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Rotational Dynamics - Energy, Torque

  1. Nov 28, 2012 #1
    Hi, I'm new to the forum, starting off with some rotational stuff that I am not grasping well...
    I have diagrams for each question too, which i feel are vital to understanding the questions

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

    A spherical ball bearing 0.025m in diameter, rolls without slipping down an inclined track which becomes circular and ends at point P where the track becomes vertical. Point P is 0.65m below the level from which the ball bearing was released from rest.

    a) what angular velocity (in r.p.m.) will the ball bearing have when it leaves the track at point P?

    b) how high will it rise above point P?


    2. Relevant equations

    conservation of energy, I = 0.5*m*r^2, v = Rω

    3. The attempt at a solution

    I'm not sure how to deal with the fact that I don't know the total height of the ball bearing - only the top half. So I can't really set one of the potential energy terms to zero... I tried working out a formula using conservation of energy:

    gh = 0.5*v^2 + 0.2*r^2*ω^2 + gh_p

    v = Rω

    gh = 0.5*r^2*ω^2 + 0.2*r^2*ω^2 + gh_p

    gh = 0.7*r^2*ω^2 + gh_p

    ω = sqrt((10g(h - h_p))/7r^2)

    with that though, providing it is correct at all... I can't get a quantitative value without knowing the total initial height. am i on the right track at all?

    for part b, I'm not sure whether or not to consider this as completely conserved, which would mean it rises to it's initial height after point P, but I can't see where I'd go otherwise..



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

    A string is wrapped around the small cylinder. If the string is pulled horizontally with a force F, what is the acceleration of the cylinder?

    r = 0.03m, R = 0.05m, F = 0.1N, m = 1kg

    2. Relevant equations

    τ = r x F / τ = rFsin(θ)
    τ = I*α
    a = r*α

    3. The attempt at a solution

    I started by finding the torque of the smaller cylinder using rFsin(θ)

    0.03m* 0.1N* sin(90) = 0.003mN

    Then using τ = I*α, where I = mr^2/2, I solved for alpha, since the torque on the small cylinder should be the total torque on the system...

    α = 2τ/MR^2

    2(0.003)/(0.05)^2 = 2.4rad/s^2

    using a = R*α, i found an acceleration of 0.12m/s^2


    Hopefully I'm doing these questions somewhat right, or am on the right track. Thanks for your time.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2012 #2

    gneill

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    Hi Stealth849, Welcome to Physics Forums.

    In future, you should make separate threads for separate problems so as to avoid a confusing mishmash of interleaved responses.

    Addressing the first problem, your approach is along the right lines. What is important about using potential energy is that the energy you gain or lose is due to a CHANGE in position. The particular initial height is not required and can be ignored. What is significant is that the specific energy left with the ball bearing when it reaches the end of the track will be equal to the g*Δh, where Δh is the difference in height between the starting point (where the ball was at rest) and the end point. This energy will be divided between linear KE and rotational KE as you've surmised.
     
  4. Nov 28, 2012 #3
    Hey, thanks for the reply and for the advice! I'll be sure to split up my questions from here on. :)

    anyway, when you say that the initial height doesn't actually matter, are you in other words saying that in my equation, (h - h_p) can just be considered the change in height Δh, which is equal to 0.65m?

    Also, just to check my algebra, can someone confirm my equation?

    thanks.
     
  5. Nov 28, 2012 #4

    gneill

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    Yup. Δh from start to finish is all you need to compute the change in energy.
    Looks okay. You might have indicated that you used the formula for the moment of inertia of a sphere to explain the "0.2" coefficient for the rotational energy.
     
  6. Nov 28, 2012 #5

    collinsmark

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    Hello Stealth849,

    Welcome to Physics Forums!

    As gneill mentioned, please limit yourself to one question per thread in the future.
    Your equation looks correct to me! :approve:

    Like gneill mentioned, the "h_p" term in the equation is zero. Assume that when the 0.65 m height was measured, it was measured from the bottom of the ball when the started at rest, to the bottom of the ball when the ball was at point P. Or it was measured at the top of the ball to the top of the ball at both locations. Either way. As long as measurement consistency is maintained, the total height difference is 0.65 m, and h_p = 0.
    If the problem doesn't mention anything about friction, then I'd assume it should be ignored.

