A spool that rotates and slides freely while pulling a mass

[AL115]

Homework Statement

In the image.

Relevant Equations

In the image.

My attempt to solve the first one.

T-mg=ma

T=m(g+a)

TR = Iα

I=(1/2)(M)(R^2)

α=a/R

TR=(1/2)(M)(R^2)(a/R)

T=(1/2)Ma

mg+ma=1/2Ma

a=(2mg)/(M-2m)

And... I got a wrong answer. As usual.

 

[vela]

First, the minus sign in the denominator of your answer points to a sign error in your analysis.

Second, your equations (once the sign error is fixed) solve the problem where the spool rotates freely but doesn't slide. You need to account for the fact that the spool accelerates both linearly and rotationally.

 

[AL115]

First, the minus sign in the denominator of your answer points to a sign error in your analysis.

Second, your equations (once the sign error is fixed) solve the problem where the spool rotates freely but doesn't slide. You need to account for the fact that the spool accelerates both linearly and rotationally.

Thanks for the reply. I already thought of the bold but I still couldn't solve it. I don't know how the linear acceleration of the spool (which is it's center of mass acceleration right?) affect the tangential acceleration (which is equal to the acceleration of the bolck with mass m right?)

Can you show me the solution please?

 

[haruspex]

Thanks for the reply. I already thought of the bold but I still couldn't solve it. I don't know how the linear acceleration of the spool (which is it's center of mass acceleration right?) affect the tangential acceleration (which is equal to the acceleration of the bolck with mass m right?)

Can you show me the solution please?

We don't show solutions, we point out errors and provide hints.

T-mg=ma

Which way are you defining as positive for a, up or down?

α=a/R

Is this the same a as above?
This is not a wheel making rolling contact with a surface. Think about the relative motions, edge of disc where the string leaves it compared with centre of disc, centre of disc compared with table.

 

[AL115]

We don't show solutions, we point out errors and provide hints.

Which way are you defining as positive for a, up or down?

Is this the same a as above?This is not a wheel making rolling contact with a surface. Think about the relative motions, edge of disc where the string leaves it compared with centre of disc, centre of disc compared with table.

-Oh. Pardon me then I am new here so I didn't know.

-Up. Isn't the spool pulling the string?

-Yes. Isn't a equal to the tangential acceleration of the disk?

-Hmmm. I never considered relative motion. I will try to solve it again with this in mind. Thanks.

 

[haruspex]

Up. Isn't the spool pulling the string?

You can choose either way as positive as long as you are consistent. If you choose up as positive but it turns out to move down you will simply get a negative answer.

I am new here so I didn't know.

Take a moment to read the guidelines https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Yes. Isn't a equal to the tangential acceleration of the disk?

Its acceleration in the lab frame, yes.

 

[AL115]

You can choose either way as positive as long as you are consistent. If you choose up as positive but it turns out to move down you will simply get a negative answer.

Take a moment to read the guidelines https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

Its acceleration in the lab frame, yes.

I tried to solve it but no result. The spool doesn't move in an uniform circular motion when it moves linearly. But it moves in a uniform circular motion relative to the acceleration of center of mass. Right? But how can that help me when I don't know the acceleration of the center of mass? And can I still use the equations of torque in this situation?

Is it strictly forbidden to show the solution? Btw it is not even a HW. Just a problem that I want to desperately know and understand it's answer.

 

[haruspex]

I don't know the acceleration of the center of mass

What forces act on the spool? What equation can you write for linear acceleration?

can I still use the equations of torque in this situation?

Yes, so what equation can you write for angular acceleration?

 

[AL115]

What forces act on the spool? What equation can you write for linear acceleration?

Yes, so what equation can you write for angular acceleration?

For linear acceleration of the spool, I suppose MAcm where Acm = the acceleration of the center of mass. And the net torque is equal to Iα which should equal the force of the tension(T) times the radius (R) in other words(TR). Right?

The angular acceleration is equal to the tangential acceleration (At) divided by radius in other words (At/R).

