Calculating Velocity of Clothes in a Dryer

[Dorothy Weglend]

A clothes dryer is designed so that when the clothes reach an angle of 68 degrees to the horizontal, they fall. Rotation is constant, in a vertical circle. The radius of the dryer is 0.330 m.

I have solved this two different ways, and I get two different answers.

Method 1: I take the component of mg along Fc, so at any angle theta I have:

N = Fc + mg sin theta.

When N = 0, Fc = -mg sin 68, using Fc = mv^2/r,

v^2 = -rg sin 68

Method 2:

All I'm interested in is the y component, along which gravity acts. If N=0, then Ny=Nx=0. So I can just deal with vector along mg:

Ny = Fc sin theta + mg

Setting Ny = 0, and solving as above, I get:

v^2 = -rg/(sin 68)

These are obviously not equivalent (although they are close, numerically).

Can someone help me see which one is right, and why?

Thank you,
Dorothy

 

[andrevdh]

M1 is "better".

The weight is acting downwards and the normal force from the drum on a clothes piece is acting inwards radially towards the centre of the drum. When an object is spun in a juice extractor it want to escape by flying outwards. The wall of the container prevents it by pushing inwards on it. So the normal force pushes radially inwards. This means that the sum of the normal force and the component of the weight provides the centripetal force.

Their sum need to stay constant (for a constant spinning speed). As the angle increases the component of the weight will increase (sin() increases). This means that the normal force will decrease (the weight is pulling the clothes down more and more). At the point where the normal force becomes zero the clothes will drop down (lose contact with the wall). Beyond that point our equation tells us that the normal force need to become negative in order to keep on spinning. This means that the drum must pull the clothes towards it in order to keep it rotating - a process that is not designed for. So it falls down beyond this point.

 

[Dorothy Weglend]

Thanks, Andrevdh. I think I understand. Basically, you are saying that M1 is better because Fc does not vary at all as the drum rotates, but mg does vary, with the sine of the angle. So it is better to use the component of the force which varies?

 

[andrevdh]

Correct, but your equation

...
Method 1: I take the component of mg along Fc, so at any angle theta I have:

N = Fc + mg sin theta.
...Dorothy

needs a little bit of work though.

Since both the normal force and the component of the weight points radially inward your equation should actually read

N + W\sin(\theta) = F_C

This means that these two forces produces the circular motion of the clothes in the dryer.

 

[Dorothy Weglend]

Correct, but your equation



needs a little bit of work though.

Since both the normal force and the component of the weight points radially inward your equation should actually read

N + W\sin(\theta) = F_C

This means that these two forces produces the circular motion of the clothes in the dryer.

I thought that the normal force would point radially outward, opposing Fc. So in the lower half of the drum, you would have:

N + mg sin theta = Fc

as you wrote. But in the upper half, wouldn't this have to change to:

N - mg sin theta = Fc

as gravity decreases the normal force caused by rotation of the drum?

Thanks a lot!

Dorothy

 

[andrevdh]

As I tried to explain with the juice extractor that the object wants to escape the motion by "flying out". So it presses up against the wall. The wall reacts by pressing back via the normal force- this is the way the normal force works under normal cicumstances - we push on the ground it responds by pushing back.

This "flying out" action is due to inertia - the object wants to move off in a straight line (tangentially to the circle) but the wall forces it into circular motion (keep on pushing it inwards). In this case the component of the weight lends a helping hand in the circular motion (In the top half of the rotation - like projectiles following curved paths. In the bottom half it works against the circular motion, but the wall is strong enough to provide for the motion anyway via the normal force). If the component of the weight becomes too large (larger than the required centripetal force) the object gets pulled downwards by its weight and it loses contact with the wall.

Anyway if the normal force was pointing upwards in the top half it would mean that the drum is attracting the clothes (or you might say that the clothes is sticking to the drum if you put wet clothes into the dryer - not a normal practise!).

