A rod sliding/falling down on a frictionless horizontal plane

[Sam Jelly]

Homework Statement

Find the acceleration of the rod’s CM with respect to the frictionless floor. The rod initial angle with the floor is θ . The rod length is L.

Relevant Equations

A = r(dw/dt)

This is the answer to this question why?

My attempt :

 

[kuruman]

Your acceleration is independent of the acceleration of gravity $g$. Does this make sense to you? I would draw a free body diagram.

 

[haruspex]

This is the answer to this question why?

Are you saying the given answer to "the question" is $a_y=\frac 12L\cos(\theta)\dot\omega$? If so, the question must be "find the relationship between the linear and angular accelerations", right?

If so, I agree with you. The given answer would be for the case where the rod is not slipping.

 

[Sam Jelly]

Are you saying the given answer to "the question" is $a_y=\frac 12L\cos(\theta)\dot\omega$? If so, the question must be "find the relationship between the linear and angular accelerations", right?

If so, I agree with you. The given answer would be for the case where the rod is not slipping.

Yes, the question is to find the linear acceleration in term of the angular acceleration. But my answer seems to be different.

 

[haruspex]

Yes, the question is to find the linear acceleration in term of the angular acceleration. But my answer seems to be different.

Wait… is the question only asking for the initial acceleration when the rod is released from rest? At that time, $\omega=0$, so the answers are the same.

 

[Sam Jelly]

Wait… is the question only asking for the initial acceleration when the rod is released from rest? At that time, $\omega=0$, so the answers are the same.

It’s the acceleration at any point in time.

 

[haruspex]

It’s the acceleration at any point in time.

Ok… but why then does it say theta is the initial angle, not the angle at an arbitrary time?
Btw, you do have a sign error, so your answer is not right either.

 

[Sam Jelly]

Ok… but why then does it say theta is the initial angle, not the angle at an arbitrary time?
Btw, you do have a sign error, so your answer is not right either.

Oh yes I forgot the -sin theta.
I was also confused, at first I thought they ignored the -L/2 w^2 sin theta because it’s like in the radial direction.

 

[haruspex]

Oh yes I forgot the -sin theta.
I was also confused, at first I thought they ignored the -L/2 w^2 sin theta because it’s like in the radial direction.

Ok.
Just out of interest, here's another way to arrive at your answer.
If we add a horizontal acceleration so that the point of contact with the ground does not move, it won’t change the kinematic relationship we are interested in. In the resulting rotation, the tangential acceleration is $\frac 12L\dot\omega$, while radial acceleration is $-\frac 12L\omega^2$. Summing the vertical components yields your (corrected) answer.