Motion of a spool of thread on an inclined plane related to inertia

[songoku]

Homework Statement

I am trying to solve question in this link:
https://www.chegg.com/homework-help/spool-thread-mass-m-rests-plane-inclined-angle-end-thread-t-chapter-8-problem-108cp-solution-9780077405731-exc

Relevant Equations

torque = F x d
torque = I . α
Newton's law
friction = μ . N

a) Describe the motion of the spool: it will roll down the plane, rotating counter clockwise.

I am confused when setting up equation of torque. If I take center of the spool as pivot, only the tension of thread produces torque but the direction will be clockwise and it makes the spool will move upwards instead of downwards.

If I take the bottom point of the spool that is in contact with the plane as the pivot, there will be two forces producing torque: tension and weight. The equation will be:
mg sin θ . R - T .(R + r) = I . α
mg sin θ . R - T .(R + r) = I . a / R (where a is the acceleration of the center of mass of the spool). Is this equation correct?

If yes, why taking different pivot will give different equations? I think no matter where I put the pivot, the equation should be the same?

b) direction of frictional force: upwards parallel to the plane (in the same direction as tension). Is this correct?

Thanks

 

[BvU]

If I take center of the spool as pivot, only the tension of thread produces torque

You sure about that ?

 

[songoku]

You sure about that ?

I draw the free body diagram and the forces acting on the spool are tension, weight and normal force (no friction for question (a)). Weight and normal force pass through the center so won't produce torque if the center is the pivot so I think I am pretty sure about it.

Thanks

 

[BvU]

My bad, in a) there is no friction, so no torque from that: you are correct.
But all you can conclude from the torque is not

makes the spool will move upwards

but 'will rotate in a clockwise direction'. For the motion you will have to do a little more...

 

[songoku]

My bad, in a) there is no friction, so no torque from that: you are correct.
But all you can conclude from the torque is not
but 'will rotate in a clockwise direction'. For the motion you will have to do a little more...

I assume it starts from rest. That's why I can't imagine the motion if the torque produced by the tension is clockwise because it will have negative angular acceleration and also negative angular speed when it moves. If it starts from rest and the torque is clockwise, won't it move upwards?

The other equation I get is from Newton's 2nd law:
mg sin θ - T = m.a

To find the acceleration, I need to substitute T from equation of torque

Thanks

 

[BvU]

Twofold reply:

1) IF it were to move upwards, where would the mechanical energy come from ?

2) What does the free body diagram tell you about the motion ?

 

[songoku]

Twofold reply:

1) IF it were to move upwards, where would the mechanical energy come from ?

I am not sure. When describing the motion, I say that it moves downwards but I can not interpret the torque. I don't know the implication of the clockwise torque to the motion.

2) What does the free body diagram tell you about the motion ?

It will accelerate down the inclined plane.

But if it moves downwards, it should rotate counter clockwise, right?

Thanks

 

[BvU]

it should rotate counter clockwise, right?

Why ? (in case of no friction)

I need to substitute T from equation of torque

What have you got (equation) that can help you here ?
Can you find a relationship between angle of rotation and position along the plane ?

 

[haruspex]

If it starts from rest and the torque is clockwise, won't it move upwards?

It will only ascend if the net force is up the slope. Torque alone is not a force.
If there is some friction, the clockwise rotation will lead to a frictional force up the slope, but it will still only ascend if this exceeds the component of gravity down the slope.

 

[songoku]

Why ? (in case of no friction)

Because it starts from rest. Is it possible for the spool to move downwards while rolling clockwise?

What have you got (equation) that can help you here ?
Can you find a relationship between angle of rotation and position along the plane ?

To find T from equation of torque, in my opinion it is either T.r = I . a/R or mg sin θ . R - T .(R + r) = I . a / R

I am not sure about relationship between angle of rotation and position along the plane. You mean like if it has rotated for one circle, how far does it move?

 

[songoku]

It will only ascend if the net force is up the slope. Torque alone is not a force.
If there is some friction, the clockwise rotation will lead to a frictional force up the slope, but it will still only ascend if this exceeds the component of gravity down the slope.

I think I can imagine it better. It can move downwards while rotating clockwise. So for equation of torque:
(i) taking center of spool as pivot: T.r = I . a/R (is this correct?)

(ii) taking bottom of the spool as pivot: T .(R + r) - mg sin θ . R = I . a / R (is this correct?)

 

[haruspex]

I think I can imagine it better. It can move downwards while rotating clockwise. So for equation of torque:
(i) taking center of spool as pivot: T.r = I . a/R (is this correct?)

(ii) taking bottom of the spool as pivot: T .(R + r) - mg sin θ . R = I . a / R (is this correct?)

Neither is correct. In each, you are supposing $R\alpha=a$, but that is only true when rolling.
What is the simple relationship between $\alpha$ and $a$ here?

 

[songoku]

Neither is correct. In each, you are supposing $R\alpha=a$, but that is only true when rolling.

