Acceleration of system related to rolling motion and pulley

[songoku]

Homework Statement

Sphere A of mass 2 kg is connected to object B (4 kg) through a pulley (0.5 kg). If sphere A rolls without slipping, find the acceleration of the system.
(Radius of pulley = 20 cm)

Relevant Equations

Στ = I . α

ΣF = m.a

There is no friction mentioned by the question so I assume the plane is frictionless but can the sphere roll without slipping if there is no friction?

This is my attempt:
Equation of translation motion of object A (assuming A moves upwards):
T_A - W_A sin θ = m_A . a_{COM (A)}
T_A = m_A . a_{COM (A)} + W_A sin θ ...(1)

Equation of rotational motion of object A:
T_A . R = $\frac{2}{5}$ m_AR² . $\frac{a_{tangential(A)}}{R}$
T_A = $\frac{2}{5}$ m_A . a_{tangential (A)} ...(2)

Since A rolls without slipping → a_{COM (A)} = a_{tangential (A)}

Equating (1) and (2), I get a_{COM (A)} = - 13 ms^-2 → is this value possible? bigger than acceleration of free fall?

So this means sphere A rolls downwards, but the acceleration of the point of contact between the sphere and inclined plane (let say point X) is not zero so the sphere is not rolling without slipping with respect to point X. Does this mean the question refers the motion of rolling without slipping actually with respect to the point where the tension acts on the sphere / top part of the sphere (let say point Y)?

But the acceleration of point Y on sphere A should be the same as the acceleration of the center of mass of object B so the acceleration of B is zero?

I am really confused. Where is my mistake?

And by "acceleration of system", what does the question refer to? Is it acceleration of center of mass of A or of B?

Thanks

EDIT: or maybe actually this question is about static condition? None is moving?

 

[Lnewqban]

It must be friction for the sphere to roll.
B is heavier than A; therefore, even if the angle is 90 degrees, A will roll up and B will move dowmwards.
B has mechanical advantage of 2 respect to the sphere; therefore, the sphere will move half as quick as B.
The rotational inertia of both, the fixed puley and the roling sphere, will slow down the movement of B.

 

[erobz]

The plane/wheel must have friction to roll without slipping. But it is static friction, so it's not doing non conservative work on the sphere.

1) I think since the pulley has mass (and size); you are going to have to account for the change in Tension around it. i.e the tension acting on the hanging mass is not the tension acting on the rolling mass.

2) The acceleration of the hanging mass is not equivalent to the acceleration of the CoM of the sphere, which is $a_{CoM} = r_s \alpha$

 

[songoku]

It must be friction for the sphere to roll.
B is heavier than A; therefore, even if the angle is 90 degrees, A will roll up and B will move dowmwards.
B has mechanical advantage of 2 respect to the sphere; therefore, the sphere will move half as quick as B.
The rotational inertia of both, the fixed puley and the roling sphere, will slow down the movement of B.

The plane/wheel must have friction to roll without slipping. But it is static friction, so it's not doing non conservative work on the sphere.

1) I think since the pulley has mass (and size); you are going to have to account for the change in Tension around it. i.e the tension acting on the hanging mass is not the tension acting on the rolling mass.

2) The acceleration of the hanging mass is not equivalent to the acceleration of the CoM of the sphere, which is $a_{CoM} = r_s \alpha$

By "acceleration of system", is it acceleration of CoM of B?

Thanks

 

[malawi_glenn]

By "acceleration of system", is it acceleration of CoM of B?

You think B and the CoM of the sphere will have different acceleration?

 

[erobz]

By "acceleration of system", is it acceleration of CoM of B?

Thanks

Put a coordinate between the point of tangency(s) of the pulley and the rolling mass along the length of the rope.. The acceleration of that coordinate is the acceleration of the hanging mass, which is not equivalent to the acceleration of the mass center of the rolling mass.

 

[songoku]

You think B and the CoM of the sphere will have different acceleration?

Yes, I think B has twice the acceleration of CoM of the sphere

Put a coordinate between the point of tangency(s) of the pulley and the rolling mass along the length of the rope.. The acceleration of that coordinate is the acceleration of the hanging mass, which is not equivalent to the acceleration of the mass center of the rolling mass.

