Disk connected to another object through a pulley on an incline

[songoku]

Homework Statement

A disk of radius r and mass m is placed on an incline which makes an angle of 30 degree. The disk is suspended by a string attached to its center and a mass 3m is hung at the other end. The surface of incline is rough with the coefficient of static and kinetic friction denoted as μ_s and μ_k respectively. Assume that the string and the pulley does not have any friction and the pulley does not rotate.
a. find the linear acceleration of the disk when the disk rolls without slipping
b. for (a) to happen, find the condition of μ_s
c. when the condition of (b) is not met, the disk starts rolling but with sliding. Find a relative acceleration between the linear and angular acceleration, i.e. sliding acceleration against the surface of incline

Homework Equations

torque = I . α
∑F = m.a

The Attempt at a Solution

a.
For the 3m object:
∑F = m.a
W - T = 3ma
3mg - T = 3ma ... (1)

For the disk:
∑F = m.a
T - f - W sin θ = m.a
T - μ_k N - mg sin θ = m.a
T - μ_k . mg cos θ - mg sin θ = m.a ... (2)

I just need to solve for a from equation (1) and (2). Is this correct?b.
Στ = I.α
f . r = 1/2 mr² . a/r (I assume moment of inertia of disk is 1/2 Mr²
μ_s . mg cos θ . r = 1/2 mr² . a/r (I am not sure about this part. The disk is moving so why don't we use μ_k instead?)

μ_s = 0.5 a / (g cos θ) then I substitute the acceleration I got from part (a). Is this correct?

c. I am not sure what the question asks me to do. Won't the equation be the same as part (a)? What is "relative acceleration between the linear and angular acceleration"? Is it linear acceleration minus angular acceleration?

Thanks

 

[haruspex]

T - μk N - mg sin θ = m.a

The disk is rolling, not sliding. What equation relates normal force to the force of static friction? Be careful, there's a very commonly made mistake.

I just need to solve for a from equation (1) and (2). Is this correct?

No, when you have corrected your equation (2) you will find you need a third equation. What else applies for a rotating disk?

The disk is moving so why don't we use μk instead?

It is moving but not sliding, so it is static friction, not kinetic.

 

[songoku]

The disk is rolling, not sliding. What equation relates normal force to the force of static friction? Be careful, there's a very commonly made mistake.

So I need to change the friction into static?

T - μ_s mg cos θ - mg sin θ = m.a

Solving for a, I get:
a = 1/4 * (3g - μ_s g cos θ - g sin θ)

No, when you have corrected your equation (2) you will find you need a third equation. What else applies for a rotating disk?

Do I need another equation? I think I already get the acceleration

It is moving but not sliding, so it is static friction, not kinetic.

Sorry, why we use static friction when it is not sliding?

Thanks

 

[haruspex]

So I need to change the friction into static?

Yes.

I think I already get the acceleration

No, as I wrote, your equation 2 is wrong.
I ask again, what equation relates normal force to the force of static friction? Hint: consider a body standing on a level surface, with no applied lateral force. What is the frictional force there? Does your equation fit that?

why we use static friction when it is not sliding?

Friction is all about relative motion of the surfaces in contact. If there is no relative motion (as in rolling without sliding) it is static friction.

 

[songoku]

No, as I wrote, your equation 2 is wrong.
I ask again, what equation relates normal force to the force of static friction? Hint: consider a body standing on a level surface, with no applied lateral force. What is the frictional force there? Does your equation fit that?

Static friction = μ_s . N

For a body standing on a level surface with no applied force (the body is stationary), the static friction will be zero?

Friction is all about relative motion of the surfaces in contact. If there is no relative motion (as in rolling without sliding) it is static friction.

How to determine whether there is relative motion or not? In my mind, the position of the disk changes with respect to the surface (just like when the disk rolls and slipping) so there is relative motion?

Thanks

 

[haruspex]

Static friction = μs . N

No, that is the common blunder.

For a body standing on a level surface with no applied force (the body is stationary), the static friction will be zero?

Right, but that shows your equation above must be wrong.

How to determine whether there is relative motion or not? In my mind, the position of the disk changes with respect to the surface (just like when the disk rolls and slipping) so there is relative motion?

By definition, rolling motion is when there is no relative motion between the surfaces in contact.

 

[songoku]

No, that is the common blunder.

You mean there is other equation or it should be just zero?

If there is other equation, certainly I do not know the equation. What I know is the formula of static or kinetic friction is the coefficient times normal force

 

[haruspex]

You mean there is other equation or it should be just zero?

If there is other equation, certainly I do not know the equation. What I know is the formula of static or kinetic friction is the coefficient times normal force

For static friction, $F_s\leq\mu_s N$
For another equation:

Homework Equations

torque = I . α

 

[songoku]

For static friction, $F_s\leq\mu_s N$
For another equation:

I am not sure if I get the hint but let me try:

For 3m object:
3mg - T = 3ma
T = 3mg - 3ma

For disk:
T - mg sin θ = ma

Is it correct until this part?

Thanks

 

[haruspex]

T - mg sin θ = ma

No, you have left out friction.
My hint quoted a torque and angular acceleration equation. What will be rotating? What is the relationship between its angular acceleration and its linear acceleration? What is the torque on it?

 

[songoku]

No, you have left out friction.
My hint quoted a torque and angular acceleration equation. What will be rotating? What is the relationship between its angular acceleration and its linear acceleration? What is the torque on it?

