# 3.3.A

Video

Simulation

• Two rigid bodies 1 and 2 are connected by a hinge at B.

• Particle A on rigid body 1 is moving vertically with velocity $$v_A>0$$.

• Particle C on rigid body 2 is moving vertically.

• Particle D on rigid body 2 is moving horizontally.

Given symbols: $$h$$. Proceed as follows.

## Steps

### 1. Velocities

Use arrows to define:

• Velocity $$v_A$$ of a particle at A (downward = positive).

• Velocity $$v_C$$ of a particle at C (downward = positive).

• Velocity $$v_D$$ of a particle at D (to the left = positive).

Solution

### 2. Angular Velocity of Body 2

Define the angular velocity $$\omega_2$$ of rigid body 2 (clockwise = positive). Draw the respective arrow at the instant centre of rotation of body 2.

Solution

### 3. Angular Velocity of Body 1

Define the angular velocity $$\omega_1$$ of rigid body 1 (counter clockwise = positive). Draw the respective arrow at the instant centre of rotation of body 1.

Solution

### 4. Comparing Angular Velocities

Show that:

$\omega_1 = \omega_2$

Solution

$\begin{split}v_B &= r \omega_1 \\ &= r \omega_2\end{split}$

So that:

$\begin{split}r \omega_1 &= r \omega_2 \\\ \omega_1 &= \omega_2\end{split}$

### 5. Velocity of A

Show that:

$v_A = 3 v_D$

Solution

Since $$\omega = \omega_1=\omega_2$$:

$\begin{split}v_A&=3 h \omega \\ v_D&=h \omega\end{split}$

Therefore:

$v_A=3 v_D$

Find $$v_A$$ for the following given quantities:

$\begin{split}v_D &= 2.5\,\tfrac{\mathrm m}{\mathrm s} \\ h &= 10\,\mathrm m\end{split}$

Solution

$\begin{split}v_A &=3 v_D \\ &= 7.5\,\tfrac{\mathrm m}{\mathrm s}\end{split}$

### 6. Angular Velocity of Body 2

Find the angular velocity of body 2 $$\omega_2$$ in $$\tfrac{1}{\mathrm s}$$ (1 per second) for the same following given quantities:

Solution

$\begin{split}\omega_2 &= \tfrac{v_A}{3 h}\\ &= \tfrac{7.5\,\tfrac{\mathrm m}{\mathrm s}}{30\,\mathrm m} \\ &= 0.25 \,\tfrac{1}{\mathrm s}\end{split}$