Dynamics Problem: Determine Velocity and Acceleration of Rigid body

[owlwise2010]

Homework Statement

Determine Velocity and Acceleration of Rigid body: omega=40 w=4rad/s alpha=2rad/s2

Relevant Equations

v=rw
a=rw2

Hello,

Wanted to know if this was the correct approach to solve this example. I uploaded the question along with an attempt.

Thanks in advance.

 

[haruspex]

Problem Statement: Determine Velocity and Acceleration of Rigid body: omega=40 w=4rad/s alpha=2rad/s2
Relevant Equations: v=rw
a=rw2

if this was the correct approach

It is a valid approach for the speeds, but I think it is easier to get them by writing the coordinates of the points as functions of the angle and differentiating.
I get a different value for CE.

But you cannot use this instantaneous centre of rotation approach for the acceleration.

 

[collinsmark]

Problem Statement: Determine Velocity and Acceleration of Rigid body: omega=40 w=4rad/s alpha=2rad/s2
Relevant Equations: v=rw
a=rw2

Hello,

Wanted to know if this was the correct approach to solve this example. I uploaded the question along with an attempt.

Thanks in advance.

Throughout this post I'm assuming that point A is constrained to the x-axis and point B is constrained to the y-axis.

Here are a few things to consider:

If we take the numerical figures given at face value, with [itex] \theta = 40^o [/itex] and [itex] \omega = 4 \ \mathrm{\frac{rad}{sec}} [/itex], and assuming [itex] \omega [/itex] is defined as the time derivative of [itex] \theta [/itex], it means that [itex] \theta [/itex] is increasing, not decreasing. The block is not sliding down by gravity under its own volition, rather it is being shoved up against the wall! It means point A is moving to the left, not the right. At least that's true if we take the given figures at face value. If instead the box were sliding down, [itex] \omega [/itex] should be negative.

Although point C is constrained, it's not constrained to the x-axis. In other words, point C's position, [itex] v_C [/itex] and [itex] a_C [/itex] all have both x- and y-components.

Acceleration is kind of tricky here. There is acceleration not just due to the box's rotation, but also the box's translation. Be prepared to use the product rule differentiating the velocity to obtain the acceleration, as this should make it relatively straightforward -- proper use of the product rule takes care of the rotational vs. translational distinction automatically.

Follow @haruspex's advice. It's good advice.

[Edit: Although I didn't mention it, you will need to use the chain rule when differentiating too.]