Chapter 12.2, Problem 12.91P

A 1-Ib ball A and a 2-Ib ball B are mounted on a horizontal rod that rotates freely about a vertical shaft. The balls are held in the positions shown by pins. The pin holding B is suddenly removed and the ball moves to position C as the rod rotates. Neglecting friction and the mass of the rod and knowing that the initial speed of A is $v_A=8$ ft/s, determine (a) the radial and transverse components of the acceleration of ball B immediately after the pin is removed, (b) the acceleration of ball B relative to the rod at that instant, (c) the speed of ball A after ball B has reached the stop at C.

  

To determine

(a)

Radial and transverse component of acceleration of ball B immediately after the pin is removed.

Answer to Problem 12.91P

Radial component:

${\left(a_B\right)}_r=0$

Transverse component:

${\left(a_B\right)}_{\theta }=0$

Explanation of Solution

Given:

Let $r$ and $\theta$ be polar coordinates with the origin lying at the shaft.

Constraint of rod:

${\theta }_B={\theta }_A+\pi radians$

${\dot{\theta }}_B={\dot{\theta }}_A=\dot{\theta }$

${\ddot{\theta }}_B={\ddot{\theta }}_A=\ddot{\theta }$

Components of acceleration:

Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them.

Owing to frictionless sliding of B along the rod,

${\left(F_B\right)}_r=0$

Radial component of acceleration of B.

$F_r=m_B{\left(a_B\right)}_r$

${\left(a_B\right)}_r=0$

Transverse components of acceleration,

${\left(a_A\right)}_{\theta }=r_A\ddot{\theta }+2{\dot{r}}_A\dot{\theta }=ra\dot{\theta }$

${\left(a_B\right)}_{\theta }=r_B\ddot{\theta }+2{\dot{r}}_B\dot{\theta }=ra\dot{\theta }$   ... (1)

Since the rod has no mass, it must be in equilibrium.

Draw its free body diagram, applying Newton’s third law:

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∑M0=0 :

$r_A{\left(F_A\right)}_{\theta }+r_B{\left(F_B\right)}_{\theta }=r_A{m}_A{\left(a_A\right)}_{\theta }+r_B{m}_B{\left(a_B\right)}_{\theta }=0$

$r_A{m}_A{r}_A\ddot{\theta }+r_B{m}_B\left(r_B\ddot{\theta }+2{\dot{r}}_B\dot{\theta }\right)=0$

$\ddot{\theta }=\frac{-2m_B{\dot{r}}_B\dot{\theta }}{m_A{r}_A^2+m_B{r}_B^2}$

At $t=0$ ,

${\dot{r}}_B=0$

So that

$\ddot{\theta }=0$

From Eq. (1),

${\left(a_B\right)}_{\theta }=r_B\ddot{\theta }+2{\dot{r}}_B\dot{\theta }=ra\dot{\theta }$

${\left(a_B\right)}_{\theta }=0$

To determine

(b)

Acceleration of ball B relative to rod.

Answer to Problem 12.91P

Acceleration of ball B relative to rod:

${\ddot{r}}_B=61.4\text{ft}/s^2$

Explanation of Solution

Given:

Acceleration of B relative to the rod,

At $t=0$ ,

${\left(v_A\right)}_{\theta }=8\text{ft}/s=96\text{in}/s$

$\dot{\theta }=\frac{{\left(v_A\right)}_{\theta }}{r_A}=\frac{96}{10}=9.6rad/s$

${\ddot{r}}_B-r_B{\dot{\theta }}^2={\left(a_B\right)}_r=0$

${\ddot{r}}_B=r_B{\dot{\theta }}^2=\left(8\right){\left(9.6\right)}^2=737.28\text{in}/s^2$

${\ddot{r}}_B=61.4\text{ft}/s^2$

To determine

(c)

Acceleration of ball B relative to rod.

Answer to Problem 12.91P

Acceleration of ball B relative to rod:

${\left(v_A\right)}_f=2.98\text{ft}/s$

Explanation of Solution

Given:

Speed of A:

Substituting $\frac{d}{dt}\left(m{r}^2\dot{\theta }\right)$ for $r{F}_{\theta }$ in each term of moment equation gives,

$\frac{d}{dt}\left(m_A{r}_A^2\dot{\theta }\right)+\frac{d}{dt}\left(m_B{r}_B^2\dot{\theta }\right)=0$

Integrating with respect to time,

$m_A{r}_A^2\dot{\theta }+m_B{r}_B^2\dot{\theta }={\left(m_A{r}_A^2\dot{\theta }\right)}_0+{\left(m_B{r}_B^2\dot{\theta }\right)}_0$

Applying to the final state with ball B moved to the stop at C,

$\left(\frac{W_A}{g}{r}_A^2+\frac{W_B}{g}{r}_C^2\right){\dot{\theta }}_f=\left(\frac{W_A}{g}{r}_A^2+\frac{W_B}{g}{\left(r_B\right)}_0^2\right){\dot{\theta }}_0$

${\dot{\theta }}_f=\frac{W_A{r}_A^2+W_B{\left(r_B\right)}_0^2}{W_A{r}_A^2+W_B{r}_C^2}{\dot{\theta }}_0$

${\dot{\theta }}_f=\frac{\left(1\right){\left(10\right)}^2+\left(2\right){\left(8\right)}^2}{\left(1\right){\left(10\right)}^2+\left(2\right){\left(16\right)}^2}\left(9.6\right)$

${\dot{\theta }}_f=3.5765rad/s$

${\left(v_A\right)}_f=r_A{\dot{\theta }}_f=\left(10\right)\left(3.5765\right)=35.765\text{in}/s$

$\begin{array}{l}\dot{\theta }=\frac{{\left(v_A\right)}_{\theta }}{r_A}=\frac{96}{10}=9.6rad/s\\ \frac{d}{dt}\left(m_A{r}_A^2\dot{\theta }\right)+\frac{d}{dt}\left(m_B{r}_B^2\dot{\theta }\right)=0\end{array}$