 

[Dorothy Weglend]

In the bottom half it works against the circular motion, but the wall is strong enough to provide for the motion anyway via the normal force). If the component of the weight becomes too large (larger than the required centripetal force) the object gets pulled downwards by its weight and it loses contact with the wall.

Thanks andrevdh. I hope you don't think I'm hopelessly stupid, but I still don't get it. Without gravity, we would just have

N = F_c (assuming a fast enough rotational speed).

In order for the laundry to fall, something must be subtracting from N, which is W sin(theta). Otherwise, N never becomes zero and the laundry would never fall.

That's why I had, in my original equation

N - W sin(theta) = F_c

If I follow your suggestion: N + W sin(theta) = F_c, then how can N ever become zero?

Dorothy

 

[andrevdh]

I changed my previous post a little Dorothy while you were typing your response - read it again and see if that clears things up.

Remember that the normal force is a reaction force - it originates due to the object pushing up against it (the ground).

 

[andrevdh]

Basically as the weight component becomes stronger in the radial direction it decreases the amount by which the clothes is pushing up against the wall of the dryer. This will lead to the wall pushing back with a reduced normal force - until the weight component is totally sufficient to supply the centripetal force in which case the normal force will become zero. Beyond this the weight component becomes so strong that it pulls the clothes away from the wall - it starts to fall in a parabolic path.

 

[Dorothy Weglend]

Basically as the weight component becomes stronger in the radial direction it decreases the amount by which the clothes is pushing up against the wall of the dryer. This will lead to the wall pushing back with a reduced normal force - until the weight component is totally sufficient to supply the centripetal force in which case the normal force will become zero. Beyond this the weight component becomes so strong that it pulls the clothes away from the wall - it starts to fall in a parabolic path.

Thanks andrevdh, for your patience.

I think I am starting to see it now. The 'parabolic' path helped me. So it will travel along the line of force, mg sin 68?

Well, I guess the velocity vector, tangential to Fc will be involved also, but I hope you understand what I am saying, which is that it will travel rightwards, along some vector of which mg sin 68 is a component?

That makes sense to me, and also helps me to understand why it should be N + mg sin 68 = Fc.

Let me know if this is right, if you have the time.

Thanks again!
Dorothy

 

[Doc Al]

I have solved this two different ways, and I get two different answers.

Method 1: I take the component of mg along Fc, so at any angle theta I have:

N = Fc + mg sin theta.

When N = 0, Fc = -mg sin 68, using Fc = mv^2/r,

v^2 = -rg sin 68

This is fine, except as andrevdh has already corrected: N and mg sin 68 both point toward the center, so I would write:
N + mg sin 68 = mv^2/r

Method 2:

All I'm interested in is the y component, along which gravity acts. If N=0, then Ny=Nx=0. So I can just deal with vector along mg:

Ny = Fc sin theta + mg

Setting Ny = 0, and solving as above, I get:

v^2 = -rg/(sin 68)

These are obviously not equivalent (although they are close, numerically).

Can someone help me see which one is right, and why?

The problem with your second solution (in addition to the sign issues already pointed out) is that you assume that the only forces acting with a vertical component are mg and N. Not so. Recall that everything is moving at constant speed, thus there must be a friction force to balance the component of weight tangent to the circle. Figure out what that friction force must be and include it when you add up the vertical components. Then you'll get the same answer with both methods.

 

[Dorothy Weglend]

Thanks Doc Al. Especially for the explanation of the problem with the second solution. That makes makes a lot of sense to me.

This physics forum is really a great place. There is never time in a class to ask questions like this. It is so helpful to my understanding. Thanks to you (again) and andrevdh.

Dorothy

 

[BSCS]

I have a couple of questions for this problem, so I figured I'd post them here. I've spent a week on this, and have been through several books, so it's not due to a lack of effort.

1.) The normal force exerted on the laundry, the centripetal acceleration, and the component of gravity that acts on the laundry all act inward... correct? What force acts "outward" to cause the normal?

2.) What exactly is causing the velocity of the laundry? The rotating of the drum?