Yes but when the question asks about the acceleration of the system, which acceleration does the question ask, the hanging mass or the CoM of the rolling sphere or both of them? I thought "acceleration of system" always refers to only one value

Thanks

 

[erobz]

Yes, I think B has twice the acceleration of CoM of the sphereYes but when the question asks about the acceleration of the system, which acceleration does the question ask, the hanging mass or the CoM of the rolling sphere or both of them? I thought "acceleration of system" always refers to only one value

Thanks

Oh...Yeah, that's a little problematic. I probably wouldn't worry about it all that much, and just continue to solve the problem. Do you have the answer(s) from the key?

 

[kuruman]

Yes but when the question asks about the acceleration of the system, which acceleration does the question ask, the hanging mass or the CoM of the rolling sphere or both of them? I thought "acceleration of system" always refers to only one value

Good question. This is an exercise in physics, not in mind-reading. The author of the problem did not specify what the "system" is so calculate both accelerations. If I saw this problem as is on a test, I would raise my hand and ask for clarification.

 

[Lnewqban]

Yes but when the question asks about the acceleration of the system, which acceleration does the question ask, the hanging mass or the CoM of the rolling sphere or both of them?

I would do a vector addition for both linear accelerations, of A and B.
We will then have magnitude and direction of that overall acceleration.

As the problem does not provide dimensions, the tail of that "acceleration of the system vector" can't be located, but we know that it would be somewhere on an imaginary line that joints the centers of mass of A and B.
As the pulley is not moving, it does not have any linear acceleration vector.

 

[Steve4Physics]

It’s worth noting that the radius of sphere A is not supplied. (From the diagram, it's not the same as the radius of the pulley.) This creates problems.

I also struggle to understand how the string is attached to the sphere. Is the end of the string fixed to the top of the sphere? Or is the string wound around the sphere?

(And whoever drew the diagram had no idea what an angle of 53º looks like.)

 

[erobz]

It’s worth noting that the radius of sphere A is not supplied. (From the diagram, it's not the same as the radius of the pulley.) This creates problems.

I also struggle to understand how the string is attached to the sphere. Is the end of the string fixed to the top of the sphere? Or is the string wound around the sphere?

(And whoever drew the diagram had no idea what an angle of 53º looks like.)

Nice catch with the sphere radius. I assume it's just been left out of the post.

I assume its wrapped around, since it rolls without slipping. If it were pinned to the top, I think there would be some problems with that condition (and much more missing information)?

(And whoever drew the diagram had no idea what an angle of 53º looks like.)

 

[kuruman]

It’s worth noting that the radius of sphere A is not supplied. (From the diagram, it's not the same as the radius of the pulley.) This creates problems.

What kind of problems? I don't think the radius of the sphere matters as long as the string connected to it is parallel to the incline. When you write the torque equation for the sphere and set $\alpha =a_{\text{cm}}/R_{\text{sphere}},~$ the radius cancels out.

 

[Steve4Physics]

Nice catch with the sphere radius. I assume it's just been left out of the post.

On reflection (which I should have done before posting!) I suspect that the sphere's radius cancels-out during the working.

I assume its wrapped around, since it rolls without slipping. If it were pinned to the top, I think there would be some problems with that condition (and much more missing information)?

Yes indeed - wrapped around seems most likely

 

[erobz]

Yeah, I stopped at finding $\alpha$, not $a$. I too see that $R$ cancels.

 

[haruspex]

I would do a vector addition for both linear accelerations, of A and B.
We will then have magnitude and direction of that overall acceleration.

That makes no sense to me. In what way would such a single vector represent an overall acceleration? Consider a system of two unequal masses suspended vertically at opposite ends of a rope passing over a pulley. Your calculation would give an "overall acceleration " of zero!

 

[kuruman]

That makes no sense to me. In what way would such a single vector represent an overall acceleration? Consider a system of two unequal masses suspended vertically at opposite ends of a rope passing over a pulley. Your calculation would give an "overall acceleration " of zero!

Yes, but the acceleration of the center of mass of the two masses is not zero unless the masses are equal. That makes sense.

 

[erobz]

Yeah, I stopped at finding $\alpha$, not $a$. I too see that $R$ cancels.

Actually, never mind...I've bungled this as usual!

In trying to isolate $T_A$ in terms of $\alpha_A$ can someone confirm that the friction force appears in both the Force balance and Torque balance, and if you get the direction of $f_{fr}$ wrong, $T_A$ disappears from the equation?

