Acceleration of a rotational and translational system

[songoku]

Homework Statement

Find the acceleration of each object. Note: the question mark in the diagram should be θ (angle of inclined plane)

2. Homework Equations
τ = F . r
Στ = I . α
α = a / R
ΣF = m . a

The Attempt at a Solution

For object A:
ΣF = m . a
T_A - W_A sin θ = m_A . a_A
T_A - m_A . g. sin θ = m_A . α_A . r₁

Στ = I_A . α_A
T_A . r₁ = I_A . α_AFor object B:
ΣF = m . a
W_B - T_B = m_B . a_B
m_B . g - T_B = m_B . α_pulley . r₃For pulley:
Στ = I_pulley . α_pulley
T_B . r₃ - T_A . r₄ = I_pulley . α_pulleyMy questions:
1. Are my equations above correct?

2. Is it correct that α_A ≠ α_pulley?

3. Can I state that a_A = α_A . r₁ = α_pulley . r₄?

Thanks

 

[TSny]

My questions:
1. Are my equations above correct?

If object A rolls without slipping, there is a force on A that helps prevent slipping which you left out.

2. Is it correct that α_A ≠ α_pulley?

Yes! That's correct.

3. Can I state that a_A = α_A . r₁ = α_pulley . r₄?

No.

For the first equality, you have a_A = α_A r₁. Is r₁ the radius that you should use here? Consider the condition for rolling without slipping of a wheel.

For the second equality, you have a_A = α_pulley r₄. The relation between a_A and α_pulley is a little more complicated than this. (I assume that a_A represents the acceleration of the center of object A.) It might help to consider the two colored spots painted on the sloping string shown below

Are the linear accelerations of the red and blue spots the same?

 

[songoku]

If object A rolls without slipping, there is a force on A that helps prevent slipping which you left out.

You mean friction between A and the plane? Can the torque for rolling being produced by tension A only?

No. No.

So, α_A will be the same as α_pulley?

For the first equality, you have a_A = α_A r₁. Is r₁ the radius that you should use here? Consider the condition for rolling without slipping of a wheel.

I am not sure I get the hint. Is it correct if I use r₂ instead of r₁ because r₂ is the radius of rolling?

For the second equality, you have a_A = α_pulley r₄. The relation between a_A and α_pulley is a little more complicated than this. (I assume that a_A represents the acceleration of the center of object A.) It might help to consider the two colored spots painted on the sloping string shown below
View attachment 109619Are the linear accelerations of the red and blue spots the same?

I don't think the linear accelerations of the red and blue spots are the same. The linear acceleration of blue spot should be higher than red spot because blue spot will cover more distance in one rotation. The ratio of their linear accelerations will be the same as the ratio of their radius? (a_red / a_blue = r₁ / r₄?Can I also state that a_A / a_B = r₁ / r₃?

Thanks

 

[TSny]

You mean friction between A and the plane?

Yes.

Can the torque for rolling being produced by tension A only?

For general values of the masses, etc., it seems unlikely that the friction force would be zero.

So, α_A will be the same as α_pulley?

No, they won't be the same. Sorry, when I read this part of your post I saw the symbol =, whereas you actually had ≠. I edited my earlier post.

I am not sure I get the hint. Is it correct if I use r₂ instead of r₁ because r₂ is the radius of rolling?

Yes, that's right.

I don't think the linear accelerations of the red and blue spots are the same.

Note that there is a fixed length of string between the two dots. We assume the string cannot stretch.

Can I also state that a_A / a_B = r₁ / r₃?

No.

 

[songoku]

Yes.
For general values of the masses, etc., it seems unlikely that the friction force would be zero.

If friction is considered, the equation will be:
ΣF = m . a
T_A - m_A . g. sin θ - friction = m_A . a_A
T_A - m_A . g. sin θ - friction = m_A . α_A . r₂

Στ = I_A . α_A
T_A . r₁ + f . r₂ = I_A . α_A

Note that there is a fixed length of string between the two dots. We assume the string cannot stretch.

No.

Are their linear accelerations the same? They both have different angular accelerations and radius so it is possible for them to have same linear acceleration?

Thanks

 

[TSny]

If friction is considered, the equation will be:
ΣF = m . a
T_A - m_A . g. sin θ - friction = m_A . a_A
T_A - m_A . g. sin θ - friction = m_A . α_A . r₂

Στ = I_A . α_A
T_A . r₁ + f . r₂ = I_A . α_A

Yes.

Are their linear accelerations the same? They both have different angular accelerations and radius so it is possible for them to have same linear acceleration?

Yes. Yes.

 

[songoku]

Yes.

Yes. Yes.

Object with radius r₃ and r₄ will have same α, so:
a_blue / a_B = r₄ / r₃ ; a_blue = a_red = α_A . r₂

a_B = α_A . r₂ . r₃ / r₄

 

[TSny]

a_blue / a_B = r₄ / r₃

OK

a_blue = a_red = α_A . r₂

a_red ≠ α_A . r₂
α_A . r₂ gives the acceleration of the red dot relative to the center of object A. But a_red denotes the acceleration of the red dot relative to the earth.

