For a mountain bike tyre at 25psi, the length b is around 5cm. Typical radius is 0.37m (29er). This would yield the ratio b/r as 0.05/.37 = 0.135 so rolling friction would equal 0.135 x weight = 0.135 x 400N = 54N (assuming 400N is total downward force). Typical rolling friction figures quoted at this pressure and weight for mountain bike tyres are between 3 and 4N. Bicycling Science by Gordon Wilson quotes the friction Rolling Force = 0.075 x W x b/r from empirical measurements. This factor of 0.075 reduces the 54N to 0.075 x 54 = 4N which is spot on. The only way I can see to rectify this is that there must be a force acting to accelerate the wheel when the rubber tyre recovers to its original shape. The squashing of the rubber tyre produces an opposing force but the relaxing of the rubber tyre must produce a propelling force. Hysteresis losses dissipate the energy and cause rubber tyres to have high rolling friction. Steel is much less compressible and without hysteresis losses too and so rolling friction is much less.
R = N + f . Since R pasing through the center ( R is opliquiely to the left) , then f should be directed horizontally to the left and not to the right.
Anybody please say... Why normal force acting against weight of the roller? In other problems reaction force only acting against weight of any body. And what is the difference between normal force and reaction force
If I have a small wheel with a radius of about 0.02m, when \( b \) is large and the angle is large, then I cannot write sin(ø)= tan(ø) So how would it be then?
What is meant by reactionary force? is it the resultant of the frictional force and the normal force (frictional force= horizontal component of reactionary force and normal force=vertical component of reactionary force)?
I use the term "reactionary" force in a general sense. The force reacts to the application of an original force. When you walk, you muscles push backwards on your rear foot. But since there is friction between the sole of your shoes and the floor, the friction force will push back with an equal force but in the opposite direction. This is what I call a reactionary force since it reacts to your force applied to your foot. It is this reactionary force that propels you forward.
@@MichelvanBiezen Thanks! I understand now. I have one more question.. I searched on the internet about "equation of rolling friction" and it gave me this. F (rolling friction) = Normal force x Crr (Crr= dimensionless rolling resistance coefficient ) Whats the difference between your equation and this one? and is Crr (dimensionless rolling resistance coefficient) different from the coefficient of rolling friction (the fancy l)
The Crr is a shortcut method to approximate the rolling friction. Each type of tire, or wheel, will receive a Crr which then can be used to approximate the rolling friction. The videos show the principal of rolling friction.
Theta or in this case "phi" should be the same for both cases, since reaction force (R) is an extension of the radius (r) means that both have the same slope and also Weight and Normal Force are both perpendicular to the ground, we can see that the gap or theta ("phi") would be the same. Hope this helps (sorry for being late too)
im sorry i still dont understand the reason why both angle should be the same. even if reaction force starts from the end of contact of wheel, the 'phi' could be different if the friction force is weaker or stronger. so i wonder why Friction force has the same length with half of length of deformed wheel.
I knew rolling friction was caused by deformation, but I always thought rolling "friction" was actually not friction at all, but a net torque counterclockwise (if the ball is rolling clockwise) on the front part of the deformed ball, where it touched the ground, since now the normal force of the ground was off-center... and then I thought the reason the distance was the coefficient of rolling friction was because that distance was the lever arm for the normal force THAT distance away from the center bottom point of the wheel. Then I thought we multiplied the normal force by that distance in order to get the counterclockwise torque slowing the wheel. However, in this case, the friction force itself is pointing opposite to the direction of MOTION, and not opposite to the angular velocity... If the friction force is drawn parallel to the path of motion of the ball, then wouldn't it just be Kinetic Friction for a ball that's slipping, and not rolling friction (or rolling resistance?)? Thank you!
It is correct that rolling friction is more complicated than is explained in the video, and that additional deformation occurs. But the fundamental principle of rolling friction is caused by the deformation at the point of contact (as seen in the video), causing the direction of the resulting forces to slow down the wheel. Note that one of the advantages of trains is that they have steel wheels and thus have very little deformation at the point of contact and thus very little rolling friction.
@@MichelvanBiezen I think that your approach is equivalent to the method @Joshua suggests; notably, applying a binomial expansion to the denominator of your final result produces W*b/r*(1+1/2(b/r)^2+...), which, to lowest order, is the well known result for rolling friction, easily derived from considering moments.
Sir ur best teacher 👨🏫 please illustrate us more with pure rolling motions how is k and r used in ratio of ratio of kinetic energy and total energy mechanical energy in mechanics
In order to calculate the minimum force required to act on the top of the wheel to make it start moving, would you treat the problem as an equilibrium situation and find the tipping point at the point where the bottom of the wheel leaves the ground?