OK let me starts over:

a)
For disk:
ΣF = m.a
T - f_s - mg sin θ = m.a ...(2)

Στ = I . α
f_s . r = 1/2 mr² . a/r
f_s = 1/2 m.a ...(3)

Subs (3) into (2): T - 1/2 m.a - mg sin θ = m.a ...(4)

Then I solve for a using equation (4) and equation (1)? (equation 1: 3mg - T = 3ma)b)
f_s = 1/2 m.a
μ_s . N = 1/2 m.a
μ_s . mg cos θ = 1/2 m.a

Solve for μ_s and then subs the acceleration from answer part (a). Is this correct?c) No idea what the question is asking

Thanks

 

[haruspex]

Then I solve for a using equation (4) and equation (1)? (equation 1: 3mg - T = 3ma)

Yes.

b)
fs = 1/2 m.a
μs . N = 1/2 m.a

You are still not getting my point here. You have again presumed f_s=μ_sN.
Look at the equation (inequality) I gave you in post #8. This is the correct formula relating normal force to frictional force in the case of static friction. It does not tell you exactly what the static frictional force is; it only sets an upper bound. The actual frictional force is anything up to μ_sN.
If you use that you will get a bound on μ_s, and that is what the question is asking for. It gives you the minimum value of the coefficient for rolling without sliding.

c) No idea what the question is asking

I find the question wording a little unusual. It distinguishes between rolling without sliding (part a) and rolling with sliding (part c). The more usual terminology is that rolling means it is not sliding, and what the question calls rolling with sliding I would call rotating with sliding.

In part a you used rα=a. That is only true for rolling without sliding, so for part c you need a different equation.
Instead, the relationship between kinetic frictional force and normal force is an equality: f_k=μ_kN.

 

[songoku]

You are still not getting my point here. You have again presumed f_s=μ_sN.
Look at the equation (inequality) I gave you in post #8. This is the correct formula relating normal force to frictional force in the case of static friction. It does not tell you exactly what the static frictional force is; it only sets an upper bound. The actual frictional force is anything up to μ_sN.
If you use that you will get a bound on μ_s, and that is what the question is asking for. It gives you the minimum value of the coefficient for rolling without sliding

Ahhh ok I see. That's what it means by condition for μ_s

I find the question wording a little unusual. It distinguishes between rolling without sliding (part a) and rolling with sliding (part c). The more usual terminology is that rolling means it is not sliding, and what the question calls rolling with sliding I would call rotating with sliding.

In part a you used rα=a. That is only true for rolling without sliding, so for part c you need a different equation.
Instead, the relationship between kinetic frictional force and normal force is an equality: f_k=μ_kN.

If it rotates, doesn't it mean that the object not sliding and if it sliding doesn't it mean that it is not rotating?
Is it possible the object rotating while sliding?
It is easy to imagine the object sliding without rotating but how to imagine rotating without sliding and with sliding? I mean how will the object move in case of
rotating without sliding and with sliding?

So for object 3m it is still the same:
3mg - T = 3ma ... (1)

For disk:
∑F = m.a
T - f_k - W sin θ = m.a
T - μ_k N - mg sin θ = m.a ...(2)

Στ = I . α
f_k . r = 1/2 mr² . α ...(3)

I can't change α to a/r and I don't know what other equation I should use

Thanks

 

[haruspex]

Is it possible the object rotating while sliding?

Yes. If a disk radius r is moving at speed v relative to the ground surface and rotating ("forwards") at angular speed ω then the point in contact with the ground is moving at speed v-rω relative to the ground. It is rolling (i.e. not sliding) if and only if v-rω=0.

I don't know what other equation I should use

Use N=mg cos(θ), as in your post #1.

 

[songoku]

Use N=mg cos(θ), as in your post #1.

OK I use that equation to find f_k in terms of μ_k but what should I change α to?

And what exactly the question in (c) asking me to find?

Thanks

 

[haruspex]

what should I change α to?

You do not change it to anything. You need to find it.
You have four equations now and five unknowns: T, f_k, a, α and N, and you have five equations. Solve to find a and α.

what exactly the question in (c) asking me to find?

It is asking for the relative acceleration between the disk and the ground at the point of contact.
If the centre of the disk is accelerating at rate a, and the (forward) angular acceleration is α, what is the linear acceleration (in the upslope direction) of the point on the disk which is touching the ground?

 

[songoku]

You do not change it to anything. You need to find it.
You have four equations now and five unknowns: T, f_k, a, α and N, and you have five equations. Solve to find a and α.

It is asking for the relative acceleration between the disk and the ground at the point of contact.
If the centre of the disk is accelerating at rate a, and the (forward) angular acceleration is α, what is the linear acceleration (in the upslope direction) of the point on the disk which is touching the ground?

I am really sorry but I still don't understand. Let say I already get the a and α, so my answer to question (c) is these two accelerations? The question asks me to find a and α separately?

Thanks

 

[haruspex]

I am really sorry but I still don't understand. Let say I already get the a and α, so my answer to question (c) is these two accelerations? The question asks me to find a and α separately?

Thanks

If the angular acceleration of the disk, in a "forward" direction, is α then the top of the disk is accelerating by αr telative to the centre of the disk, and the bottom of the disk is accelerating by -αr relative to the centre.
If the disk centre is accelerating at rate a relative to the ground then the bottom of the disk is accelerating at rate a-αr relative to the ground. The question is asking you to find a-αr.

 

[songoku]

I am really sorry for late reply. Thank you very much for the help