the frame

$a_{C/F}=\frac{v_{C/F}{}^2}{\rho }=\frac{{\left(0.09m/s\right)}^2}{0.1}=0.081m/s^2\downarrow$

The Coriolis acceleration $a_c$

$\begin{array}{l}{a}_c=2\omega u\\ =2\left(3rad/s\right)\left(0.09m/s\right)\\ =0.54m/s^2\downarrow \end{array}$

The acceleration $a_c$ of pin P at point C

$\begin{array}{l}{a}_C=a_{C^1}+a_{C/F}+a_c\\ =\left[1.8m/s^2\downarrow \right]+\left[0.081m/s^2\downarrow \right]+\left[0.54m/s^2\downarrow \right]\\ =\left[2.421m/s^2\downarrow \right]\end{array}$

Conclusion:

The acceleration of pin P at point C is $2.421m/s^2$ downwards.