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Homework Help: Toppling problem

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data
    This is a solved problem I am trying to understand, I have attached the problem.

    2. Relevant equations

    3. The attempt at a solution
    In the solution of i), I don't understand why is it ok to apply the torque equation about O. The question does not state that block does not slide before toppling and ##\tau=I\alpha## can be applied only about CM or a fixed point, right?

    Its been quite some time I have done these kind of problems, I am very sorry for any silly questions.

    Any help is appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Oct 18, 2013 #2


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    The equation T = Iα applies only when you are interested in trying to figure out the angular acceleration produced by a net torque applied to a body. Your problem merely asks you to determine the amount of force required to topple the body, not how fast the body topples once it starts to rotate.

    This is a quasi-static analysis which is solved using the equations of statics alone.
  4. Oct 18, 2013 #3
    Okay, got the first part.

    What about the second one? How did they conclude that ##\mu_{min}=0##, the reasoning doesn't make sense. Can I have a few hints for this part?

  5. Oct 18, 2013 #4
    Hey, Pranav!
    The question doesn't mention how it topples. It topples on the front or the back edge depending on the conditions. Think about it this way, if it has to topple forward (fall towards front), then friction must be present but what about the back edge (fall backwards) and remember force F can be varied to give desired results. Another question you can ask yourself is, can the block topple even if friction is absent. If the friction isn't present then as you know you'll have to calculate torque about its COM to avoid other technicalities. I hope this gives you clues!!!
  6. Oct 19, 2013 #5
    Why? The solution to the part ii) says that friction is not necessary for toppling. :confused:

    What technicalities? Why can't I take moments about the point O? :confused:
  7. Oct 19, 2013 #6
    read the complete thing!!!!

    it can topple backwards, without any frictional force whatsoever, it would just need a specific force for that to happen.

    If friction isn't present then the body must be accelerating, right???.....if you take moment about point O, then you'll have to take acceleration of body into account otherwise you'll get incorrect equation!!!
  8. Oct 19, 2013 #7


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    Hi Pranav-Arora! :smile:
    If the block can topple with µ = 0, clearly that is the minimum!

    So all you have to show is that it can.

    Hint: if this was a billiard ball, if you hit it slowly enough (below the centre of mass), it will move forward, rolling

    but if you hit it hard enough, it will move forward with backspin. :wink:
  9. Oct 19, 2013 #8
    Hi tiny-tim! :)

    Can you please help me in forming the equations for the case of billiard ball? I understand better by doing the maths.

    Our question is that what is the minimum force so that the billiard ball move forward with backspin. Lets assume that the billiard ball is a solid sphere of mass M and radius R.

    How should I begin here? Should I take the moments about CM?

    Okay, lets make the equations.

    Torque about O: ##\tau=mg(a/2)-F(a/4)## where anticlockwise is considered to be positive.

    Newton's second law: ##F=mb## where b is the acceleration of block.

    How do I relate the angular acceleration ##\alpha## and b? Is it ##b=(a/\sqrt{2})\alpha##?

    I am very sorry if these are dumb questions but I honestly can't proceed, I don't know what I am missing here. :(
  10. Oct 19, 2013 #9
    You have not included fictitious torque!!!

    First of all if friction isn't acting the block will topple backwards, so normal force would act along back edge. Try to see that, block can topple either way forward or backward.

    Consider it a advice: Its always good to take torque about COM (for at least accelerating body) unless situation forces you to do otherwise.
    if you introduce fictitious torque, you'll only increase calculations and hence complications.

    For toppling condition: taking torque about COM, we see that normal force acts on the back edge, so we see that normal force gives clockwise torque about com and external force F gives anticlockwise torque. If you apply any higher force, it would simply topple backward. This situation satisfies toppling condition. And you can even see that friction was not even required to show that.
  11. Oct 19, 2013 #10
    Can you please write the missing term?

    So if there had been friction, how would I determine the direction of toppling?

    In the present case, how did you conclude that it will topple backwards? I am sorry if this is obvious. It feels like it should topple about back edge given the force is acting below the CM but is it somehow possible to prove it?

