Calculating Track Acceleration Due to Gravity

[robax25]

Homework Statement

Consider the situation when the ruler is to hit the ground. Which force is driving the rotation? What is the track acceleration of the bottom, in the middle of the ruler and at the top end of the ruler? Calculate it based on the individual parameters of your ruler.

Relevant Equations

g=9.81 m/s² ruler =30cm.

The force is driving the rotation is gravity. My question is that what is the track acceleration? How do I calculate it?

 

[berkeman]

Is there a drawing or picture that goes with this question? What ruler, what track, what rotation?

 

[robax25]

Is there a drawing or picture that goes with this question? What ruler, what track, what rotation?

I uploaded the picture.

 

[robax25]

I have calculated so far,
Ruler length L=30 cm. r=15cm
m=300 g. θ=90°
As gravity produces Torque, we write

Mg sinθ*r = (1/3) ML²*α
α= 49.05 rad /s²

The track acceleration at the top end of the ruler = 14.715 rad/s²

 

[haruspex]

I uploaded the picture.

I see a ruler standing at an angle on a wooden plank, and presumably falling.
I do not see anything I would describe as a track.
Is there any information regarding friction?

 

[robax25]

no.It is just angular acceleration in respect of length.The angular acceleration is at Top end and at bottom end and at midpoint. I think that at bottom, the angular acceleration is zero.

 

[haruspex]

no.It is just angular acceleration in respect of length.The angular acceleration is at Top end and at bottom end and at midpoint. I think that at bottom, the angular acceleration is zero.

I don't understand. Your post #1 asks about "track acceleration". What does that mean?
If the ruler is rigid it undergoes the same angular acceleration about its centre at all points.

 

[robax25]

it is just a tricky question. He asks that Does it happen?. Track acceleration means you just put something down . It means acceleration occurs due to gravity. Now calculate acceleration in respect of the length of the ruler. It is at top end, at mid point and at bottom. I agree with you that you get one angular acceleration at top end and others are same. Actually, from my poin of view that you are right. you get only one value and others are same.

 

[haruspex]

Track acceleration means you just put something down .

That explains nothing to me. I'll just ignore all that and assume the question wants the linear acceleration, as a vector, at those various points.
I gather that this for when the ruler is just about to hit the table, and we are to assume the ruler does not slip.

Please explain how you get that value for alpha in post #4.
Your value for "track acceleration" at the top is expressed in units of angular acceleration, but is a different number. Did you mean m/s^2? If so, what about centripetal acceleration?

Or.. maybe by track acceleration you mean tangential acceleration?

 

[robax25]

Torque = I α .yes. you get acceleration due to Torque, Torque happens as ruler falls down. I also at first think centripetal acceleration but it does not happen in this case.

 

[haruspex]

Torque = I α .yes. you get acceleration due to Torque, Torque happens as ruler falls down. I also at first think centripetal acceleration but it does not happen in this case.

You have an object rotating about a fixed point that is not its mass centre. How can there not be centripetal acceleration?

 

[robax25]

that i calculate

 

[robax25]

it is just a free fall due to gravity.Then How Do I calculate it ? IF I did mistake, show me pls.

 

[haruspex]

it is just a free fall due to gravity.Then How Do I calculate it ? IF I did mistake, show me pls.

I'm on an iPad and viewing pdf files is nontrivial. Please take the trouble to type your working.
It is not free fall. There are contact forces from the ground: normal force and, if it is not slipping, friction.

 

[robax25]

I upload again photo

 

[robax25]

L=30 cm. =0.3 m
L/2=r= 0.15 (distance of center of gravity)
m=300gm =0.3kg
θ= 90 degree.
As force due to gravity produces Torque,
Mgsinθ*r = 1/3 ML²*α
α = gsinθ*r/(1/3)L²
α= 49.05 rad/s²

 

[haruspex]

Mgsinθ*r = 1/3 ML²*α
α = gsinθ/(1/3)L²

What happened to r?

 

[robax25]

What happened to r?

I correct it.

 

[haruspex]

I correct it.

Ok, so what about

The track acceleration at the top end of the ruler = 14.715 rad/s²

Did you mean 14.715 m/s²?

 

[robax25]

I mean 49.05 rad/s²

 

[haruspex]

I mean 49.05 rad/s²

Ok, but that is the angular acceleration at the top of (anywhere along) the ruler.
We still haven't nailed down what "track acceleration" means, but asking for its value at different points along the ruler strongly suggests it is a linear acceleration that is wanted. That could be the linear tangential acceleration, the radial (i.e. centripetal) acceleration, or the overall linear acceleration (the vector sum of the two).
Do you understand that there will be centripetal acceleration? To find it, you need to know the angle at which the ruler was released.

 

[robax25]

angel is 90 degree

 

[haruspex]

angel is 90 degree

So it starts almost vertical, right?
Then what will be the centripetal acceleration when it is about to hit the ground if there is no slipping?

 

[robax25]

centripetal acceleration = V²/r, initially, V0=0 and final velocity is also V=0, it falls down because of gravitational force. How do I calculate it?. No Idea.

 

[haruspex]

final velocity is also V=0

Only after it has landed. You need the value just before landing.
What will be conserved?