Is the spool not rolling? If the pivot is at the center of the spool then the tension will produce clockwise torque so it should be rolling clockwise while moving downwards?

What is the simple relationship between $\alpha$ and $a$ here?

I am not sure. The only equation I know relating α and a is a = α.r

 

[haruspex]

Is the spool not rolling?

It is rotating, not rolling. Rolling means not sliding, i.e. if there is friction it is static, not kinetic.

The only equation I know relating α and a is a

You need to think about how that equation comes about. If a wheel radius R is rolling at angular rate $\omega$ then in a short period of time dt it rotates through angle $d\theta=\omega dt$ and advances distance $Rd\theta$. Hence $v=\frac{Rd\theta}{dt}=R\omega$.
In the present case, it is not rolling on the ground, so this does not apply. Think about the string instead.

 

[songoku]

You need to think about how that equation comes about. If a wheel radius R is rolling at angular rate $\omega$ then in a short period of time dt it rotates through angle $d\theta=\omega dt$ and advances distance $Rd\theta$. Hence $v=\frac{Rd\theta}{dt}=R\omega$.
In the present case, it is not rolling on the ground, so this does not apply. Think about the string instead.

Sorry I don't get the hint. The tension will produce tangential acceleration at the tip of inner circle. Is this tangential acceleration equal to angular acceleration times r?

 

[haruspex]

The tension will produce tangential acceleration at the tip of inner circle.

Tangential acceleration of what? The straight part of the string is static, and the bit of the spool in contact with that is instantaneously stationary, so its acceleration is radial.

For a wheel rolling on the ground, the bit of wheel in contact with the ground is instantaneously stationary. Compare that with the above.

 

[songoku]

Tangential acceleration of what?

Tangential acceleration of every point on the inner circle.

The straight part of the string is static, and the bit of the spool in contact with that is instantaneously stationary, so its acceleration is radial.

By static, you mean the length of the string is always constant? I thought when it moves downwards the string will become longer but the extension comes from the spoil not from like extensible string / elastic string.

And the point where the string touches the spool is instantaneously stationary so no tangential acceleration? I don't understand this part because I thought all points on the inner circle will have tangential acceleration.

For a wheel rolling on the ground, the bit of wheel in contact with the ground is instantaneously stationary. Compare that with the above.

For this case, point of the spool that is in contact with ground is not instantaneously stationary so speed of center of mass is not equal to tangential speed of the point and acceleration of center of mass is not equal to tangential acceleration of the point. But I don't know how to find the relation between acceleration of center of mass and tangential acceleration of the point.

Thanks

 

[jbriggs444]

And the point where the string touches the spool is instantaneously stationary so no tangential acceleration?

The point where the string touches the spool is not moving down-slope. It is restrained by the string. Zero velocity. Zero acceleration.

If one were to adopt a frame of reference anchored to the center of the spool then in this frame, all points on the inner ring would be in accelerating circular motion. However, @haruspex is intending that you describe things from the ground frame.

 

[songoku]

The point where the string touches the spool is not moving down-slope. It is restrained by the string. Zero velocity. Zero acceleration.

If one were to adopt a frame of reference anchored to the center of the spool then in this frame, all points on the inner ring would be in accelerating circular motion. However, @haruspex is intending that you describe things from the ground frame.

Sorry I still don't understand. I thought the spool moving downwards while rotating clockwise so shouldn't all parts of the spool moving downwards the inclined plane and having acceleration (linear and tangential) with respect to the ground?

 

[jbriggs444]

Sorry I still don't understand. I thought the spool moving downwards while rotating clockwise so shouldn't all parts of the spool moving downwards the inclined plane and having acceleration (linear and tangential) with respect to the ground?

The part of the spool where the string is attached cannot possibly move down slope. The string tied to the wall prevents it.

 

[BvU]

The part of the spool where the string is attached cannot possibly move down slope. The string tied to the wall prevents it.

Well 'spoken' ! So for @songoku : that 's the point about which the thing rotates. Not a fixed point in space, nor a fixed point on the spool, though !

A long time ago (#6) I asked for the FBD ... ??

 

[haruspex]

By static, you mean the length of the string is always constant?

No. Initially, some of the string is straight and some of it is wound on the spool. As the wheel rotates, some that was wound joins the straight section. For any bit of the string, once it is in the straight section it does not move again. No velocity, no acceleration.

thought all points on the inner circle will have tangential acceleration.

For a wheel rotating at constant rate about a fixed centre, the points on the periphery have radial acceleration, towards the centre, but no tangential acceleration.
If we now let the wheel roll at a steady velocity, this remains true.
For a wheel rolling and accelerating, the acceleration of a point on the periphery can be thought of as the sum of a linear acceleration and a radial one. So in that case it would have a tangential component for most points (all except those at the same height as the wheel's centre), but I don’t think that's helpful here.