My logic was that, the a is a combination of the pure rotational motion and pure linear motion.

So TR=Iα --->T=(1/2)(M)(At)

mg+ma=(1/2)(M)(At)

From the answers in the image (the one that has the problem) we can see that At = 2Acm. And Acm = 3a. Which means At=6a.

If we substitute we find that my equations are wrong.

 

[haruspex]

The angular acceleration is equal to the tangential acceleration (At) divided by radius in other words (At/R).

This is one place you are going wrong. As I mentioned, that is true for a wheel in rolling contact, and that is because the point of contact is instantaneously stationary, and its motion relative to the wheel's centre is $-r\alpha$, so $a-r\alpha=0$.
But that does not apply here. Think instead about the relationship between
- the linear acceleration of the disc's centre
- the angular acceleration of the disc
- the linear acceleration of the point on the disc's circumference where the string is being unwound.
What equation can you write connecting those? How does the last of them relate to the acceleration of the suspended mass?

You will also need to be careful with signs. These are often problematic in questions involving both linear and rotational motion. Since you have taken up as positive for a as acceleration of m, it will be positive to the left for the horizontal section of string.

 

[AL115]

This is one place you are going wrong. As I mentioned, that is true for a wheel in rolling contact, and that is because the point of contact is instantaneously stationary, and its motion relative to the wheel's centre is $-r\alpha$, so $a-r\alpha=0$.
But that does not apply here. Think instead about the relationship between
- the linear acceleration of the disc's centre
- the angular acceleration of the disc
- the linear acceleration of the point on the disc's circumference where the string is being unwound.
What equation can you write connecting those? How does the last of them relate to the acceleration of the suspended mass?

You will also need to be careful with signs. These are often problematic in questions involving both linear and rotational motion. Since you have taken up as positive for a as acceleration of m, it will be positive to the left for the horizontal section of string.

For linear acceleration of the spool, I suppose MAcm where Acm = the acceleration of the center of mass. And the net torque is equal to Iα which should equal the force of the tension(T) times the radius (R) in other words(TR). Right?

The angular acceleration is equal to the tangential acceleration (At) divided by radius in other words (At/R).

My logic was that, the a is a combination of the pure rotational motion and pure linear motion.

View attachment 254950

So TR=Iα --->T=(1/2)(M)(At)

mg+ma=(1/2)(M)(At)

From the answers in the image (the one that has the problem) we can see that At = 2Acm. And Acm = 3a. Which means At=6a.

If we substitute we find that my equations are wrong.

I just made a dumb mistake. Acm is not equal to 3a. a is equal to 3Acm. which means At is equal to (2/3)a
which means

mg+ma=(1/2)M(2/3)a --->a = 3mg/(M-3m)

which means I have a problem with the sign. So the equation should be T=mg-ma so that means that the mass is accelerating downward and not upward. right?

 

[AL115]

This is one place you are going wrong. As I mentioned, that is true for a wheel in rolling contact, and that is because the point of contact is instantaneously stationary, and its motion relative to the wheel's centre is $-r\alpha$, so $a-r\alpha=0$.
But that does not apply here. Think instead about the relationship between
- the linear acceleration of the disc's centre
- the angular acceleration of the disc
- the linear acceleration of the point on the disc's circumference where the string is being unwound.
What equation can you write connecting those? How does the last of them relate to the acceleration of the suspended mass?

You will also need to be careful with signs. These are often problematic in questions involving both linear and rotational motion. Since you have taken up as positive for a as acceleration of m, it will be positive to the left for the horizontal section of string.

This is what I came up with.

I am not sure if this is how the problem should be solved. I also relied on the fact that At = 2Acm which I got from looking at the answers. Now I need to know why At = 2Acm.

 

[haruspex]

which means I have a problem with the sign

Yes, the hanging mass will accelerate downwards, but as I wrote, it is perfectly fine to take up as positive. The sign error came in because of inconsistencies elsewhere.

