Chapter 15.7, Problem 15.244P

A square plate of side 2r is welded to a vertical shaft that rotates with a constant angular velocity ${\omega }_1$ . At the same time, rod AB of length r rotates about the center of the plate with a constant angular velocity ${\omega }_2$ with respect to the plate. For the position of the plate shown, determine the acceleration of end B of the rod if $(a)\theta =0,(b)\theta =90°,(c)\theta =180°$ .

Fig. P15.244

To determine

(a)

Acceleration of end B if $\theta =0$.

Answer to Problem 15.244P

The acceleration $a_B$ of point B is $r{\omega }_2{}^2\sin 30°j-\left(r{\omega }_2{}^2\cos 30°+2r{\omega }_1{\omega }_2\right)k$.

Explanation of Solution

Given information:

Constant angular velocity of vertical shaft is ${\omega }_1$.

The relative angular velocity of rod AB with respect to the plate is ${\omega }_2$.

The acceleration in a three-dimensional motion of a particle relative to rotating frame is defined as:

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

In the above equation,

$a_P$ - 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:

Angular velocity $\Omega$ of rotating frame of reference

$\Omega ={\omega }_{1}j$

The relative angular velocity ${\omega }_2$ with respect to the reference frame

${\omega }_2={\omega }_2\left(\cos 30°j+\sin 30°k\right)$

The relative position vector $r_{A/O}$ of point A with respect to point O

$r_{A/O}=r\left(\sin 30°j-\cos 30°k\right)$

The relative position vector $r_{B/O}$ of point B with respect to point O

$r_{B/O}=r_{A/O}+r_{B/A}$

Assume $B^1$ as the coinciding point on frame with point B.

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

$a_{B^1}=\Omega \times v_{B^1}=\Omega \times \left(\Omega \times r_{B/O}\right)$

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

$a_{B/F}={\omega }_2\times v_{B/F}=-{\omega }_2{}^2{r}_{B/A}$

At $\theta =0$ ,

The relative position vector $r_{B/A}$ of point B with respect to point A

$r_{B/A}=r\left(-\sin 30°j+\cos 30°k\right)$

The relative position vector $r_{B/O}$ of point B with respect to point O

$r_{B/O}=0$

Therefore,

$a_{B^1}=\Omega \times \left(\Omega \times r_{B/O}\right)=0$

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

$\begin{array}{l}{v}_{B/F}={\omega }_2\times r_{B/A}\\ ={\omega }_2\left(\cos 30°j+\sin 30°k\right)\times r\left(-\sin 30°j+\cos 30°k\right)\\ =r{\omega }_{2}i\end{array}$

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

$a_{B/F}={\omega }_2\times v_{B/F}=r{\omega }_2{}^2\left(\sin 30°j-\cos 30°k\right)$

The Coriolis acceleration

$\begin{array}{l}2\Omega \times v_{B/F}=2\left({\omega }_{1}j\right)\times \left(r{\omega }_{2}i\right)\\ =-2r{\omega }_1{\omega }_{2}k\end{array}$

Acceleration $a_B$ of point B

$\begin{array}{l}{a}_B=a_{B^1}+a_{B/F}+2\Omega \times v_{B/F}\\ =0+r{\omega }_2{}^2\left(\sin 30°j-\cos 30°k\right)-2r{\omega }_1{\omega }_{2}k\\ =r{\omega }_2{}^2\sin 30°j-\left(r{\omega }_2{}^2\cos 30°+2r{\omega }_1{\omega }_2\right)k\end{array}$

Conclusion:

The acceleration $a_B$ of point B is $r{\omega }_2{}^2\sin 30°j-\left(r{\omega }_2{}^2\cos 30°+2r{\omega }_1{\omega }_2\right)k$.

To determine

(b)

Acceleration of end B if $\theta =90°$

Answer to Problem 15.244P

The acceleration $a_B$ of point B is $-r\left({\omega }_1{}^2+{\omega }_2{}^2+2{\omega }_1{\omega }_2\cos 30°\right)+r{\omega }_1{}^2\cos 30°k$.

Explanation of Solution

Given information:

Constant angular velocity of vertical shaft is ${\omega }_1$.

The relative angular velocity of rod AB with respect to the plate is ${\omega }_2$.

