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Ball rolling up a ramp with friction -- Find max height (Unsolved)

  1. Jun 20, 2015 #1
    Hello,

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

    A spherical continuous ball is sliding with a constant velocity v along a frictionless lane. Thereafter it enters an inclined surface (the angle between the surface and the horizontal plane is α) with the coefficient of friction µ between the ball and the surface.

    2. Relevant equations
    Find the maximal height it may reach if it is known that before full stop the ball was rolling.

    3. The attempt at a solution
    I solved it using energies but i get an extra dependence on velocity when the ball starts rolling, and my professor told me to repeat the exercice using torque, where the solution is nicer.
    I divided the motion in two parts:
    1-First the ball is sliding, when it starts going up the ramp the frictional force causes it to start rotating and lose velocity until v(t1)=wR (rolling condition).
    2-From rolling until it stops at v(t2)=0

    1: I use v(t1)=v(0)+at => (wR)=v(0)+at, but a depends on t as well, so I don't know how to continue.
     
    Last edited: Jun 20, 2015
  2. jcsd
  3. Jun 20, 2015 #2

    Doc Al

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    Staff: Mentor

    Your equation does not seem dimensionally correct. (You have velocity squared terms added to velocity terms--that won't work.)

    Apply Newton's 2nd law to translation and rotation.
     
  4. Jun 20, 2015 #3
    Sorry it was a typo, i edited it.
     
  5. Jun 20, 2015 #4

    Doc Al

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    They still show a mix of velocity and velocity2 terms mixing together.
     
  6. Jun 20, 2015 #5
    Omg im mixing up equations. Now should be correct. Anyway I tried doing torque=Friction*Radius=Inertia*alpha. Solving for alpha and substituting at equation w=w0+alpha*t for w=v1/r (rolling condition) i get t1=(2*v1)/(5*g*cos(z)*µ), t1 is time it takes to start rolling. v1 velocity when it starts rolling.
     
  7. Jun 20, 2015 #6
    I still don't know how to solve the problem. Some help would be much apreciated.
     
  8. Jun 20, 2015 #7

    Doc Al

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    Good. Now you're on the right track. This gives you an expression for alpha.

    Now find an expression for the translational acceleration. (What forces act on the ball?)

    Then, by combining the two, you can set up a kinematic equation to find out when the ball begins rolling without slipping.
     
  9. Jun 20, 2015 #8
    I did Fg*sin(z)+Ffriction=ma, solved for a, substituting at x1=x0+vt+1/2at^2, then substituting the t for the t1 that i found and i should get the displacement for the first part?.

    For the second part, when the ball is rotating im not sure what to do. I guess now the torque changes sign, but same magnitud? Then find alpha again, this time i want to find the time for when w is 0,again, use the same displacement equation and the total displacement times sin(z) should give me the final height right?
     
  10. Jun 20, 2015 #9

    Doc Al

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    It's not clear to me how you solved for the time. You need both a and alpha to solve for t1. Remember: the angular velocity increases from zero while the translational velocity decreases from the initial speed.

    But once you get t1, you can use that kinematic equation to get the distance traveled.

    Once it starts rolling without slipping, things change. For one thing, there's no more kinetic friction. But you can solve for the translational acceleration.

    You can also use conservation laws.
     
  11. Jun 21, 2015 #10
    This is what I did so far, could you tell me if it is correct?
    First I will consider the motion from ##t_{0}## until the ball starts rolling at ##t_{1}##. We have that:
    \begin{gather*}
    \tau = I\alpha \\
    \tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r
    \end{gather*}
    Solving for ##\alpha## we get
    \begin{equation}
    \alpha = \frac{mg\cos(\theta)\mu_{d}\cdot r}{I}
    \end{equation}
    From Newton second law we also have
    \begin{gather*}
    \sum F = ma \\
    F_{g}+F_{f}= mg\sin(\theta) + mg\cos(\theta)\mu_{d} = mg(\sin(\theta) + \cos(\theta)\mu_{d}) =ma
    \end{gather*}
    Solving for a
    \begin{equation}
    a= g(\sin(\theta) + \cos(\theta)\mu_{d})
    \end{equation}
    Now using the kinematics equations
    \begin{gather*}
    \omega_{f}-\omega_{i}=\alpha t \\
    v_{f}-v_{i}=a t
    \end{gather*}
    Solving t and ##v_{f}## for ##v_{i}=v, \omega_{i}=0, \omega_{f}=\frac{v_{1}}{r}## (rolling condition):
    \begin{gather}
    t_{1} = \frac{v_{1}}{r\alpha} \\
    v_{1} = v+at_{1}
    \end{gather}
    The displacement is
    \begin{gather}
    \Delta s_{1} = \frac{v_{1}^{2}-v^{2}}{2a}
    \end{gather}
    ##\Delta s_{1}## is a HUGE fraction.

    Is it correct up to here?
     
  12. Jun 21, 2015 #11

    Doc Al

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    Looks good but I would be careful with signs when you get to the kinematics step. I recommend that you let ##a## and ##\alpha## represent the magnitudes of the accelerations. I would set up the condition for rolling without slipping like so: ##\alpha r t = v - a t##

    But you're doing great!
     
  13. Jun 22, 2015 #12
    Dont you need the portion of the drag force which causes rotational torque and the remaining portion that turns linear velocity into heat ( and reduces the linear velocity) ?
    This way you can calculate the point in time when rotational surface velocity = linear velocity ( rolling without slipping )
     
  14. Jun 22, 2015 #13

    Doc Al

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    The kinetic friction has been included in the force analysis. That's all that's needed for this part of the problem. (We are using dynamics, not energy methods here.)
     