The diagram on the left shows the resistance force (friction) from the road being applied to the tire. (That force acts in the opposite direction to the motion of the tire). The next diagram shows the friction force of the tire pushing against the road. (That force propels the car forward). Note that without friction force the car could not be pushed forward.
Is this equation for rolling frictional force is same for a wheel of wheelchair moving up a slope? And what is an average value of coefficient rolling frictional force ? Can you please help, this is for a project I am working.
Actually it depends on whether the wheel is moving clockwise or anticlockwise. I guess he assumed that the wheel is moving anticlockwise. But yeah, he should have mentioned that thing.
@@Shubhamkumar-cq5wt why the reaction force would be right of line of normal force passing thru centre when moving anticlockwise?... Frictional force direction is correct
That is almost the case for a wheel of a railroad car, which means the rolling friction is almost zero. (That makes freight trains a lot more efficient)
The reaction line does NOT in general pass through the center of the rim on an actual free-rolling wheel. On a free-rolling wheel, the moment due to the offset normal force is greater than the moment due to friction, which is what causes the wheel to spin at a slowing rate while translating at a slowing rate as well. For the reaction force to pass through the rim center, a forward push force must be applied exactly equal to the ground-contact normal force, times the horizontal offset distance it acts from the center of the wheel, divided by the wheel radius. Then, the friction force on the wheel will equal your push force, and the moment due to the normal force will be equal and opposite to the moment from friction. The result will be a force and moment balanced wheel. If you push with a greater forward force, then the friction force will increase further, but by a lesser amount than the increase of your push force, and the moment due to friction will exceed the moment due to the offset normal force, causing the wheel's spin to accelerate. The actual orientation of the reaction line is of little significance in evaluating rolling resistance of a wheel, and the edge of the contact zone is not actually where the reaction force acts, which is very misleading in this video. The most important thing to remember is that any wheel, no matter what material it is made of, will deform on any surface, creating a normal pressure distribution over the contact area, with greater pressures occurring in front of the center (wheel loading) than behind (unloading). This is a property of materials called hysteresis, and is responsible for the presence of rolling resistance. A 1m diameter semi-truck wheel having a rolling resistance coefficient of 0.001 will have the resultant of its normal stress distribution acting half a millimeter in front of the wheel axis. For every additional 1000 kgf of load placed on the wheel, an additional 5 Nm of rolling resistance moment must be overcome to keep the wheel rolling. A propelling torque can be applied directly to the wheel via the axle to counteract the rolling resistance moment, as in a driven wheel, and there will be no increase in rearward friction acting on the wheel. For a non-driven wheel, an increase in rearward friction supplies the necessary counter-moment, and a corresponding increase in push force is required to keep the wheel rolling steadily. In the case of the semi truck wheel, the 5 Nm moment will be supplied by a 10 N (1 kgf) friction force acting 0.5m from the wheel center. Whether the wheel is driven rotationally, or pushed, equal amounts of energy are consumed in moving the wheel along to overcome rolling resistance.
I don't see how we are "counting friction force again". Note that the rolling friction is simply a function of how much the wheel or tire is indented by the load placed upon it.
This is NOT correct.. the magnitude of the friction force varies with the degree of deformation, but is primarily a factor of the hysteresis characteristics of the material being deformed!
Sir ur best teacher 👨🏫 please illustrate us more with pure rolling motions how is k and r used in ratio of ratio of kinetic energy and total energy mechanical energy in mechanics
For a mountain bike tyre at 25psi, the length b is around 5cm. Typical radius is 0.37m (29er). This would yield the ratio b/r as 0.05/.37 = 0.135 so rolling friction would equal 0.135 x weight = 0.135 x 400N = 54N (assuming 400N is total downward force). Typical rolling friction figures quoted at this pressure and weight for mountain bike tyres are between 3 and 4N. Bicycling Science by Gordon Wilson quotes the friction Rolling Force = 0.075 x W x b/r from empirical measurements. This factor of 0.075 reduces the 54N to 0.075 x 54 = 4N which is spot on. The only way I can see to rectify this is that there must be a force acting to accelerate the wheel when the rubber tyre recovers to its original shape. The squashing of the rubber tyre produces an opposing force but the relaxing of the rubber tyre must produce a propelling force. Hysteresis losses dissipate the energy and cause rubber tyres to have high rolling friction. Steel is much less compressible and without hysteresis losses too and so rolling friction is much less.
R = N + f . Since R pasing through the center ( R is opliquiely to the left) , then f should be directed horizontally to the left and not to the right.