 

[kuruman]

Actually, never mind...I've bungled this as usual!

In trying to isolate $T_A$ in terms of $\alpha_A$ can someone confirm that the friction force appears in both the Force balance and Torque balance, and if you get the direction of $f_{fr}$ wrong, $T_A$ disappears from the equation?

The friction will not appear in the torque balance equation if, as I did, you calculate the torque about the point of contact. If you calculate the torque about the center of the sphere, friction will appear in both equations but it should not disappear if you try to isolate $T_A$ just because you put friction in the wrong direction.

 

[erobz]

The friction will not appear in the torque balance equation if, as I did, you calculate the torque about the point of contact. If you calculate the torque about the center of the sphere, friction will appear in both equations but it should not disappear if you try to isolate $T_A$ just because you put friction in the wrong direction.

I went with the latter approach...I'm now quite concerned I have no idea what I'm doing! Would you mind if I PM'd you my equations for that part?

 

[kuruman]

I went with the latter approach...I'm now quite concerned I have no idea what I'm doing! Would you mind if I PM'd you my equations for that part?

Go ahead.

 

[jbriggs444]

My inclination is to skip most of the equations and algebra and, instead, tot up the change in potential energy per unit rise of B and the total kinetic energy as a function of the velocity of B. It is then a quick intuitive leap to a solution.

 

[Lnewqban]

That makes no sense to me. In what way would such a single vector represent an overall acceleration? Consider a system of two unequal masses suspended vertically at opposite ends of a rope passing over a pulley. Your calculation would give an "overall acceleration " of zero!

For this 2-D problem, the overall acceleration of the system could have more than one interpretation, it seems.
Perhaps it is a wrong approach to that question, but it occurred to me that the system of translating masses A and B could be considered as the ends of one enlarging or stretching link in a mechanism.

In designing mechanisms, it is important to determine instantaneous accelerations of the links, because the inertia forces put stresses on its components.

The COM of that imaginary link should have instantaneous linear velocity and acceleration, which, to me, could be considered to be "the acceleration of the system" requested in the OP.

Another approach would be to ignore the change of direction over the pulley and to align both masses in a FBD.
The acceleration of the system would have the direction of the resultant driving force, and a value between the linear accelerations of masses A and B.

 

[kuruman]

For this 2-D problem, the overall acceleration of the system could have more than one interpretation, it seems.

What is your understanding of the system in this problem? Specifically, what entities, do you think, are included in it? My point is that one cannot start looking for the acceleration of a system unless one first defines what this system consists of.

 

[haruspex]

Yes, but the acceleration of the center of mass of the two masses is not zero unless the masses are equal. That makes sense.

Yes, that would be at least reasonable.

 

[songoku]

Oh...Yeah, that's a little problematic. I probably wouldn't worry about it all that much, and just continue to solve the problem. Do you have the answer(s) from the key?

No I don't

Actually, never mind...I've bungled this as usual!

In trying to isolate $T_A$ in terms of $\alpha_A$ can someone confirm that the friction force appears in both the Force balance and Torque balance, and if you get the direction of $f_{fr}$ wrong, $T_A$ disappears from the equation?

This is what I did:

Equation of translation of A:
T_A + f - W_A sin θ = m_A . a_{COM (A)} (where f is the friction) ... (1)

Equation of rotational motion of A:
T_A . r - f . r = I_A . α_A ... (2)

Equation of rotational motion of pulley:
T_B . R - T_A . R = I_pulley . α_pulley ...(3)

Equation of B:
W_B - T_B = m_B . a_{COM (B)} ... (4)

Equation relating a_{COM (A)} and a_{COM (B)}:
a_{COM (B)} = 2 a_{COM (A)} ... (5)

There are 5 equations and 5 unknowns (T_A , T_B , f , a_{COM (A)} and a_{COM (B)}) so I am able to solve it.