 

[songoku]

Well 'spoken' ! So for @songoku : that 's the point about which the thing rotates. Not a fixed point in space, nor a fixed point on the spool, though !

A long time ago (#6) I asked for the FBD ... ??

I have answered about FBD in post (#7)? For question (a), the forces acting on the spool are normal force, weight and tension.

The part of the spool where the string is attached cannot possibly move down slope. The string tied to the wall prevents it.

No. Initially, some of the string is straight and some of it is wound on the spool. As the wheel rotates, some that was wound joins the straight section. For any bit of the string, once it is in the straight section it does not move again. No velocity, no acceleration.

Ah I see. Ok I understand about it and now I also understand what jbriggs444 tried to tell me

For a wheel rotating at constant rate about a fixed centre, the points on the periphery have radial acceleration, towards the centre, but no tangential acceleration.
If we now let the wheel roll at a steady velocity, this remains true.
For a wheel rolling and accelerating, the acceleration of a point on the periphery can be thought of as the sum of a linear acceleration and a radial one. So in that case it would have a tangential component for most points (all except those at the same height as the wheel's centre), but I don’t think that's helpful here.

For this case, the spool is rolling and accelerating (is this called rolling with slipping?).

So the tension will produce clockwise torque and there will be angular acceleration on the spool. Is it correct to say there will be tangential acceleration on all points on inner an outer circle of spool, except at the point where the spool in contact with the tension?

If yes, the tangential acceleration will be α.r and α.R?

If this is rolling with slipping, then the point of spool in contact with inclined plane will have speed of v_CM + ω.r and will have tangential acceleration of a_CM - α.R

This is as far as I can get. Not sure whether it is correct and I still do not know how to relate acceleration of center of mass to angular acceleration in this case.

Thanks

 

[haruspex]

is this called rolling with slipping?

Rolling means not slipping. This is rotating while slipping on the surface beneath it.
(What is it not slipping with respect to?)

Is it correct to say there will be tangential acceleration on all points on inner an outer circle of spool

As I wrote, it is not helpful to think in terms of tangential acceleration here. You seem to be confusing tangential acceleration with tangential velocity.
Better to think of the motion as the sum of a linear acceleration (of the wheel's centre) and the radial (centripetal) accelerations of the points of the wheel.

except at the point where the spool in contact with the tension?

That is certainly an important point of the wheel to think about, but not with regard to "tangential acceleration". Think about its velocity, and what that tells you about the relationship between the linear velocity of the wheel and its angular velocity.

 

[songoku]

Rolling means not slipping. This is rotating while slipping on the surface beneath it.
(What is it not slipping with respect to?)

As I wrote, it is not helpful to think in terms of tangential acceleration here. You seem to be confusing tangential acceleration with tangential velocity.
Better to think of the motion as the sum of a linear acceleration (of the wheel's centre) and the radial (centripetal) accelerations of the points of the wheel.

That is certainly an important point of the wheel to think about, but not with regard to "tangential acceleration". Think about its velocity, and what that tells you about the relationship between the linear velocity of the wheel and its angular velocity.

Maybe I get the hint. The point where the string attached to the spool is static so the velocity is zero. This means that it is the point where the spool is not slipping. The velocity of center of mass (center of the spool) can be taken as ω.r with respect to the point. Can I say that α = a/r instead of a/R?

Thanks

 

[BvU]

Can I say that α = a/r instead of a/R?

Yes. keep the axis of rotation in mind and go looking for an expression for $\alpha$

 

[songoku]

Yes. keep the axis of rotation in mind and go looking for an expression for $\alpha$

Question (a)
Torque = I . α
T . r = I . a/r
T = I . a/r²

mg sin θ - T = m.a
mg sin θ - I . a/r² = m.a
a = (mg sin θ) / (m + I/r²)

Is that correct?Question (b)
Direction of frictional force is upwards parallel to the plane because if the bottom point of the spool slips, it will slip in direction downwards parallel to the plane.

So, now this is case of rolling (no slipping). The instantaneous velocity of the bottom point of the spool that is in contact with the ground is zero and velocity of center of mass can be taken as ω.R

But the point where the string attached to the spool is still static so its instantaneous velocity is also zero and velocity of center of mass can be taken as ω.r

Where is the mistake in my reasoning? I think inner and outer circle should rotate with same angular velocity since they are concentric but that does not make sense since the velocity of center of mass will be different.

Thanks

 

[haruspex]

Is that correct?

Yes.

I think inner and outer circle should rotate with same angular velocity

They do.
If the point in contact with the plane is stationary, and the point in contact with the string is stationary, what do you conclude?

 

[songoku]

They do.
If the point in contact with the plane is stationary, and the point in contact with the string is stationary, what do you conclude?

The spool is stationary?

 

[haruspex]

The spool is stationary?

Yes. It's a statics question.

 

[songoku]

I think I can do the rest. Thank you very much for all the help BvU, haruspex, jbriggs444