With a as positive up for the hanging mass it must be positive to the left in the diagram for the spool.
With tension as positive up for the hanging mass it must be positive to the right on the spool.
Your diagrams imply you are taking anticlockwise (as viewed from above) as positive for the spool's rotation.
It remains to decide which way is positive for a_c, the linear acceleration of the spool's centre. You choose, and state your choice.

With all that in place, post the equations you get for:
- the linear acceleration of the point where the string leaves the spool, relative to the spool's centre, in terms of the angular acceleration and radius
- that same linear acceleration, but now relative to the table
- the equation relating that last to a, the upward acceleration of the hanging mass
- the equation relating the tension to the angular acceleration
- the equation relating the tension to the linear acceleration a_c

Please do not post working as images. Per forum guidelines, that is intended for diagrams and textbook extracts etc. Typing in your working makes it not only easier to read but also easier to refer to specific steps when commenting.

 

[AL115]

Yes, the hanging mass will accelerate downwards, but as I wrote, it is perfectly fine to take up as positive. The sign error came in because of inconsistencies elsewhere.

With a as positive up for the hanging mass it must be positive to the left in the diagram for the spool.
With tension as positive up for the hanging mass it must be positive to the right on the spool.
Your diagrams imply you are taking anticlockwise (as viewed from above) as positive for the spool's rotation.
It remains to decide which way is positive for a_c, the linear acceleration of the spool's centre. You choose, and state your choice.

With all that in place, post the equations you get for:
- the linear acceleration of the point where the string leaves the spool, relative to the spool's centre, in terms of the angular acceleration and radius
- that same linear acceleration, but now relative to the table
- the equation relating that last to a, the upward acceleration of the hanging mass
- the equation relating the tension to the angular acceleration
- the equation relating the tension to the linear acceleration a_c

Please do not post working as images. Per forum guidelines, that is intended for diagrams and textbook extracts etc. Typing in your working makes it not only easier to read but also easier to refer to specific steps when commenting.

I chose the downward direction in the image (the direction of the acceleration of the center of mass) to be the negative direction.

-α R
-α R+a_c
-α R=(2/3)a @bold wouldn't the hanging mass accelerate downward?
-T=(1/2)MRα
-T=MRa_c

are they correct? Also can you tell me why α R=2a_c

 

[haruspex]

α R=(2/3)a

How do you get that?
My question was the relationship between α R+a_c and a.

 

[AL115]

How do you get that?
My question was the relationship between α R+a_c and a.

I got it from the answers in the first image.

Question 13- Answer C-
α =(4mg)/(d(M+3m))

R=d/2

α R=(2mg)/(M+3m)=(2/3)a

a = α R+a_c

α R is the a_t in the last image.

 

[haruspex]

I got it from the answers in the first image.

Question 13- Answer C-
α =(4mg)/(d(M+3m))

R=d/2

α R=(2mg)/(M+3m)=(2/3)a

a = α R+a_c

α R is the a_t in the last image.

Yes, but surely you should be getting it by your own efforts.
If you follow the steps I laid out in post#13 you will arrive at this.

 

[AL115]

Yes, but surely you should be getting it by your own efforts.
If you follow the steps I laid out in post#13 you will arrive at this.

I just realized that I made a dumb mistake (again) in the post #14. I wrote the equation relating the tension to the linear acceleration a_c as T=MRa_c which is obviously wrong. It is T=Ma_c. And we also know that
T = (1/2)MRα = (1/2)Ma_t

We get that a_t=2a_c. Now everything makes sense.

My problem was that I thought that there was some external forces. One causing the spool to rotate in anti-clockwise direction (thus pulling the sting which makes the mass accelerate upward). And another causing the linear motion.

But that's wrong. The force that is causing both the linear motion and the rotational motion in the spool was the tension force!
T=Ma_c.
T = (1/2)MRα

Thank you very very much for the help.

Just one more question. I am still confused. Why could the tension force cause a linear motion when it was applied on the edge of the disk?

 

[haruspex]

Why could the tension force cause a linear motion when it was applied on the edge of the disk?

It is the only force applied, so it must cause linear acceleration according to ΣF=ma.