    But looking at your attached figure, once the ball reaches point P it runs off the track, and shoots up vertically. There is no longer anything to keep the ball from slipping, and it is free to keep rotating. So no, it does not climb back up to its original height. Some of the energy is maintained as rotational energy. You'll have to do a separate conservation of energy calculation to find the final height.
    Is there any more information given in the problem statement? For example, are we to assume that there is no slipping between R and the ground below? Or is the cylinder on a frictionless, horizontal plane? It makes a difference for this problem.
    Are you sure about I = mr2/2 ? That's the correct formula if all the significant mass is contained in a solid cylinder of radius r, and all the mass between r and R is negligible.

    But if the object is constructed more like a yo-yo of outer radius R, and only containing a tiny section in the middle of radius r for the string to wrap around, a better equation for the moment of inertia might be I = mR2/2.

    Any more information about the object?
    If we are to assume that the object does not slip on the ground (i.e. no movement where the object at radius R touches the ground), then there is another force and another torque, caused by static friction.
     
    Last edited: Nov 28, 2012
  7. Nov 28, 2012 #6
    hey thanks for the replies everyone!

    for my first question, plugging in numbers I get:

    ω = sqrt(10gΔh/(7r^2))

    ω = sqrt(10*9.8*0.65/(7*0.0125^2))

    which gives me 241 rad/s as rotational velocity... part of me feels like this is too large a value to be correct, but maybe my intuition is getting in the way here. does that seem like a reasonable value? it means that the ball is rotating at 38 full rotations per second, and 0.64 rpm...


    And for the second question, that was my bad on saying I = mr^2/2. It should have been mR^2/2 as you said collin. just missed the capitalization typing it out.

    there is no other info on the object though, or the conditions. I would assume that there is no slipping involved. I guess the thing that gets me is what I really solved for.

    If I did everything right, I think I would have solved for a_tan of the whole thing. But I don't know if that's what I really want or if I wanted the translational acceleration a_cm, which I don't know how to find...

    Saying that there is another force and torque involved, how could I include that with the limited information given?

    Ʃτ = Iα

    τ_t - τ_f = Iα

    But there I only know the tension force and the moment of inertia...

    Sorry if that is confusing.
     
  8. Nov 28, 2012 #7

    haruspex

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    Yes
    Er.. no.

    I'm unclear where you are up to with part (b). It's in free fall now, and you know its launch velocity so you know how it will behave. Is there still a problem?
     
  9. Nov 28, 2012 #8
    my bad. meant to multiply that for 2301 r.p.m. but i'm only wondering if this seems like a reasonable value for rotational velocity. It just seems very high to me intuitively. I guess I just wanted to know if this is a value that would be observed in real life under these conditions.
     
  10. Nov 28, 2012 #9

    collinsmark

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    Your answer of 241 rad/s looks correct to me. :smile: (Using g = 9.8 m/s.)

    Don't forget, you'll still need to find the final height.
    I would also assume that no slipping is involved. If the object was on a frictionless plane then a = F/m, regardless of where the string is attached, and rotational stuff wouldn't make a difference. Oh sure, it might rotate, but that wouldn't change the fact that the linear acceleration would still be a = F/m. And that would make for a pretty silly problem, being that the author of the problem went through the trouble of specifying r and R.

    But perhaps more importantly, by looking at the attached figure it's really unclear to me what the moment of inertia is supposed to be.
    • If the object is like a solid, heavy, iron cylinder of radius r with a couple of flimsy CDs of radius R glued on at each end, then I = mr2/2 makes for a great approximation.
    • If the object is like a heavy cylinder of radius R with a narrow notch cut in the center of radius r, then I = mR2/2 is the better formula.
    It's not possible to tell which to use from the information given. Is I = mR2/2 specified somewhere in the original problem statement?
    I don't think you'll need any trigonometric functions to solve this problem.
    The assumption is that the force of friction is keeping the bottom of the cylinder of radius R stuck to the ground. This force points in a direction parallel to the ground, in the opposite direction as F

    It will be easier to visualize how the forces and torques relate once you draw your free body diagram (FBD).
    Good! :approve: That's one equation.

