# Acceleration at two points on a rigid body

The rod AB is falling while sliding on the friction-less wall and the floor. I need to find the acceleration of the points A and B.

I am sure that the point A is freely falling, so its acceleration will be g. But what about the point B. Common sense says that it should also be g as the rod is rigid and cannot be deformed. But how do I prove this using laws of physics ?

• Why are you so sure that A is free falling? For instance would this seem correct if B would be positioned almost directly underneath point A? – fibonatic Feb 2 '14 at 15:01
• If you claim to know the motion of A, then you don't need any physics to get the motion of B. It is purely geometry. – Brian Moths Feb 2 '14 at 15:07
• Doing this by forces is relatively difficult. Given the frictionless assumption, doing it by energy conservation becomes attractive. Only, you'll have to work it with an eye on @NowIGet 's geometric constraints. – dmckee --- ex-moderator kitten Feb 2 '14 at 15:10

Let $a_1$ be the acceleration of point $A$ and $a_2$ be that of $B$.
Let $x$ and $y$ be the coordinates of $B$ and $A$ respectivly.
By pythagorean theorem $x^2+y^2=L^2 \tag{1}$
Where $L$ is the length of the rod.
Now to find a relation b/w $a_1$ and $a_2$ you need to differentiate eqn 1 twice w.r.t time.
Differentiating once gives
$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$
again differentiating gives
$2x\dfrac{d^2x}{dt^2}+2[\dfrac{dx}{dt}]^2+2y\dfrac{d^2y}{dt^2}+2(dy/dt)^2=0$
Let $v_1$ be the velocity of $A$ and $v_2$ be that of $B$.
Then will will have
$xa_1+{v_1}^2+ya_2+{v_2}^2=0$.

In general $a_1$ is a variable so it is not equal to $g$. For details read this and some other related posts.

As the author of this answer, I recommend you track the center of mass C with the equations of motion for rigid body and then find the end point motion at A and B.

Once the linear and angular acceleration of the center of mass $\vec{a}_C$ and $\vec{\alpha}$ is calculated then follow the formula for rigid body kinematics to get the acceleration at the end points.

For example for point A use:

\begin{aligned} \vec{v}_A & = \vec{v}_C+ \vec{\omega}\times(\vec{r}_A-\vec{r}_C) \\ \vec{a}_A & = \vec{a}_C + \vec{\alpha}\times(\vec{r}_A-\vec{r}_C) + \vec{\omega}\times(\vec{v}_A-\vec{v}_C) \end{aligned}

Alternatively you can start from the end point velocities

$$v_A = \dot{\theta} \ell \sin \theta \\ v_B = \dot{\theta} \ell \cos \theta$$

where $\theta$ is the rod angle from vertical and take the total derivative with respect to time to get the the accelerations at the end. To find the result you will need the angular speed and acceleration which you get from the rigid body equations of motion.