Anybody please say... Why normal force acting against weight of the roller? In other problems reaction force only acting against weight of any body.
And what is the difference between normal force and reaction force
If I have a small wheel with a radius of about 0.02m, when \( b \) is large and the angle is large, then I cannot write sin(ø)= tan(ø)
So how would it be then?
Unless the wheel is compressed a lot under the load, the principles should be the same.
What is meant by reactionary force? is it the resultant of the frictional force and the normal force (frictional force= horizontal component of reactionary force and normal force=vertical component of reactionary force)?
I use the term "reactionary" force in a general sense. The force reacts to the application of an original force. When you walk, you muscles push backwards on your rear foot. But since there is friction between the sole of your shoes and the floor, the friction force will push back with an equal force but in the opposite direction. This is what I call a reactionary force since it reacts to your force applied to your foot. It is this reactionary force that propels you forward.
@@MichelvanBiezen Thanks! I understand now.
I have one more question.. I searched on the internet about "equation of rolling friction" and it gave me this.
F (rolling friction) = Normal force x Crr (Crr= dimensionless rolling resistance coefficient )
Whats the difference between your equation and this one? and is Crr (dimensionless rolling resistance coefficient) different from the coefficient of rolling friction (the fancy l)
The Crr is a shortcut method to approximate the rolling friction. Each type of tire, or wheel, will receive a Crr which then can be used to approximate the rolling friction. The videos show the principal of rolling friction.
How did you reached on conclusion that angle "phi" in both the cases will be same.
Theta or in this case "phi" should be the same for both cases, since reaction force (R) is an extension of the radius (r) means that both have the same slope and also Weight and Normal Force are both perpendicular to the ground, we can see that the gap or theta ("phi") would be the same.
Hope this helps (sorry for being late too)
im sorry i still dont understand the reason why both angle should be the same.
even if reaction force starts from the end of contact of wheel, the 'phi' could be different if the friction force is weaker or stronger.
so i wonder why Friction force has the same length with half of length of deformed wheel.
Very impressive sir...got it in first attempt 😀
Great! 🙂
Is there a difference between kinetic vs. static rolling friction? Like force required to move a car vs. keeping it moving.
Rolling friction is associated with rolling motion, (kinetic).
How do you get the coefficient of rolling friction (tire from the central point to the point where it no longers makes contact) ?
Best I have ever seen!
Thanks!
are you native in english... can you help me to improve my english and i will help you to learn arabic??
@@abdalalaamer8359Hii
I'm not a native speaker, I'm just a brazilian trying to learn too. Sorry.
I knew rolling friction was caused by deformation, but I always thought rolling "friction" was actually not friction at all, but a net torque counterclockwise (if the ball is rolling clockwise) on the front part of the deformed ball, where it touched the ground, since now the normal force of the ground was off-center... and then I thought the reason the distance was the coefficient of rolling friction was because that distance was the lever arm for the normal force THAT distance away from the center bottom point of the wheel. Then I thought we multiplied the normal force by that distance in order to get the counterclockwise torque slowing the wheel.
However, in this case, the friction force itself is pointing opposite to the direction of MOTION, and not opposite to the angular velocity...
If the friction force is drawn parallel to the path of motion of the ball, then wouldn't it just be Kinetic Friction for a ball that's slipping, and not rolling friction (or rolling resistance?)?
Thank you!
If my question isn't clear please tell me, I'll ask it better. Thanks for all your videos, you're awesome!
It is correct that rolling friction is more complicated than is explained in the video, and that additional deformation occurs. But the fundamental principle of rolling friction is caused by the deformation at the point of contact (as seen in the video), causing the direction of the resulting forces to slow down the wheel. Note that one of the advantages of trains is that they have steel wheels and thus have very little deformation at the point of contact and thus very little rolling friction.
@@MichelvanBiezen I think that your approach is equivalent to the method @Joshua suggests; notably, applying a binomial expansion to the denominator of your final result produces W*b/r*(1+1/2(b/r)^2+...), which, to lowest order, is the well known result for rolling friction, easily derived from considering moments.
That's perfect explaintion I have ever heard
you are a legend!
Absolutely nice explanation. Thanks legend.
Glad you liked it!
Sir ur best teacher 👨🏫 please illustrate us more with pure rolling motions how is k and r used in ratio of ratio of kinetic energy and total energy mechanical energy in mechanics
Ruler is back
In order to calculate the minimum force required to act on the top of the wheel to make it start moving, would you treat the problem as an equilibrium situation and find the tipping point at the point where the bottom of the wheel leaves the ground?
We have videos that cover that in the physics section: PHYSICS 11.1 RIGID BODY ROTATION
@@MichelvanBiezen Thanks Michel!
