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Homework Help: Rolling Friction vs Velocity

  1. Mar 12, 2012 #1


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    Hi there! First off, I am new to these forums so be nice! The question I have isn't homework, but rather personal curiosity.

    1. The problem statement, all variables and given/known data

    Imagine that you have some type of cart on a conveyer belt. The cart is held stationary with respect to the world by a rope. The conveyer belt moves at a velocity with respect to the cart. Assuming we can neglect all air resistance (since the cart is stationary with respect to the surrounding air), how does the force on the rope (in Newtons) change with increasing conveyer belt velocity? (see attached diagram)

    2. Relevant equations
    The focus here is how the force changes with rolling friction and velocity. Since I am not actually in any physics class I become stumped at this point as how to represent this problem using an equation.

    3. The attempt at a solution
    I assume that the force on the rope would increase at some rate proportional to the coefficient of rolling friction, and the velocity of the conveyer belt.

    Attached Files:

  2. jcsd
  3. Mar 12, 2012 #2
    Good question.Really good question. Let us suppose function of velocity is v=g(t)

    The role of friction between any two bodies is to prevent relative motion between them i.e to make the relative velocity between the two zero.
    To be more precise the role of friction is to keep relative motion between the common (or touching portion) of the two bodies to be 0.

    Since the wheels can spin there can be a scenario where the cart's center of mass is at rest wrt to ground(but moving wrt belt) but also the cart's lowermost point i.e the wheels is at rest wrt the belt(but moving wrt ground). {Can you picturise this scenario? }

    Initially since the cart (rather the point of contact on wheels) and belt have relative velocity of v wrt each other the rolling friction acts with its limiting value. (Why??)

    Now the role of rolling friction is to make relative velocity v between the wheels' lowermost point and and the belt to be 0.To do this, the friction creates a torque on the wheels (why?) and makes the wheels turn.

    If at a certain time the velocity of belt is v and the wheel is rotating with angular velocity w (anticlockwise), for no slipping to occur between the wheels and belt

    v=rw. (why?)
    differentiate this wrt to time to get the relationship between g'(t)=r(alpha) where alpha is angular acceleration.

    Use torque equation about centre of mass of wheel (COM) to write r*f=I*(alpha).(r is radius of wheel, f is value of friction, I is value of moment of inertia about COM)

    Substitute for alpha to obtain a relationship between f and g'(t).
    Now can you write an equation between N and f? (Its fairly simple)(N is value of string tension)

    What happens if limiting value of friction exceeds the maximum tension the string can provide?
    What happens to the cart, the wheels and the string then? (why?)
    Last edited: Mar 12, 2012
  4. Mar 12, 2012 #3


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    I get

    f= [ g'(t) I ] / [r^2]

    now am I correct in saying that g'(t) would be considered the acceleration of the belt? I don't believe that this statement is correct seeing as that would mean that there would be no tension on the rope if the belt is moving at a constant velocity.

    Now if our rope was of limited strength, it would break when it's maximum strength was equal to what is shown in the above equation.
  5. Mar 12, 2012 #4
    Correct f= [ g'(t) I ] / [r^2]
    Your last part answer is also right.and g'(t) is indeed acceleration of belt as it is differential of its velocity wrt time.

    But how are N and f related?
    (N=f= [ g'(t) I ] / [r^2]) as acceleration of COM of cart is 0)

    If acceleration of belt is 0, tension is indeed 0.
    Think of it this way, when velocity of belt is constant the friction acts initially with full value umg (for few seconds or less) and generates a torque which makes the wheel role with w such that rw=v. (v, r is constant so this w has a fixed value )

    Now since the wheel is not moving wrt the belt , friction no longer exits and the wheel continues to roll with this w and 0 angular acceleration.

    Also COM of cart is at rest wrt ground (infact the entire cart except the wheels is at rest wrt ground) .So net force on cart is 0.Since no other force apart from tension acts in the horizontal direction (we just saw friction is 0) Tension in string is 0.
    Actually when car is at rest(wrt ground) it is not pulling on the string to create the elastic force of tension.
    Last edited: Mar 12, 2012
  6. Mar 12, 2012 #5


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    But isn't there still going to be some amount of friction between the wheels and the belt, even with a constant velocity. With zero belt acceleration there must still be some force on the string. If there wasn't the cart would remain stationary wrt the ground even if we changed the belt velocity to an infinite (but constant) velocity, and then cut the rope. This thought experiment just doesn't add up in my head.
  7. Mar 12, 2012 #6
    The cart remains stationary to ground.the wheel doesnt.(it turns).

    infact the role of friction is to keep the lowermost contact point (with belt) of wheel travelling at the speed of v so that it is at rest wrt the belt.

    i.e. role of friction is to keep the touching points of two bodies relatively at rest.
    This is an important point.I hope you got this.

    Once the wheel starts turning such that the lowermost contact point is always travelling with v(and thus at rest wrt to belt) no friction acts on the wheel and thus the cart.(its like having a block lying on a table moving with velocity v.Both the block and table move with v and so No friction acts between table and block).(remember friction only acts when there is relative motion between points in contact)

    if the belt was travelling at high velocity and we cut the string.the car can still be at rest wrt ground.the wheels however will be turning v fast as w=v/r for the wheels to remain at rest wrt to belt.(think of this as a car stuck in wet mud.the wheels are rolling and rolling but the cart is not moving forward making the cart at rest wrt ground. However the wheels are not at rest wrt ground.infact they are in motion. the lowermost point of wheel is in motion such that it is at rest wrt belt).

    Remember the key concept.
    Cart can be at rest wrt ground always and moving wrt belt. The wheels on the other hand are turning keeping the contact points of wheels at rest wrt to belt(and thisnin motion wrt ground)

    In essence part of the cart (its solid body is at rest wrt ground) and part of the cart its wheels is in motion wrt ground.

    i would also like to point out that in our earlier solution we have ignored the fact in calculation g'(t) that when belt applies friction on the wheels, the wheels apply the reaction force on cart(newtons third law). This will change the function g(T).However, if belt is of large mass , then g'(t) will be large in comparison to small change in acceleration produced by friction force. ( A more rigorous derivation can be done by taking dv/dt as g'(t) + f/M where M is mass of belt ). All these results are valid after the steady period is reached i.e wheels are turning at that w which prevents slipping or relative motion between the belt and wheels)
    Last edited: Mar 12, 2012
  8. Mar 12, 2012 #7
    I think that what you are missing is the fact that role of friction is to prevent relative motion and not motion. (so if wheels contact point and belt are always moving with same velocity they are at rest wrt each other and no friction acts)

    also answer why is wr=v? (if you can answer this 75% of your doubts will go)

    also do you realise that the cart has no translational velocity wrt ground? i.e it is at rest wrt ground

    however its wheels are turning and the wheels lowermost point which is the only point which touches the belt is at rest wrt to belt.
    Last edited: Mar 12, 2012
  9. Mar 13, 2012 #8
    I made a few typos in my second last post.
    Tje reaction friction force acts on belt not cart .(Sorry.God knows what I was thinking)

    also thisinin is actually "thus in"

    And it will be g'(t) -f/M as friction on belt by wheels acts leftwards
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