  15. Jun 22, 2015 #14
    Alright, it took me a while (I decided to learn some Latex) but I think I have the answer, or something close. Here is the entire Latex document as I will send it to my professor. The entire mark of the semester depends on this exercice, so feel free to comment And many thanks to Dol Al for the help!

    First I will consider the motion from $t_{0}$ until the ball starts rolling at $t_{1}$. We have that:
    \begin{gather*}
    \tau = I\alpha \\
    \tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r
    \end{gather*}
    Solving for $\alpha$ we get
    \begin{equation}
    \alpha = \frac{mg\cos(\theta)\mu_{d}\cdot r}{I}
    \end{equation}
    From Newton second law we also have
    \begin{gather*}
    \sum F = ma \\
    F_{g}+F_{f}= mg\sin(\theta) + mg\cos(\theta)\mu_{d} = mg(\sin(\theta) + \cos(\theta)\mu_{d}) =ma
    \end{gather*}
    Solving for a
    \begin{equation}
    a= g(\sin(\theta) + \cos(\theta)\mu_{d})
    \end{equation}
    Now using the kinematics equations
    \begin{gather*}
    \omega_{f}-\omega_{i}=\alpha t \\
    v_{f}-v_{i}=a t
    \end{gather*}
    Solving t and $v_{f}$ for $v_{i}=v, \omega_{i}=0, \omega_{f}=\frac{v_{1}}{r}$ (rolling condition):
    \begin{gather}
    t_{1} = \frac{v_{1}}{r\alpha} \\
    v_{1} = v+at_{1} = v(\frac{\sin(\theta)+\frac{3}{5}\cos(\theta)\mu_{d}}{\sin(\theta)+\cos(\theta)\mu_{d}})
    \end{gather}
    The displacement for time $t_{1}$ is
    \begin{gather}
    \Delta s_{1} = \frac{v_{1}^{2}-v^{2}}{2a}
    \end{gather}
    Substituting (5) with $v_{1}=(4),a=(2)$
    \begin{gather}
    s_{1} = \frac{-2v^{2}\cos(\theta)\mu_{d}(4\cos(\theta)\mu_{d}+5\sin(\theta))}{25g(\cos(\theta)^{3}\mu_{d}^{3}+3\sin(\theta)\cos(\theta)^{2}\mu_{d}^{2}-3\cos(\theta)^{3}\mu_{d}-\sin(\theta)\cos(\theta)^{2}+3\cos(\theta)\mu_{d}+\sin(\theta))}
    \end{gather}
    And that's our displacement until the ball starts rolling.\\ \\
    Now for the displacement until it stops, we consider Newton's second law
    \begin{gather*}
    \sum F = ma \\
    F_{g}-F_{f}= mg\sin(\theta) - mg\cos(\theta)\mu_{s} = mg(\sin(\theta) - \cos(\theta)\mu_{s}) =ma_{1}
    \end{gather*}
    Solving for $a_{1}$
    \begin{equation}
    a_{1}= g(\sin(\theta) - \cos(\theta)\mu_{s})
    \end{equation}
    Now using:
    \begin{gather*}
    \Delta s_{2} = \frac{v_{2}^{2}-v_{1}^{2}}{2a_{1}}
    \end{gather*}
    Solving $\Delta s_{2}$ for $v_{2}=0$:
    \begin{gather*}
    \Delta s_{2} = \frac{-(v(\sin(\theta)+\frac{3}{5}\cos(\theta)\mu_{d}))^{2}}{2g(\sin(\theta)+\cos(\theta)\mu_{d})^{2}(\sin(\theta)-\cos(\theta)\mu_{s})}
    \end{gather*}
    The total height is
    \begin{gather*}
    h = (s1+\Delta s2)sin(\theta)
    \end{gather*}
     
  16. Jun 22, 2015 #15

    Doc Al

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    I'll try to take a look at the details of your first part when I get some time today. But I wanted to point out an error here (after the ball rolls without slipping):
    Careful! Realize that once the ball rolls without slipping, static friction acts. And that value is whatever it needs to be to prevent slipping. It is not simply ##\mu N##. (That would be the maximum value, but that's not relevant here.)
     
  17. Jun 22, 2015 #16
    I see what you mean, but I think it still has to be constant so it shouldn't be too important. I can't believe this was 1 problem of 12 in the exam. Not that it's so hard, but the computations are so lengthy. Anyway many thanks, if you notice something wrong let me know :)

    Ps: I did most of the calculations with Maple, so if the equations are right the results should be correct.
     
  18. Jun 22, 2015 #17

    Doc Al

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    Yes, the static friction is constant. But it's not equal to ##\mu mg \cos\theta##.
     
  19. Jun 22, 2015 #18

    Doc Al

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    All this looks OK.

    Again, careful with signs. (##a## is opposite in direction to ##v##.)

    Not sure how you got this expression for ##v_1##. Do a sanity check: What happens if friction is removed?
     
  20. Jun 22, 2015 #19
    The expression for ##v_1## comes from ##\omega_{f}-\omega_{i}=\alpha t## for ##\omega_{i}=0## since it's not rolling first and ##\omega_{f}=\frac{v_{1}}{r}## rolling condition.

    ##v_1## looks good to me since it has the same units as v. Do you see a mistake?
     
  21. Jun 22, 2015 #20

    Doc Al

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    I have no problem with that starting point, but I don't see how you got your final expression for ##v_1##.

    Well, it certainly has the correct units. ;) Did you do the sanity check when ##\mu \rightarrow 0##?
     
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