    You also have the linear version of Newton's second law, of course,
    F - Ff = ma.
    That's the second equation.

    You'll then need to relate how each torque relates to its corresponding force using R or r.

    Finally, you'll need one more equation that relates α and a, based on the fact that bottom of the cylinder (of radius R) is stuck to the ground. (It's a pretty simple relationship).

    You should be able to combine those equations (plus the formula for moment of inertia), to create a formula that gives a as a function of F, m, R, and r.
     
    Last edited: Nov 28, 2012
  11. Nov 28, 2012 #10
    I feel that it is modeled more like the second option, which is like a compound wheel. I figured that once I found the torque acting on the smaller portion, that it would ultimately be the same torque acting on the whole cylinder, and I could just solve for a using the moment of inertia of the entire cylinder
     
  12. Nov 28, 2012 #11
    a_tan is just the tangential acceleration, nothing to do with trig. But, the tangential acceleration isn't the same as how the center of mass of the cylinder is accelerating horizontally with translational motion.

    Also, if I relate those equations of torque, force and acceleration, I can find the frictional torque...? Or in other words, solve that equation that has two unknowns?
     
  13. Nov 28, 2012 #12
    Oh, and for the max height, I got the equation

    mgh + 1/2mv^2 + 1/2Iω^2 = mgh + 1/2Iω^2

    because rotational kinetic energy is no longer changing, those terms cancel leaving

    mgh - mgh = 1/2mv^2

    mg(Δh) = 1/2mv^2

    Δh = v^2/(2g)

    v = Rω = 3m/s

    h = 9/2*9.8 = 0.46m.

    hope that's right...
     
  14. Nov 28, 2012 #13

    collinsmark

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    If the cylinder is allowed to freely -- completely freely -- sliding at ease; or say if the whole contraption was floating in space; then there is only a single force F acting on the object and there is only one torque involved. The linear acceleration would be a trivial a = F/m, and the angular acceleration would be something that wouldn't be too difficult to calculate.

    I didn't think of this 'till now, but if the cylinder was locked in place at its center of mass (think of a workman's tool like a table saw or a stationary grinding wheel), then there would also be a single torque involved. And angular acceleration would be something that would be calculated in the same way as if it were freely floating. But since the cylinder is just rotating around an axis that's locked in place, the linear acceleration would be zero.

    In my previous posts, I was thinking that the cylinder was allowed to roll on the ground with the restriction that no slipping is allowed. The reason I assumed that the wheel is allowed to roll on the ground (as opposed to just spinning in place around a fixed axle) is because in the attached figure there is a long horizontal line at the bottom of the cylinder that I assumed to be the ground.

    Do you have any other possible explanation as to why that long horizontal line is there at the bottom of the figure? (This isn't a rhetorical question nor a hint in question form. I honestly don't know the answer to this.)
    Okay, I understand what you mean now. :blushing:
    Well, assuming we're supposed to go with the rolling on the ground option, yes the frictional torque is

    τf = Ff R.

    And you can use Newton's second law of motion to express Ff in terms of F, m and a.

    That said, now I'm not so certain that my rolling on the ground assumption was correct. This problem is very ambiguous in its description. Is there any chance you can ask for clarification from your instructor about what is moving and how it's allowed to move?
     
  15. Nov 28, 2012 #14
    Hey, sorry for the confusion in the question. It definitely does not give enough info to avoid ambiguity. :(

    That said, the problem should be about the cylinder rolling on the ground I believe.

    So when finding the frictional torque, I'm not sure what I would use as the acceleration value a.

    Looking at ƩF = ma, we get

    F_t - F_f = ma

    0.1 - F_f = 1*a

    Can I use the value of a that was initially found using tension torque? (see first post) the value was 0.12m/s^2.