    Thank you for your time.
  12. Oct 19, 2013 #11
    I personally avoid non-inertial frame specially for rotation, sorry can't help you with that equation.

    I mentally calculated torque about COM, and check the tendency of rotation if I increase the force beyond the balance condition.

    it purely depends on the situation. This intuition develops with problem solving, it isn't obvious at first to anyone.

    You can determine the direction by balancing the torque, and seeing the tendency of block's rotation if Force F is increased. If the block is accelerating, calculate torque about COM to avoid fictitious torque.

    Edit: Read the detailed post of tiny-tim below!!! It has been very well explained.
    Last edited: Oct 19, 2013
  13. Oct 19, 2013 #12


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    Hi Pranav-Arora! :smile:

    For the block:

    Assume no friction: then there are only three forces, the applied force F, the normal force N, and the weight mg.

    Let's assume that forwards is clockwise.

    If the block is not rotating, then the moment of N must balance the moment of F about the centre of mass (we're using the centre of mass so that we can ignore the weight, mg :wink:).

    Since F is anticlockwise about the centre of mass, that means N must be clockwise, ie the normal force is shifted to the rear of the block.

    So if the block topples at all, it must do so when N reaches the rear edge!

    (if F was above the centre of mass, it would be the other way round: N would be shifted forward, and the block would topple if N reached the front edge)

    For a ball:

    (i really only mentioned this to give you an intuitive idea of why there would be "backspin" on the block)

    Take moments about the instantaneous centre of rotation (which didn't exist in the block case), and combine F = ma (or F minus friction = ma) with τ = Ic.o.rα. :wink:
  14. Oct 19, 2013 #13
    Nicely explained tiny-tim! Thanks a lot! I can finally make some sense of what's going on. :smile:

    And thank you NihalSh! :)

    Let the force act at distance R/4 below the centre of mass of the ball.

    Torque about the instantaneous centre of rotation is ##\tau=F(3R/4)=I_{c.o.r}\alpha##
    where ##I_{c.o.r}=7MR^2/5##.

    From Newton's second law: ##F-f=Ma##.

    Angular acceleration is related to acceleration of CM of ball by ##\alpha=a/R##. I guess friction is known here, right?
    Last edited: Oct 20, 2013
  15. Oct 20, 2013 #14
    you might wanna clear that out, instantaneous center of rotation is different from center of rotation (usually taken to be COM for rolling ball).


    P is the instantaneous center of rotation whereas M is COM.
  16. Oct 20, 2013 #15
    I do know what is an instantaneous centre of rotation. Is there some error in my equations?
  17. Oct 20, 2013 #16
    I believe you have taken the distance (moment arm) ##\frac{R}{4}## incorrectly.

    Edit: I believe moment arm should be ##\frac{3R}{4}## if the distance (force is applied below COM) from COM is ##\frac{R}{4}##
    Last edited: Oct 20, 2013
  18. Oct 20, 2013 #17
    Ah yes, you are right. Don't know what I was thinking while making the equations. :redface:

    I have edited the post, thank you.

    Can you answer my other query?
  19. Oct 20, 2013 #18
    That relation is only valid for pure rolling (no sliding). You shouldn't calculate torque about ICR, if the body is accelerating (remember what I said previously about fictitious torque).

    friction in this case(back spin case) would be kinetic. and after a while if friction is present then, rotation rate would decrease (and change direction, if situation permits) and because friction acts in the rear direction, velocity of COM would decrease. So given sufficient time the ball would eventually start to pure roll.

    Again depends on situation, if force is steadily increased then you must know friction to tell when the object will start to backspin. On the other hand if impulse of high magnitude acts, then you can ignore friction (assuming impulse acts for such a short time that friction doesn't get a chance to set in). And you can use linear and angular momentum conservation to determine the rate of rotation, just after impact.
    Last edited: Oct 20, 2013
  20. Oct 20, 2013 #19
    I don't agree with this. In the backspin case, the particle in contact with the ground has forward velocity so the friction acts backwards.