The direction of the friction force in diagram 1 and the direction of friction force in the triangle are different?
The diagram on the left shows the resistance force (friction) from the road being applied to the tire. (That force acts in the opposite direction to the motion of the tire). The next diagram shows the friction force of the tire pushing against the road. (That force propels the car forward). Note that without friction force the car could not be pushed forward.
Please illustrate The rotational mechanics that is associated with rotation energy based problems
Thank you so much for this sir. Very helpful!
Most welcome!
@@MichelvanBiezen sir, what if the object have 4 wheels for example? what differ it from 1 wheel used? or is it the same rolling friction?
Is this equation for rolling frictional force is same for a wheel of wheelchair moving up a slope? And what is an average value of coefficient rolling frictional force ? Can you please help, this is for a project I am working.
best explanation I had for this topic! thank you very much good sir!
Wrong direction of friction in the force triangle (second diagram). Friction force must be from right to left
Actually it depends on whether the wheel is moving clockwise or anticlockwise. I guess he assumed that the wheel is moving anticlockwise. But yeah, he should have mentioned that thing.
@@Shubhamkumar-cq5wt why the reaction force would be right of line of normal force passing thru centre when moving anticlockwise?... Frictional force direction is correct
thank you so much , sir
so is there an mistake on wikipedia? i think so
what if there is no deformation on either the ground surface or the wheel?
That is almost the case for a wheel of a railroad car, which means the rolling friction is almost zero. (That makes freight trains a lot more efficient)
WHY DOES THE REACTION LINE PASS THROUGH THE CENTER OF THE RIM??
The reaction line does NOT in general pass through the center of the rim on an actual free-rolling wheel. On a free-rolling wheel, the moment due to the offset normal force is greater than the moment due to friction, which is what causes the wheel to spin at a slowing rate while translating at a slowing rate as well. For the reaction force to pass through the rim center, a forward push force must be applied exactly equal to the ground-contact normal force, times the horizontal offset distance it acts from the center of the wheel, divided by the wheel radius. Then, the friction force on the wheel will equal your push force, and the moment due to the normal force will be equal and opposite to the moment from friction. The result will be a force and moment balanced wheel. If you push with a greater forward force, then the friction force will increase further, but by a lesser amount than the increase of your push force, and the moment due to friction will exceed the moment due to the offset normal force, causing the wheel's spin to accelerate. The actual orientation of the reaction line is of little significance in evaluating rolling resistance of a wheel, and the edge of the contact zone is not actually where the reaction force acts, which is very misleading in this video. The most important thing to remember is that any wheel, no matter what material it is made of, will deform on any surface, creating a normal pressure distribution over the contact area, with greater pressures occurring in front of the center (wheel loading) than behind (unloading). This is a property of materials called hysteresis, and is responsible for the presence of rolling resistance. A 1m diameter semi-truck wheel having a rolling resistance coefficient of 0.001 will have the resultant of its normal stress distribution acting half a millimeter in front of the wheel axis. For every additional 1000 kgf of load placed on the wheel, an additional 5 Nm of rolling resistance moment must be overcome to keep the wheel rolling. A propelling torque can be applied directly to the wheel via the axle to counteract the rolling resistance moment, as in a driven wheel, and there will be no increase in rearward friction acting on the wheel. For a non-driven wheel, an increase in rearward friction supplies the necessary counter-moment, and a corresponding increase in push force is required to keep the wheel rolling steadily. In the case of the semi truck wheel, the 5 Nm moment will be supplied by a 10 N (1 kgf) friction force acting 0.5m from the wheel center. Whether the wheel is driven rotationally, or pushed, equal amounts of energy are consumed in moving the wheel along to overcome rolling resistance.
isn't R is the vector sum of friction force and normal force then why are we counting the friction f0orce again
I don't see how we are "counting friction force again". Note that the rolling friction is simply a function of how much the wheel or tire is indented by the load placed upon it.
@@MichelvanBiezen Thank you so much i got it now. And your videos are really helpful
Any time. Glad they are helping
This is NOT correct.. the magnitude of the friction force varies with the degree of deformation, but is primarily a factor of the hysteresis characteristics of the material being deformed!
That is an added factor, ignored in this calculation. Other factors are vehicle speed, inflation of the tires, etc.
@@MichelvanBiezen yup, fair.. there are multiple factors!
Turkish?
Are you asking if I am Turkish or would you like to see videos translated in Turkish?
Please put videos on tamil
Worst i have ever seen
Sir ur best teacher 👨🏫 please illustrate us more with pure rolling motions how is k and r used in ratio of ratio of kinetic energy and total energy mechanical energy in mechanics