3.) Why can the component of gravity not acting toward the center be disregarded? In doing a problem such as this, how do I know that I shouldn't just be taking mg into account? It would seem that at the very least, mg*cos(theta) would come into play with friction.

4.) As I understand it, there needs to be friction so that the laundry doesn't just stay at the bottom. Is that correct, if so, why doesn't this come into play?

Perhaps something is missing in my understanding of these concepts in general...

Thanks!

 

[Dorothy Weglend]

I have a couple of questions for this problem, so I figured I'd post them here. I've spent a week on this, and have been through several books, so it's not due to a lack of effort.

As the perpetrator of the thread, I guess I should try to answer.

1.) The normal force exerted on the laundry, the centripetal acceleration, and the component of gravity that acts on the laundry all act inward... correct? What force acts "outward" to cause the normal?

All of these counteract the laundry's natural motion in a straight line. I think this is the same idea in all circular motion, such as the famous bucket of water rotated in a vertical circle.

2.) What exactly is causing the velocity of the laundry? The rotating of the drum?

There is no other source.

3.) Why can the component of gravity not acting toward the center be disregarded? In doing a problem such as this, how do I know that I shouldn't just be taking mg into account? It would seem that at the very least, mg*cos(theta) would come into play with friction.

I don't think friction is relevant because the laundry falls when N=0, which means the frictional force must be zero because Friction = mu * N.

Of course, now you have started me wondering again why my two initial solutions differ. I think Doc Al said it was because frictional forces came into play when I focused only on Ny = 0, but that shouldn't be true. So now I guess I am confused again on this point.

Real laundry drums have "wedges" that catch the laundry and carry it upwards. This allows the drum to have a smaller speed and less friction than it would otherwise need if the surface was smooth. Well, at least the laundry machines that I use have these, and the picture of the one in my textbook has them too. So presumably the upper bit of the clothing would start to move in a parabolic pathway as it hit the critical angle. This was a huge part of my problem in understanding this, thanks again to andrevdh for straightening me out here, since I was thinking the laundry must fall vertically.

4.) As I understand it, there needs to be friction so that the laundry doesn't just stay at the bottom. Is that correct, if so, why doesn't this come into play?

See above.

Perhaps something is missing in my understanding of these concepts in general...

Probably you just need to do more laundry

Dorothy

 

[Doc Al]

3.) Why can the component of gravity not acting toward the center be disregarded? In doing a problem such as this, how do I know that I shouldn't just be taking mg into account? It would seem that at the very least, mg*cos(theta) would come into play with friction.

I don't think friction is relevant because the laundry falls when N=0, which means the frictional force must be zero because Friction = mu * N.

Of course, now you have started me wondering again why my two initial solutions differ. I think Doc Al said it was because frictional forces came into play when I focused only on Ny = 0, but that shouldn't be true. So now I guess I am confused again on this point.

When I said "friction forces" I should have said "tangential forces". My point was that the drum must exert some tangential force on the laundry to balance the tangential component of the laundry's weight. Without such forces the laundry would not be pulled along with the drum.

To answer BSCS's question directly: You can ignore the component of gravity acting tangent to the circle as well as the counteracting drum forces if you analyze forces in the radial direction. (Dorothy's Method 1) But if you choose to analyze forces in some other direction, such as the vertical (which Dorothy does in her Method 2), then you must include both gravity and the tangential drum forces to get a correct solution.

Real laundry drums have "wedges" that catch the laundry and carry it upwards. This allows the drum to have a smaller speed and less friction than it would otherwise need if the surface was smooth.

Exactly! The wedges help the drum to exert those tangential forces that pull the laundry along.

 

[BSCS]

Dorothy and Doc Al,

Thanks for the replies. My original reply was lost...

Question: For UCM problems like this in general, is it generally safe to ignore forces that would act in the tangential direction? (in this case mg*cos(theta))

Thanks again!

 

[Doc Al]

I'd put it like this: Depending upon the exact problem you are trying to solve, you may only need to consider forces in the radial direction.