Chapter 15.5, Problem 15.176P

To determine

Angular velocity and angular acceleration of rod attached at point B.

Answer to Problem 15.176P

The angular velocity ${\omega }_{BP}$ of rod BP is $7.8612rad/s$ counter clockwise.

The angular acceleration ${\alpha }_{BP}$ of rod BP is $81.146rad/s^2$ counter clockwise.

Explanation of Solution

Given information:

Angular velocity of rod AP is $5rad/s$ counter clockwise.

Angular acceleration of rod AP is $2rad/s^2$ clockwise.

Plane motion of a particle relative to a rotating frame is defined as,

$v_P=v_{P^1}+v_{P/F}$

In the above equation,

$v_P$ - Absolute velocity of particle P.

$v_{P^1}$ - Velocity of point $P^1$ of moving frame F Coinciding with P.

$v_{P/F}$ - Velocity of P relative to moving frame F.

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:

Apply sine rule for triangle ABP

$\frac{0.2}{\sin 45°}=\frac{AP}{\sin 110°}=\frac{BP}{\sin 25°}$

Therefore,

$\begin{array}{l}AP=\frac{0.2}{\sin 45°}\sin 110°=0.265785m\\ BP=\frac{0.2}{\sin 45°}\sin 25°=0.119534m\end{array}$

The relative position $r_{P/A}$ of point P relative to point A

$r_{P/A}=0.265785\left(\sin 25°i+\cos 25°j\right)=0.112325i+0.24088j$

The relative position $r_{P/A}$ of point P relative to point A

$r_{P/B}=0.119534\left(\sin 70°i+\cos 70°j\right)=0.112325i+0.04088j$

Velocity $v_P$ of point P

$\begin{array}{l}{v}_P={\omega }_{AP}\times r_{P/A}\\ =\left(5k\right)\times \left[0.112325i+0.24088j\right]\\ =-1.2044i+0.561625j\end{array}$

Acceleration $a_P$ of point P

$\begin{array}{l}{a}_P={\alpha }_{AP}\times r_{P/A}-{\omega }_{AP}{}^2{r}_{P/A}\\ =\left(-2k\right)\times \left[0.112325i+0.24088j\right]-\left(5^2\right)\left[0.112325i+0.24088j\right]\\ =-0.22465j+0.48176i-2.808125i-6.022j\\ =-2.3265i-6.2467j\end{array}$

The relative velocity $v_{P/BP}$ of collar P with respect to rod BD

$v_{P/BP}=u\left(\cos 20°i+\sin 20°j\right)$

The relative acceleration $a_{P/BP}$ of collar P with respect to rod BD

$a_{P/BP}=\dot{u}\left(\cos 20°i+\sin 20°j\right)$

Assume rod BP as the rotating frame of reference.

The velocity $v_{P^1}$ of coinciding point $P^1$ with collar P

Assume ${\omega }_{BP}$ as the angular velocity of rod BP.

$\begin{array}{l}{v}_{P^1}={\omega }_{BP}\times r_{P/B}\\ ={\omega }_{BP}k\times \left[0.112325i+0.04088j\right]\\ =-0.04088{\omega }_{BP}i+0.112325{\omega }_{BP}j\end{array}$

The acceleration $a_{P^1}$ of coinciding point $P^1$

Assume ${\alpha }_{BP}$ as the angular acceleration of rod BP

$\begin{array}{l}{a}_{P^1}={\alpha }_{BP}\times r_{P/B}-{\omega }_{BP}{}^2{r}_{P/B}\\ ={\alpha }_{BP}k\times \left[0.112325i+0.04088j\right]-{\omega }_{BP}{}^2\left[0.112325i+0.04088j\right]\\ =-0.04088{\alpha }_{BP}i+0.112325{\alpha }_{BP}j-0.112325{\omega }_{BP}{}^{2}i-0.04088{\omega }_{BP}{}^{2}j\end{array}$

The velocity $v_P$ of collar P

$\begin{array}{l}{v}_P=v_{P^1}+v_{P/BP}\\ -1.2044i+0.561625j=-0.04088{\omega }_{BP}i+0.112325{\omega }_{BP}j+u\left(\cos 20°i+\sin 20°j\right)\end{array}$

Equate components

$\begin{array}{l}i:-1.2044=-0.04088{\omega }_{BP}+u\cos 20°\\ j:0.561625=0.112325{\omega }_{BP}+u\sin 20°\end{array}$

Solve above equations

$\begin{array}{l}u=-0.93971m/s\\ {\omega }_{BP}=7.8612rad/s\end{array}$

Therefore, the relative velocity $v_{P/BP}$

$v_{P/BP}=\left(-0.93971m/s\right)\left(\cos 20°i+\sin 20°j\right)$

The Coriolis acceleration $a_C$

$\begin{array}{l}{a}_C=2{\omega }_{BP}\times v_{P/BP}\\ =2\left(7.8612rad/s\right)k\times \left(-0.93971m/s\right)\left(\cos 20°i+\sin 20°j\right)\\ =\left(5.0532m/s^2\right)i-\left(13.8835m/s^2\right)j\end{array}$

Acceleration $a_P$ of point P

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

Equate components

$\begin{array}{l}i:-2.3265=-0.04088{\alpha }_{BP}-0.112325{\omega }_{BP}{}^2+\dot{u}\cos 20°+5.0532\\ j:-6.2467=0.112325{\alpha }_{BP}-0.04088{\omega }_{BP}{}^2+\dot{u}sin20°-13.8835\end{array}$

Solve above equations

$\begin{array}{l}\dot{u}=3.0641m/s^2\\ {\alpha }_{BP}=81.146rad/s^2\end{array}$

Conclusion:

The angular velocity ${\omega }_{BP}$ of rod BP is $7.8612rad/s$ counter clockwise.

The angular acceleration ${\alpha }_{BP}$ of rod BP is $81.146rad/s^2$ counter clockwise.