    But then, it means I can't use the instantaneous centre of rotation. :confused:
  21. Oct 20, 2013 #20
    yeah, my mistake for overlooking that!!!...edited the original post....thanks for pointing that out!!!.....btw I meant backward but wrote forward the rest of it was correct....silly me!!!:redface:

    But you get the point, the ball will eventually start to roll. If translation kinetic energy is high enough it would continue in the forward direction. otherwise it will change direction and start moving in the backward direction, and at some point it will start rolling. But if translation and rotation kinetic energy both become zero simultaneously then the ball will stop moving.
    Last edited: Oct 20, 2013
  22. Oct 20, 2013 #21
    Use COM as it can always be used.
  23. Oct 20, 2013 #22


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    Hi Pranav-Arora! :smile:
    your basic equation is ##\tau=Fh =I_{c.o.r}\alpha##

    to solve this, you need one more equation

    if it's rolling, that equation is the rolling constraint ##a = r\alpha##

    if it's sliding, that equation is F - µkmg = ma :wink:
    technically, you have to use a fixed point (for the torque)

    however, you can usually get away with using the (moving) centre of mass …

    the fundamental angular momentum equation is τP = dLP/dt about any fixed point P

    since LP = Ic.o.mω + rc.o.m x mvc.o.m,

    we can ignore the last term only if d/dt(rc.o.m x mvc.o.m) = 0

    fortunately, for eg a cylinder (or a ball rolling in a straight line, but not a cone) rolling on a flat surface, if P is an "initial" position of the c.o.m, then rc.o.m is always parallel to mvc.o.m, so that's d/dt(0)

    and (same case), if P is an "initial" position of the c.o.r, then rc.o.r is always parallel to mvc.o.m, so that's d/dt((rc.o.m - rc.o.r) x mvc.o.m) = mr2α, which we can add to Ic.o.mω to get Ic.o.rω

    and for a cylinder rolling up a step, when it is rotating about the edge P of the step, rc.o.m x mvc.o.m is of magnitude mr|v| = mr2ω in a constant direction (parallel to the edge), and so d/dt of it is mr2α, which again we can add to Ic.o.mω to get Ic.o.rω !

    conclusion: it's not as simple as you might think …

    technically, you should use a fixed point P, and the complete equation LP = Ic.o.mω + rc.o.m x mvc.o.m

    but there are many situations where you can get away with just Ic.o.mω or Ic.o.rω :smile:
  24. Oct 20, 2013 #23
    Hi tiny-tim! :smile:

    In the backspin case, the ball slides so we can't use the rolling constraint. So I have these equations:

    ##\tau=Fh =I_{c.o.r}\alpha##
    ##F-\mu_k mg=ma##

    But I still need one more equation. I want to find the minimum force required for backspin. Do I need some more information to solve the problem?

    Superb explanation, thanks a lot tiny-tim! :smile:

    But I am still stuck at the following point:
    How do you get "d/dt((rc.o.m - rc.o.r) x mvc.o.m) = mr2α"? I have attached an image describing the situation. I hope it will help you to explain things easily.

    Attached Files:

  25. Oct 20, 2013 #24


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    Hi Pranav-Arora! :smile:
    if you know what F is, the first equation gives you α and the second equation gives you a

    the minimum force will be when a = rα :wink:
    rc.o.m - rc.o.r is a constant, the vertical vector of length r

    so you have d/dt(mrv) = mrdv/dt = mra = mr2α :smile:

    (btw, i've decided to use the forum's Fixedsys font for α … it comes out much better o:)

    but can everybody see it? maybe the more popular Georgia font would be better: α ? :confused:)​
  26. Oct 20, 2013 #25
    I honestly don't get this, why the force is minimum when you use the rolling constraint? :confused:

    Thank you once again tiny-tim! :smile:

    The fixedsys one looks quite small but is still readable. Georgia one is ok, looks more like an "a" but I usually use LaTeX to write out the symbols. :)
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