    Otherwise, I don't really see how I can find a value of a.
     
  16. Nov 29, 2012 #15

    collinsmark

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    Okay, :smile: for now we'll just go with the rolling on the ground option.

    (You still might want to talk with your instructor if you get the chance, before you turn the problem in, just to be sure. In the mean time though, this is at least a great learning experience!)
    Yes, that's right! :approve: With that equation, solve for Ff in terms of a. Keep it in your back pocket. We'll use it later for substitution.
    0.12 m/s2 is the correct tangential acceleration if the situation is a cylinder attached to a fixed axis, such as a stationary grinding wheel type of setup.

    But it's not correct if we're supposed to assume the situation is a cylinder rolling on the ground (and if we're supposed to be finding the linear acceleration of the cylinder's center.)
    We'll there's still the angular version of Newton's 2nd law to be put to use:

    τt - τf = Iα

    I'll give you a hint: τf = Ff R, and you already know what Ff is (in terms a), which you can substitute into the above equation (thus getting rid of the Ff variable). See if you can get everything in terms of F, R, r, m, and a. Then solve for a.
     
  17. Nov 29, 2012 #16
    Okay...

    Good hint. ;)

    Had a lot of trouble solving for a before I realized that 'r' should be used instead of R for the τ_t variable...

    working things out,

    F_f = F_t - ma

    F_t*r - F_t*R + Rma = Iα

    F_t*r - F_t*R + Rma = Rma/2

    a = F_t*r - F_t*R/(-0.5mR)

    Providing I didn't screw up the algebra there, I got

    a = 0.1(0.03) - 0.1(0.05)/(-0.5*1*0.05)

    a = 0.08m/s^2

    It would also be moving in the direction of the pull F_t.

    Not feeling at all confident about my answer, though, mostly due to the 2am algebra. :(
     
  18. Nov 29, 2012 #17

    collinsmark

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    Ooo, almost! :smile: (Seriously though, that was a really good effort!)

    The only problem is the direction of a and α relationship. And I wasn't any help on getting our direction definitions consistent.

    We defined our main equation as:

    τt - τf = Iα

    So if each side of the equation is positive, it means α is in the direction dominated by τt, which is in the clockwise direction. You need to look at the figure to see this. The force from the tension is on the bottom of the smaller cylinder, causing the torque from that force to be in the clockwise direction. (By itself, there's nothing wrong with this definition.)

    But we also had

    F - Ff = ma

    So if both sides are positive, it means that a is in the same direction as F. But if a is in the same direction as F it means that the cylinder rolls in the same direction toward F. And if that happens, the cylinder rolls counter-clockwise.

    So with the conventions we chose, a = -Rα. And that negative sign turns out to be pretty important for this problem. :cry:

    Alternately, we could have kept the a = Rα the way it was, and set up our main equation as

    τf - τt = Iα.

    That would have worked too.

    The truth is, we didn't pay too much attention to our conventions when working on this problem, but now I'm thinking we should have.
     
  19. Nov 29, 2012 #18
    Damn ahaha. So close, yet ironically, in the complete opposite direction. :P

    I realize now that I just kind of tacked on the direction due to intuition and had nothing in my calculation to back it up... Your explanation makes much more sense of it though.

    Is that to say that my algebra otherwise is alright and 0.08m/s^2 is indeed the linear acceleration of the center of mass?

    Thanks again for all the time you've put in to helping me learn these concepts.. :)
     
  20. Nov 29, 2012 #19

    collinsmark

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    Yes, otherwise alright. :approve:
    No, don't take it that far. The '-' sign affects more than the final direction.

    It's more like a
    -1 - 1/2 = -3/2
    instead of
    1 - 1/2 = 1/2
    sort of thing.
    Good luck! :smile:
     
  21. Nov 29, 2012 #20
    Ah I see. Tried it again and came up with:

    a = 2(F_t*R - F-t*r) / 3Rm

    got a value of 0.027m/s^2. Does this seem reasonable for an acceleration value in this scenario...? Seems rather small..
     
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