# Friction and rolling resistance, and work done queries

by sgstudent
Tags: rolling, work energy theorem
P: 645
 Quote by tiny-tim it's simply torque = moment of inertia times angular acceleration … the torque has to be in the same direction as the angular accelerationif you push the cart, then looking from the right-hand side, the wheels will be accelerating clockwise, and the only external forces on the wheel are the weight (irrelevant), some friction from the axle (negligible), and the friction from the road so the friction from the road has to be clockwise, which means it's backwards
oh okay! So the assumptions i took were right? Thanks for the help tiny-tim, thanks for the patience with me

But then again, when the driving wheel turns looking at it from the right hand side, it turns clockwise, but my force is the opposite direction... thanks for the help tiny-tim! btw, i also have a dynamics question to ask hope you can help out there too! Thanks
P: 645
 Quote by tiny-tim hi sgstudent! the engine exerts a torque on the back axle, which rotates the back wheels the static friction from the road makes the car move forward opposing this are the friction in the bearings, and the rolling resistance, which isn't friction but is basically the energy lost in the continual deforming of the tyre the free body diagram of the back wheel has a torque at the axle, and a forward friction force from the road, combining to give a forward acceleration (we don't normally put the rolling resistance or the friction in the bearings onto the free body diagram … the assumption is that they are already subtracted from the engine torque to give a net torque, which does go on the diagram) the work done against gravity is the same, thw work done against friction is different the work energy equation always works
Hi tiny Tim I don't quite understand the net work done=final KE-initial KE. There's another formula which is net work done=change in energy so even when an object moves up an incline with constant speed, the two formulas collide. Thanks for the help!
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P: 26,148
 Quote by sgstudent I don't quite understand the net work done=final KE-initial KE.
i don't think i said that

i said that in this case the work done against gravity is the same …

so in the general case the (difference in) gravitational energy would have to be included
P: 645
 Quote by tiny-tim i don't think i said that i said that in this case the work done against gravity is the same … so in the general case the (difference in) gravitational energy would have to be included
Oh, so will I have to include GPE when using the work energy theorem? Doc Al was trying to explain why not to here: http://www.physicsforums.com/showthread.php?t=92470 also when I use the work energy theorem, must the forces be constant force, or can it be a situation like I kick a ball so it starts from rest to rest so no net work done? Thanks for the help tiny Tim!
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P: 26,148
hi sgstudent!
 Quote by sgstudent Oh, so will I have to include GPE when using the work energy theorem? Doc Al was trying to explain why not to here: http://www.physicsforums.com/showthread.php?t=92470
if height is changing, then you must include gravity in the work energy theorem

but you can either include it as a force (on the work done side), or you can include its potential energy (on the energy side), but not both!!

as jtbell and Doc Al said seven(!) years ago
 Quote by jtbell If you include potential energy along with the kinetic energy, then you must exclude from the net force, the force that is associated with the potential energy. In your situation, if you take "net force" to mean the sum of all forces except gravity, then the work done by that net force equals the change in the sum of kinetic and (gravitational) potential energy.
 Quote by Doc Al The purpose of gravitational potential energy is to account for the effect of gravity, so if you consider the object's weight as an external force then you don't also include potential energy--you'd be counting it twice!
 … also when I use the work energy theorem, must the forces be constant force, or can it be a situation like I kick a ball so it starts from rest to rest so no net work done?
(i don't understand your kicking example, but …)

if the force F is not constant (or if the displacement is not straight), then instead of F.x we must use ∫ F.dx
 P: 645 Oh so if an object moves up an incline with constant speed, is the network done 0 or is it the gain in GPE? Thanks for the help!
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P: 26,148
exam questions are usually about the work done by a particular force

