- #1

thegreengineer

- 54

- 3

__vector mechanics.__The subtopic dealt is called

__kinematics of rigid bodies__.

## Homework Statement

The wheel rotates around an axis. This point of reference of the center of the wheel is labeled as "A". This wheel rotates clockwise direction with an angular velocity of 6 rad/s and a acceleration of 3 rad/s^2. Determine the angular velocity of the connecting rod labeled as "BC" and the velocity of the piston labeled as "C". The picture depicting the problem is shown below.

https://scontent-lax3-1.xx.fbcdn.net/hphotos-xap1/v/t1.0-9/12341313_1665682687049602_7296849227295710243_n.jpg?oh=6fcab686340e0f038e8fcdf7f2140baa&oe=56EEF735

## Homework Equations

The equations of relevance are:

1. [itex]\vec{V_B}=\vec{V_A}+\vec{V_{B/A}}[/itex]

2. [itex]\vec{V_{B/A}}=\vec{\omega_{AB}}\times\vec{r_{B/A}}[/itex]

3. [itex]\vec{a_B}=\vec{a_A}+\vec{a_{B/A}}[/itex]

4. [itex]\vec{a_{B/A}}=\left[\vec{\alpha_{AB}}\times\vec{r_{B/A}}\right]-\left[\vec{\omega_{AB}}^{2}\times\vec{r_{B/A}}\right][/itex]

B/A means "point B with respect to A". For example in V_B/A it means velocity of B relative with A (as an example).

## The Attempt at a Solution

First.[/B] What I did first of all was calculating the velocity at point B with the formula #1. In this case since the point A is stationary its velocity is zero. Therefore:

[itex]\vec{V_B}=\vec{V_{B/A}}[/itex]

[itex]\vec{V_B}=\vec{\omega_{AB}}\times\vec{r_{B/A}}[/itex]

I proceeded to use the values that the problem gives me. Since the rotation is clockwise, then both angular velocity and acceleration are negative on the direction of

**k**unit vector. So the angular velocity in vector notation is -6 rad/s

**k**. And for the position vector, since it's point B relatively with A then it is to measure a vector originating from point A to point B. In this case this vector goes upwards in the y-direction so it is positive on the direction of

**j**unit vector. Therefore position vector r_B/A is 0.2 m

**j.**Replacing on the previous equation we have:

[itex]\vec{V_B}=(-6\frac{rad}{s}\hat{k})\times(0.2m\hat{j})=1.2\frac{m}{s}\hat{i}[/itex]

NOTE: The i direction is obtained by cross product rules among unit vectors.

**Second**. One of the things the problem asks us to find is the angular velocity on BC segment. So similarly I used the previous equation to find the velocity now in point C.:

[itex]\vec{V_C}=\vec{V_B}+\vec{V_{C/B}}[/itex]

We already calculated velocity V_B on the previous section as we obtained that such velocity is 1.2 m/s

**i**. Then the only thing we have to find is V_C/B which is calculated by:

2. [itex]\vec{V_{C/B}}=\vec{\omega_{BC}}\times\vec{r_{C/B}}[/itex]

The angular velocity is what we are trying to find so it will remain as a unknown in the problem (however in vector mechanics when working with angular velocity and angular acceleration those vectors always go to the

**k**unit vector direction because of the right hand rule). We also know that r_C/B means position vector to C relative with B so in this case this vector goes from point B to point C and doing so goes rightwards on the x direction so in this case r_C/B= 0.8 m

**i**. Finally replacing this on the equation we have:

[itex]\vec{V_C}=1.2\frac{m}{s}\hat{i}+[\omega_{BC}\hat{k}\times0.8m\hat{i}]=\vec{V_C}=1.2\frac{m}{s}\hat{i}+0.8\omega_{BC}\hat{i}[/itex]

Now this is where my doubt comes. Someone told me that in order to find the value of angular velocity ω_BC we have to "equate" this by finding corresponding terms of the equation with their respective unit vector values so we could find a system of linear equations of chunks of this of those that are in terms of

**i**,

**j**, and

**k**unit vectors. The problem is that the piston C (which corresponds to velocity V_C) doesn't go only either on x or y direction; so it goes with an angle of 30° according to the problem.

Thanks.