friction and rolling resistance, and work done queries


by sgstudent
Tags: rolling, work energy theorem
tiny-tim
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#37
Apr13-12, 01:34 PM
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it depends what the angular speed was just before the net moment was zero

for the trap door, you had to get it moving at that angular speed by using a slightly greater applied moment, and then when you reckoned that the angular speed was just right, you reduced the applied moment so that the net moment was zero
sgstudent
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#38
Apr15-12, 01:03 AM
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Quote Quote by tiny-tim View Post
it depends what the angular speed was just before the net moment was zero

for the trap door, you had to get it moving at that angular speed by using a slightly greater applied moment, and then when you reckoned that the angular speed was just right, you reduced the applied moment so that the net moment was zero
Oh so it depends on the angular speed before I reduce the moment to 0 does it work the same as for forces. Like if I apply a force then when I stop applying force (assuming no friction or retarding force) then it will stay at that speed forever? Thanks for the help!
tiny-tim
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Apr15-12, 02:16 AM
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Quote Quote by sgstudent View Post
… Like if I apply a force then when I stop applying force (assuming no friction or retarding force) then it will stay at that speed forever?
that's absolutely correct

also, if you apply a force to a body that is impeded by friction, then if you stop increasing the force when it balances the friction force, it will stay at that speed forever
sgstudent
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#40
Sep14-12, 11:57 PM
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Quote Quote by tiny-tim View Post
yes


not following you

at constant speed, there is no turning force from the engine (i'm assuming we're ignoring air resistance etc)


if the net force is zero, the speed stays whatever it was in the first place
Hi tiny tim! After discussing this with my friend using our O level knowledge, we came to a conclusion. But we have some queries.

So firstly, we have our driving wheel on the air and the engine is turned on. So the Free Body Diagram would look like this: http://imgur.com/xVk0w

So in this case the net force of the wheel is 0N as F2 and F1 are equal in magnitude. However, about F1, there is a net moment which turns clockwise (F1XL).

Then now we put it on the ground and and experiences a friction oppsoing F1 as friction opposes force in a single plane of contact. So now, the Free Body Diagram would look like this: http://imgur.com/Pw1oh

From this, since F2-F1=0N, so the net force=friction only. While the wheel still has to maintain its clockwise moment so F1L > FrictionL and since L is a constant, so F1>Friction.
So from here, it all makes sense to us, but then after thinking again since that friction force that we have been dealing up to now is static friction so shouldn't F1 be equal to friction? But if F1 is equal to the friction, the the wheel would not be able to turn in the first place.

So we were stuck at this part with no solution in mind. Could you help us out with this? Thanks 7
tiny-tim
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Sep15-12, 03:46 AM
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hi sgstudent!
Quote Quote by sgstudent View Post
So firstly, we have our driving wheel on the air and the engine is turned on. So the Free Body Diagram would look like this: http://imgur.com/xVk0w

So in this case the net force of the wheel is 0N as F2 and F1 are equal in magnitude. However, about F1, there is a net moment which turns clockwise (F1XL).
first, can i check that we mean the same thing by "driving wheel"?

a driving wheel is one that is forced to turn by a torque from another part of the vehicle

a non-driving wheel is one that simply receives a (linear) force from another part of the vehicle, or another vehicle (or animal)

so we normally draw the free body diagram of a non-driving wheel with a straight arrow, usually from the centre of the wheel, representing the force

while we draw the free body diagram of a driving wheel with a small circular arrow (anywhere), representing the torque

alternatively, we can represent that torque by two equal and opposite straight arrows, but although i agree we could place them anywhere, it would be normal to place them equal distances above and below the centre of the wheel …

your arrow at the bottom of the wheel may be confusing you, since the bottom of the wheel has no significance
of course, as shown, the centre of the wheel will stay where it is, but the wheel's rotation will accelerate
Then now we put it on the ground and and experiences a friction oppsoing F1 as friction opposes force in a single plane of contact. So now, the Free Body Diagram would look like this: http://imgur.com/Pw1oh
again, the position of F1 at the bottom of the wheel has no significance, it could be anywhere (so long as FL equals the torque)
in this case, the wheel's rotation will accelerate slower than before (because the friction is an opposing torque about the centre of mass), but the centre of the wheel will (linearly) accelerate
From this, since F2-F1=0N, so the net force=friction only. While the wheel still has to maintain its clockwise moment so F1L > FrictionL and since L is a constant, so F1>Friction.
So from here, it all makes sense to us, but then after thinking again since that friction force that we have been dealing up to now is static friction so shouldn't F1 be equal to friction? But if F1 is equal to the friction, the the wheel would not be able to turn in the first place.
i did the calculation for a non-driving wheel in post #14

do the calculation for this driving wheel, and you should see that there is no difficulty
i don't understand why you say "since that friction force that we have been dealing up to now is static friction so shouldn't F1 be equal to friction?"

static friction is not µN, or any other formula, it is simply the horizontal component of the total reaction force at the surface …

and the reaction force is always calculated (even in non-rotational cases, such as a static ladder) as the "missing" force that has to be inserted to make the equation (between the known forces and the geometric constraint of the surface) balance …

so the static friction can be anything between 0 and µN
sgstudent
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#42
Sep15-12, 07:47 AM
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Hi Tiny Tim, it has been a while!