 

[haruspex]

No I don'tThis is what I did:

Equation of translation of A:
T_A + f - W_A sin θ = m_A . a_{COM (A)} (where f is the friction) ... (1)

Equation of rotational motion of A:
T_A . r - f . r = I_A . α_A ... (2)

Equation of rotational motion of pulley:
T_B . R - T_A . R = I_pulley . α_pulley ...(3)

Equation of B:
W_B - T_B = m_B . a_{COM (B)} ... (4)

Equation relating a_{COM (A)} and a_{COM (B)}:
a_{COM (B)} = 2 a_{COM (A)} ... (5)

There are 5 equations and 5 unknowns (T_A , T_B , f , a_{COM (A)} and a_{COM (B)}) so I am able to solve it.

That all looks fine, except that I see no equation relating $\alpha_{pulley}$ to other accelerations, and the OP does not give the radius of the sphere.

 

[songoku]

That all looks fine, except that I see no equation relating $\alpha_{pulley}$ to other accelerations, and the OP does not give the radius of the sphere.

α_A = a_{COM (A)} / r

α_pulley = a_{COM (B)} / R

 

[Lnewqban]

What is your understanding of the system in this problem? Specifically, what entities, do you think, are included in it? My point is that one cannot start looking for the acceleration of a system unless one first defines what this system consists of.

I understand it as essentially similar to the system shown in the attached picture, but considering the rotational inertia of the fixed pulley and the rolling sphere (a second pulley, if you wish).

 

[kuruman]

I understand it as essentially similar to the system shown in the attached picture, but considering the rotational inertia of the fixed pulley and the rolling sphere (a second pulley, if you wish).

View attachment 305490

You did not answer my question probably because I didn't make myself clear. Let's go back to the picture of the sphere on the inclined plane. It shows a bunch of items that interact with each other, i.e. exert forces on one another. These items are

1. Mass $m_A$.
2. Pulley of mass $m_C$.
3. Hanging mass $m_B$.
4. Massless, inextensible string that connects $m_A$ and $M_B$ over the pulley.
5. The incline, which is a 3-4-5 triangle, firmly attached to the Earth (not shown.)

For purposes of drawing free body diagrams, you can pick one or more of the above and proceed to write Newton's 2nd law for your chosen system according to the standard FBD recipe. What I wanted to know is which particular combination of the above would you pick as the system the acceleration of which the problem is asking you to find.

 

[Steve4Physics]

Please forgive me for going off-topic but the image in @Lnewqban ’s Post #29
https://www.physicsforums.com/attachments/manlift-jpg.305490/
immediately reminded me of an (IMO funny) Irish folk song.

For anyone so inclined - and having three minutes to spare - here it is (with lyrics).
Edit: The only song I know about Atwood machines!

Otherwise, please ignore.

 

[Lnewqban]

Thank you, @kuruman.
I would pick items 1 and 3, because those are the only ones having linear accelerations (here assuming that that is the type of acceleration requested to be calculated by the problem).

Item 5 must be considered as well because it determines the fraction of the weight of item 1 that is resisting the weight of item 3 (driving force in the system), or tension in item 4.

The rotational inertia of items 1 and 2 also resist the effect of the driving force; therefore, both need to be considered.

 

[haruspex]

Thank you, @kuruman.
I would pick items 1 and 3, because those are the only ones having linear accelerations (here assuming that that is the type of acceleration requested to be calculated by the problem).

Item 5 must be considered as well because it determines the fraction of the weight of item 1 that is resisting the weight of item 3 (driving force in the system), or tension in item 4.

The rotational inertia of items 1 and 2 also resist the effect of the driving force; therefore, both need to be considered.

You originally proposed taking the vector sum of the two accelerations, which is bizarre.

In post #23, you mention COM, so you might be proposing the acceleration of the mass centre of the system consisting of the two linearly moving masses. If so, @kuruman already offered that in post #17, and I agreed that was defensible.
But I could also interpret your post as meaning the weighted average of the magnitudes of the two linear accelerations, which also has some validity. That interpretation seems to be supported by posts #29 and #32.

But the whole discussion is fruitless. The question does not define "the system" (why exclude the massive pulley?), it does not mention magnitudes and does not mention COM. If we're playing guess the question setter's intent, it is clear s/he overlooked the ambiguity, so may well have just been thinking of the descending mass.

@songoku has solved the problem as far as is possible.

 

[Lnewqban]

You originally proposed taking the vector sum of the two accelerations, which is bizarre.

Would you mind explaining why do you believe so?

 

[haruspex]

Would you mind explaining why do you believe so?

Because it has no physical meaning, as illustrated by my example in post #16.