they usually aren't interested in the net work done (since that will usually be zero, except in the case of friction, when net work done will equal loss of mechanical energy, ie gain in thermal energy)
 Quote by sgstudent Oh so if an object moves up an incline with constant speed, is the network done 0 or is it the gain in GPE? Thanks for the help!
an exam question wouldn't leave it ambiguous like that, it would specify whether the work done by gravity was to be included (usually by asking what is the work done by gravity, or what is the work done by the applied force)
P: 645
 Quote by tiny-tim exam questions are usually about the work done by a particular force they usually aren't interested in the net work done (since that will usually be zero, except in the case of friction, when net work done will equal loss of mechanical energy, ie gain in thermal energy) an exam question wouldn't leave it ambiguous like that, it would specify whether the work done by gravity was to be included (usually by asking what is the work done by gravity, or what is the work done by the applied force)
Oh, for example if part of the question already allows me to get the net work done by minusing away all regarding work done against friction and gravity. Then in another part if they ask for the speed at the end assuming that the original speed is 0. Then can I use the work energy theorem this way? Thanks Tim!
 P: 645 I think it should work right? Since when I do work done by applied force and minus away all the negative works like wd against friction and gravity so the gravity part is cleared away by minus it already. Is this right? Thanks for the help!
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 P: 645 thanks tiny tim! I think it makes more sense to include the work done against gravity rather than to leave it to a gain in GPE. Thanks again for the help!
 P: 645 hi tiny tim, i was thinking about the problem with the wheels again so for the non-driving wheels shouldn't the friction be forwards also? Since by torque they are both having the same direction ( as in turning direction) also, when principle of moments occurs where there is no net moment where clockwise moment=anticlockwise moment can there be rotation? Like even constant speed rotation or will there be always no net moment? thanks for the help!
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P: 26,148
 Quote by sgstudent hi tiny tim, i was thinking about the problem with the wheels again so for the non-driving wheels shouldn't the friction be forwards also? Since by torque they are both having the same direction ( as in turning direction)
the angular acceleration and the linear acceleration have to be balanced …
the direction of friction is decided by which one is too big for the other
for the driving wheels, there was angular acceleration from the engine

for the non-driving wheels, there is no angular acceleration from the engine
 also, when principle of moments occurs where there is no net moment where clockwise moment=anticlockwise moment can there be rotation? Like even constant speed rotation or will there be always no net moment?
at constant angular speed, the net moment is zero
 P: 645 Hi tiny Tim! thanks for the help all this while so for the driving wheel is positive work being done? Doc Al said no here: http://www.physicsforums.com/showthread.php?t=92895 but in your previous post it was otherwise.. Also I came across a weird moments question. They asked me to find the moment about a point to lift a trap door. I'm not sure if I can use principles of moment here cos lifting it requires a net moment but then again constant speed moment can also occur. So I'm sure how does this work.. Thanks for the help tiny Tim you rock!
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P: 26,148
 Quote by sgstudent Hi tiny Tim! thanks for the help all this while so for the driving wheel is positive work being done? Doc Al said no here: http://www.physicsforums.com/showthread.php?t=92895 but in your previous post it was otherwise..
no, i think Doc Al said that kinetic friction always does negative work, but static friction can do positive work

(eg static friction on a rolling driving wheel is what accelerates the car)
 Also I came across a weird moments question. They asked me to find the moment about a point to lift a trap door. I'm not sure if I can use principles of moment here cos lifting it requires a net moment but then again constant speed moment can also occur. So I'm sure how does this work..
they're talking about constant speed moment!!

the minimum force necessary is the force that gives zero acceleration (anything less doesn't open the trap door, anything more isn't the minimum)

(the net moment is of course the applied moment plus the moment of the weight)
 P: 645 Hi Tim! it makes more sense this way. So when a car travels at constant speed is the forward friction equal to the backwards friction (non driving wheels) then when the forward wheel's friction is equal to the turning of the wheel such that the forward friction is equal to the turning force at the point of contact? Will the backwards resistance change accordingly since it is still static friction at the backwheels? Also how do I tell when a moment will travel at constant speed or stay motionless when a equilibrium occurs similarly how do i tell when net force is 0 whether it is constant or 0 speed. Thanks for the help
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P: 26,148
 Quote by sgstudent So when a car travels at constant speed is the forward friction equal to the backwards friction (non driving wheels)
yes
 … then when the forward wheel's friction is equal to the turning of the wheel such that the forward friction is equal to the turning force at the point of contact? Will the backwards resistance change accordingly since it is still static friction at the backwheels?
not following you

at constant speed, there is no turning force from the engine (i'm assuming we're ignoring air resistance etc)
 Also how do I tell when a moment will travel at constant speed or stay motionless when a equilibrium occurs similarly how do i tell when net force is 0 whether it is constant or 0 speed.
if the net force is zero, the speed stays whatever it was in the first place
 P: 645 Like when I pull that trapdoor the net moment is 0 but there is constant speed while when a ruler is place on a pivot and two objects resting on both sides and moment is 0 it there is no movement of the plank even though in both cases net moment is 0 so how do I tell if an plane will move with constant speed or 0 speed at all. Thanks for the help!

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