Yup, we have the same defination of the wheel whereby the force is actually turning it. But again for the engine i was thinking the force turing it looks like this: http://imgur.com/8doMX
whereby the engine has an axle to push against to turn it. So in air/ice one force (by the engine) would be at the wheel's end and one at the centre of the wheel (equal in magnitude) which prevents it from moving forward? So it is something like this: http://imgur.com/h4wg3 I think this picture would work even if its not a car as wheelchairs use the same pinciples where the person sitting would push the wheel for it to turn.

So after googling about this topic for a long while i came across a old post by you haha. This was what you said: A car moves forward because the engine forces the back axle to turn.

If the car was on ice, the back wheels would spin, but the rest of the car would be still.

The torque from the engine causes a force at the bit of the tyre in contact with the road. So long as that force does not exceed the maximum static friction, that force will equal the actual friction force, and the bit of the tyre in contact with the road will not move.

(Newton's first law on that bit of the tyre: zero total force means zero change in movement.)

The car will move, because the only external horizontal force on it is the actual friction force.

So i understand now that the F engine=Static friction (without slipping though). But if F engine=Static friction, the won't the net moment be 0Nm and as a result, the wheel should not even be able to start turning (if we consider the wheel to be at rest initially). Even if the wheel has a net force which is the friction (http://imgur.com/uMQjk).

So I don't understand how the wheel can rotate. The car and wheel experiences a net force but no net moment. So I'm pretty confused about this.

Also, another thing that I'm confused is that in the wheelchair/driving wheel example, the wheelchair bound person would push the wheel only at the top, so would that mean the static friction at the bottom would have the same magnitude (http://imgur.com/Kafh6)? What allows this phenomenon to happen since the force is not at the same point of each other?

Thanks Tiny Tim, you rock!
tiny-tim
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Sep15-12, 05:43 PM
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Quote Quote by sgstudent View Post
… But again for the engine i was thinking the force turing it looks like this: http://imgur.com/8doMX
whereby the engine has an axle to push against to turn it. So in air/ice one force (by the engine) would be at the wheel's end and one at the centre of the wheel (equal in magnitude) which prevents it from moving forward? So it is something like this: http://imgur.com/h4wg3 I think this picture would work even if its not a car as wheelchairs use the same pinciples where the person sitting would push the wheel for it to turn.
…
So i understand now that the F engine=Static friction (without slipping though).
no, even at zero acceleration, you also have to include the reaction force from the load
in all cases, the equations come from two free body diagrams: one for the wheel, and one for the load (or for the wheel-and-load combined) …

in particular, the first one will include the reaction force between the wheel and the load …

i think it's your failure to include that that is confusing you
… in the wheelchair/driving wheel example, the wheelchair bound person would push the wheel only at the top, so would that mean the static friction at the bottom would have the same magnitude (http://imgur.com/Kafh6)? What allows this phenomenon to happen since the force is not at the same point of each other?
try writing out the actual equations, one for the wheel and one for the person (including the reaction force between them in each), and using the same linear acceleration for both …
what do you get?
sgstudent
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#44
Sep15-12, 11:23 PM
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Quote Quote by tiny-tim View Post
no, even at zero acceleration, you also have to include the reaction force from the load
in all cases, the equations come from two free body diagrams: one for the wheel, and one for the load (or for the wheel-and-load combined) …

in particular, the first one will include the reaction force between the wheel and the load …

i think it's your failure to include that that is confusing you

try writing out the actual equations, one for the wheel and one for the person (including the reaction force between them in each), and using the same linear acceleration for both …
what do you get?
Hi tiny tim because we haven't learned any theory on this so we don't really know any equations besides F=ma and moment=Fd.. Also, what was that reaction force you mentioned? Thanks!
tiny-tim
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Sep16-12, 05:03 AM
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Quote Quote by sgstudent View Post
… we haven't learned any theory on this so we don't really know any equations besides F=ma and moment=Fd.. Also, what was that reaction force you mentioned? Thanks!
the only other equations you need are τ = Iα (total moment of force = moment of inertia times angular acceleration), and the "rolling constraint" v = ωr, a = αr …

just do F = ma for the person, and τ = Iα for the wheel (about the centre of the wheel), remembering to include the reaction force ħR in both (it'll be + for one and - for the other)
the reaction force is the "internal" (linear) force between the axle-and-wheels and the body of the wheelchair …

it's what pulls the body! …

for each part separately, it's an external force, of course

(technically, there's also a reaction torque, from the friction between the axle and the body)
sgstudent
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#46
Sep16-12, 07:22 AM
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Hi Tiny Tim, are there any pictures because i can't virtualise that. I was thinking the forces should look like this: http://imgur.com/WNSV7

Firstly, will my F friction = F push? Because of static friction.