 

[songoku]

OK

a_red ≠ α_A . r₂
α_A . r₂ gives the acceleration of the red dot relative to the center of object A. But a_red denotes the acceleration of the red dot relative to the earth.

So a_A = α_A . r₂ that is used in the equation of Newton's law (ΣF = m.a) is the acceleration with respect to the center of object A, not with respect to earth?

a_B = α_pulley . r₃
a_blue . r₃ / r₄ = α_pulley . r₃
a_blue = α_pulley . r₄
a_red = α_pulley . r₄

a_red / a_A = r₁ / r₂ ?

Thanks

 

[TSny]

So a_A = α_A . r₂ that is used in the equation of Newton's law (ΣF = m.a) is the acceleration with respect to the center of object A, not with respect to earth?

a_A is the acceleration of the center of object A relative to the earth. For rolling without slipping, a_A = α_A . r₂ .

a_B = α_pulley . r₃
a_blue . r₃ / r₄ = α_pulley . r₃
a_blue = α_pulley . r₄
a_red = α_pulley . r₄

All of this looks good.

a_red / a_A = r₁ / r₂ ?

This is not correct.
One way to get the relation between a_red and a_A is to use the idea of relative motion. You can think of the (tangential) acceleration of the red point with respect to the Earth as the (tangential) acceleration of the red point relative to the center of object A plus the acceleration of the center of object A with respect to the earth:
a_red = a_red/A + a_A.

 

[songoku]

This is not correct.
One way to get the relation between a_red and a_A is to use the idea of relative motion. You can think of the (tangential) acceleration of the red point with respect to the Earth as the (tangential) acceleration of the red point relative to the center of object A plus the acceleration of the center of object A with respect to the earth:
a_red = a_red/A + a_A.

a_red/A = α_A . r₁

a_red = a_red/A + a_A
a_red = α_A . r₁ + a_A
a_red = (a_A / r₂) . r₁ + a_A
a_A = a_red . r₂ / (r₁ + r₂) ; a_red = a_blue = a_B . r₄ / r₃
a_A = a_B . r₂ . r₄ / ((r₁ + r₂) . r₃)

Is this correct?

Thanks

 

[TSny]

a_A = a_B . r₂ . r₄ / ((r₁ + r₂) . r₃)

Is this correct?

Yes. Good.

 

[songoku]

Yes. Good.

and the relation between α_A and α_pulley:

α_pulley = α_A . (r₁ + r₂) / r₄ ?

I have another question similar to this one. I will do it first and then post it again. Thanks

 

[songoku]

Let the red spot and blue spot the same as question before:
a_red = a_blue = a_C

a_red = a_A . (r₁ + r₂) / r₂
a_C = a_A . (r₁ + r₂) / r₂

a_B / r₃ = a_C / r₄
a_B = a_A . (r₁ + r₂) . r₃ / (r₂ . r₄)

α_pulley = α_A . (r₁ + r₂) / r₄

Is this good?

Thanks

 

[TSny]

and the relation between α_A and α_pulley:

α_pulley = α_A . (r₁ + r₂) / r₄ ?

Looks right.

 

[TSny]

Let the red spot and blue spot the same as question before:
a_red = a_blue = a_C

a_red = a_A . (r₁ + r₂) / r₂
a_C = a_A . (r₁ + r₂) / r₂

a_B / r₃ = a_C / r₄
a_B = a_A . (r₁ + r₂) . r₃ / (r₂ . r₄)

α_pulley = α_A . (r₁ + r₂) / r₄

I think that's all correct.

 

[songoku]

Let red spot is the center of object and blue spot is the point where the tension touches object B

a_red = a_blue = a_A

a_A = α_A . r_A

a_blue = 2 a_B , so a_A = 2 a_B

Then:
a_A = 2 a_B
α_A . r_A = 2 . α_B . r_B

and: α_pulley = α_A . r_A / r_PIs this correct? Sorry for the trouble. My teacher didn't teach about this type of question but this is given as practice for the test today. Thanks

 

[TSny]

Let red spot is the center of object and blue spot is the point where the tension touches object B

a_red = a_blue = a_A

a_A = α_A . r_A

a_blue = 2 a_B , so a_A = 2 a_B

Then:
a_A = 2 a_B
α_A . r_A = 2 . α_B . r_B

and: α_pulley = α_A . r_A / r_PIs this correct?

It all looks correct to me. Good luck on your test!

 

[songoku]

It all looks correct to me. Good luck on your test!

The question that came out was the third picture I posted and unfortunately the teacher a_A = a_B without any further explanation why...Thank you very much for your help TSny

 

[conscience]

The question that came out was the third picture I posted and unfortunately the teacher a_A = a_B without any further explanation why...

Because of string constraint , the acceleration of center of A is equal to the acceleration of the topmost point of B .

 

[TSny]

The question that came out was the third picture I posted and unfortunately the teacher a_A = a_B without any further explanation why...

As conscience stated, and as you showed in post #17, the acceleration of the center of A equals the acceleration of the point where the string meets the top of B. If the teacher stated that a_A = a_B, could it be that "B" refers to the top of m_B? Anyway, I still think your analysis in post 17 is correct.