The acceleration in a three dimensional motion of a particle relative to rotating frame is defined as:

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

In the above equation,

$a_P$ - 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:

At $\theta =90°$ ,

The relative position vector $r_{B/A}$ of point B with respect to point A

$r_{B/A}=ri$

The relative position vector $r_{B/O}$ of point B with respect to point O

$r_{B/O}=ri+r\sin 30°j-r\cos 30°k$

Therefore,

$\begin{array}{l}{a}_{B^1}=\Omega \times \left(\Omega \times r_{B/O}\right)\\ ={\omega }_{1}j\times \left[{\omega }_{1}j\times \left(ri+r\sin 30°j-r\cos 30°k\right)\right]\\ =-r{\omega }_1{}^{2}i+r{\omega }_1{}^2\cos 30°k\end{array}$

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

$\begin{array}{l}{v}_{B/F}={\omega }_2\times r_{B/A}\\ ={\omega }_2\left(\cos 30°j+\sin 30°k\right)\times ri\\ =r{\omega }_2\left(\sin 30°j-\cos 30°k\right)\end{array}$

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

$\begin{array}{l}{a}_{B/F}={\omega }_2\times v_{B/F}\\ ={\omega }_2\left(\cos 30°j+\sin 30°k\right)\times r{\omega }_2\left(\sin 30°j-\cos 30°k\right)\\ =-r{\omega }_2{}^{2}i\end{array}$

The Coriolis acceleration

$\begin{array}{l}2\Omega \times v_{B/F}=2\left({\omega }_{1}j\right)\times \left(r{\omega }_2\left(\sin 30°j-\cos 30°k\right)\right)\\ =-2r{\omega }_1{\omega }_2\cos 30°i\end{array}$

Acceleration $a_B$ of point B

$\begin{array}{l}{a}_B=a_{B^1}+a_{B/F}+2\Omega \times v_{B/F}\\ =-r{\omega }_1{}^{2}i+r{\omega }_1{}^2\cos 30°k-r{\omega }_2{}^{2}i-2r{\omega }_1{\omega }_2\cos 30°i\\ =-r\left({\omega }_1{}^2+{\omega }_2{}^2+2{\omega }_1{\omega }_2\cos 30°\right)+r{\omega }_1{}^2\cos 30°k\end{array}$

Conclusion:

The acceleration $a_B$ of point B is $-r\left({\omega }_1{}^2+{\omega }_2{}^2+2{\omega }_1{\omega }_2\cos 30°\right)+r{\omega }_1{}^2\cos 30°k$.

To determine

(c)

Acceleration of end B if $\theta =180°$

Answer to Problem 15.244P

The acceleration $a_B$ of point B is $-r{\omega }_2{}^2\sin 30°j+r\left(2{\omega }_1{}^2\cos 30°+{\omega }_2{}^2\cos 30°+2{\omega }_1{\omega }_2\right)k$.

Explanation of Solution

Given information:

Constant angular velocity of vertical shaft is ${\omega }_1$.

The relative angular velocity of rod AB with respect to the plate is ${\omega }_2$.

The acceleration in a three dimensional motion of a particle relative to rotating frame is defined as:

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

In the above equation,

$a_P$ - 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:

At $\theta =180°$

The relative position vector $r_{B/A}$ of point B with respect to point A

$r_{B/A}=r\left(\sin 30°j-\cos 30°k\right)$

The relative position vector $r_{B/O}$ of point B with respect to point O

$r_{B/O}=2r\left(\sin 30°j-\cos 30°k\right)$

Therefore,

$\begin{array}{l}{a}_{B^1}=\Omega \times \left(\Omega \times r_{B/O}\right)\\ ={\omega }_{1}j\times \left[{\omega }_{1}j\times \left(2r\left(\sin 30°j-\cos 30°k\right)\right)\right]\\ =2r{\omega }_1{}^2\cos 30°k\end{array}$

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

$\begin{array}{l}{v}_{B/F}={\omega }_2\times r_{B/A}\\ ={\omega }_2\left(\cos 30°j+\sin 30°k\right)\times r\left(\sin 30°j-\cos 30°k\right)\\ =-r{\omega }_{2}i\end{array}$

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

$a_{B/F}={\omega }_2\times v_{B/F}=r{\omega }_2{}^2\left(-\sin 30°j+\cos 30°k\right)$

The Coriolis acceleration

$\begin{array}{l}2\Omega \times v_{B/F}=2\left({\omega }_{1}j\right)\times \left(-r{\omega }_{2}i\right)\\ =2r{\omega }_1{\omega }_{2}k\end{array}$

Acceleration $a_B$ of point B

$\begin{array}{l}{a}_B=a_{B^1}+a_{B/F}+2\Omega \times v_{B/F}\\ =2r{\omega }_1{}^2\cos 30°k+r{\omega }_2{}^2\left(-\sin 30°j+\cos 30°k\right)+2r{\omega }_1{\omega }_{2}k\\ =-r{\omega }_2{}^2\sin 30°j+r\left(2{\omega }_1{}^2\cos 30°+{\omega }_2{}^2\cos 30°+2{\omega }_1{\omega }_2\right)k\end{array}$

Conclusion:

The acceleration $a_B$ of point B is $-r{\omega }_2{}^2\sin 30°j+r\left(2{\omega }_1{}^2\cos 30°+{\omega }_2{}^2\cos 30°+2{\omega }_1{\omega }_2\right)k$.