So using the formula, the net F=F friction for the wheel but for the person on the seat the net F= 0?

As for moment of the wheel, it is 0Nm?

I'm getting realise confused about this now. Is there a explanation for this concept? Because I'm really curious about this haha. Thanks
tiny-tim
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Sep30-12, 02:42 PM
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hi sgstudent!

i'm sorry for the delay
Quote Quote by sgstudent View Post
Hi Tiny Tim, are there any pictures because i can't virtualise that. I was thinking the forces should look like this: http://imgur.com/WNSV7
first, it's a lot better if you stop using "F" for all your forces: use eg P for push, F for friction, R for reaction etc

for the free body diagram of the person, there's two horizontal forces …

(i'm ignoring all vertical forces)

… the push force (same as the reaction force) from the wheel onto the person's arm, and the friction force from the seat
together, they make the mass of the person times acceleration
for the free body diagram of the wheelchair, there's three horizontal forces …

… the push force (same as the reaction force) from the person's arm onto the wheel, the friction force from the person onto the seat, and the friction force from the ground
together, they make the mass of the wheelchair times acceleration
of course, you can also do a free body diagram for the wheel on it own, or for the person and the chair combined (excluding the wheel), or for the person the chair and the wheel all combined

(for the wheel on its own, there's also a small friction torque from the axle of the wheel onto the wheel)
Firstly, will my F friction = F push? Because of static friction.
there's only two forces, so Fpush - Ffriction = ma (which is 0 only if a = 0)
So using the formula, the net F=F friction for the wheel but for the person on the seat the net F= 0?
if a = 0, yes
As for moment of the wheel, it is 0Nm?
if α = 0, yes (otherwise, it's Iα)
sgstudent
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#48
Oct2-12, 05:34 AM
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Quote Quote by tiny-tim View Post
hi sgstudent!

i'm sorry for the delay


first, it's a lot better if you stop using "F" for all your forces: use eg P for push, F for friction, R for reaction etc

for the free body diagram of the person, there's two horizontal forces …

(i'm ignoring all vertical forces)

… the push force (same as the reaction force) from the wheel onto the person's arm, and the friction force from the seat
together, they make the mass of the person times acceleration
for the free body diagram of the wheelchair, there's three horizontal forces …

… the push force (same as the reaction force) from the person's arm onto the wheel, the friction force from the person onto the seat, and the friction force from the ground
together, they make the mass of the wheelchair times acceleration
of course, you can also do a free body diagram for the wheel on it own, or for the person and the chair combined (excluding the wheel), or for the person the chair and the wheel all combined

(for the wheel on its own, there's also a small friction torque from the axle of the wheel onto the wheel)


there's only two forces, so Fpush - Ffriction = ma (which is 0 only if a = 0)


if a = 0, yes


if α = 0, yes (otherwise, it's Iα)
Hi Tiny Tim

But i thought the Fpush must always be equal to Ffriction as the Ffriction is static friction. So does it mean that the torque of the wheel will be just the friction from the person to the seat? Thanks
tiny-tim
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Oct2-12, 05:43 AM
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Quote Quote by sgstudent View Post
But i thought the Fpush must always be equal to Ffriction as the Ffriction is static friction.
if the rider is accelerating, then the friction force from the seat must be more than the force on his arm from the wheel, mustn't it?
sgstudent
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#50
Oct2-12, 07:04 AM
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Quote Quote by tiny-tim View Post
if the rider is accelerating, then the friction force from the seat must be more than the force on his arm from the wheel, mustn't it?
But how is that possible? Isn't the push force equal to the static friction?
tiny-tim
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Oct2-12, 09:00 AM
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Quote Quote by sgstudent View Post
But how is that possible? Isn't the push force equal to the static friction?
which static friction?

the push force from his arm on the wheel equals the static friction force on the top of the wheel (it's just another name for it)

but that doesn't equal either of the other two friction forces (on the seat, and on the ground) unless the acceleration is zero
sgstudent
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#52
Oct9-12, 06:47 AM
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Quote Quote by tiny-tim View Post
which static friction?

the push force from his arm on the wheel equals the static friction force on the top of the wheel (it's just another name for it)

but that doesn't equal either of the other two friction forces (on the seat, and on the ground) unless the acceleration is zero
Hi tiny tim! I thought the static friction of the push is the friction from the ground?
tiny-tim
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Oct9-12, 09:55 AM
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Quote Quote by sgstudent View Post
Hi tiny tim! I thought the static friction of the push is the friction from the ground?
well, you push the wheel of a wheelchair using static friction also (and there's static friction on the seat)
anyway, nothing equals anything else unless the acceleration is zero
sgstudent
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#54
Jan4-13, 08:57 AM
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Quote Quote by tiny-tim View Post
well, you push the wheel of a wheelchair using static friction also (and there's static friction on the seat)
anyway, nothing equals anything else unless the acceleration is zero
Oh but since the friction acting on the wheels by the ground is also static friction, then shouldn't there be a push force that is equal to the friction on the ground? What would that push force be and since the push force is equal to the static friction on the wheel how can there be a net force?


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