1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ball doing a loop

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A ball with radius ##r## is inside a hollow cylinder with radius ##r+R##.
    mk7hulja.png

    In the first part of the assignment, one has to calculate the minimum kinetic energy the ball has to have at the bottom in order to complete a full loop in the cylinder. It turns out to be ##E_{\text{kin, min}}=\frac{27}{10}mgR##.

    Now we suppose the kinetic energy of the ball is 10% less than needed to do the loop. We want to compute the angle ##\theta## at which the ball will lose contact to the wall of the cylinder, using the angle ##\alpha = \pi-\theta##.

    2. Relevant equations


    3. The attempt at a solution
    My idea was to calculate the speed ##v(\alpha)## the ball has for a given angle ##\alpha##. Then I would calculate the vertical component of the centripetal acceleration and equate it with the gravitational acceleration:
    \begin{align*}
    E_{\text{tot}}& = E_{\text{kin}}+E_{\text{rot}}+E_{\text{pot}} \\
    & = \frac12mv(\alpha)^2+\frac12 I\omega(\alpha)^2+mgh\\
    &=\frac12mv(\alpha)^2+\frac12 I\left(\frac{v(\alpha)}{r}\right)^2+mg(R+R\cos\alpha)\\
    &=\frac12mv(\alpha)^2+\frac12 \cdot\frac25mr^2\left(\frac{v(\alpha)}{r}\right)^2+mgR(1+\cos\alpha)\\
    &=\frac12mv(\alpha)^2\left(1+\frac25\right)+mgR(1+\cos\alpha)\\
    &=\frac7{10}mv(\alpha)^2+mgR(1+\cos\alpha).
    \end{align*}
    Now use ##E_{\text{tot}}=0.9\cdot\frac{27}{10}mgR=\frac{243}{100}mgR##:
    \begin{align*}
    \frac{243}{100}mgR& = \frac7{10}mv(\alpha)^2+mgR(1+\cos\alpha)\\
    \frac7{10}v(\alpha)^2&=gR(\frac{243}{100}-1-\cos\alpha)\\
    v(\alpha)^2&=\frac{10}7gR(\frac{143}{100}-\cos\alpha)
    \end{align*}
    Now the vertical component of the centripetal acceleration would be ##\cos\alpha\cdot\frac{v(\alpha)^2}R## and has to be smaller than ##g##, which gives us
    \begin{align*}
    \frac{\cos\alpha}{R}\cdot\frac{10}7gR(\frac{143}{100}-\cos\alpha)&<g\\
    \cos\alpha\cdot\frac{10}7(\frac{143}{100}-\cos\alpha)&<1\\
    \frac{143}{100}-\cos\alpha&<\frac{7}{10\cos\alpha}
    \end{align*}
    so in the end we get the inequality ##\cos\alpha+\frac{7}{10\cos\alpha}>\frac{143}{100}##. If I regard this as an equality to find the minimum angle, however, the equality does not have any solutions. What am I doing wrong?
     
  2. jcsd
  3. Jan 29, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That doesn't work. Tangential acceleration also has a vertical component.
    What do you know about the forces when it loses contact?
     
  4. Jan 30, 2017 #3
    There is the gravitational force ##mg## which pulls the ball down, and the centripetal force which pushes it up and to the right. I just thought that the component to the right should be irrelevant and I should only focus on the vertical components.
    What do you mean by that? Doesn't centripetal acceleration only have a radial component towards the center of the of the circle?
     
  5. Jan 30, 2017 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Centripetal force is not an applied force. It is that component of the resultant of the applied forces that is normal to the velocity. There only two applied forces here, the normal force and gravity.
    I wrote tangential acceleration, not centripetal.

    So, again, what do you know about the forces at the point of loss of contact?
     
  6. Jan 30, 2017 #5
    In that case, we have the gravitational force and the normal force with its two components down and to the left.
     
  7. Jan 30, 2017 #6

    jbriggs444

    User Avatar
    Science Advisor

    At the point where the ball is just about to drop free of the wall, what is the magnitude of the normal force?
     
  8. Jan 30, 2017 #7
    As far as I know, the normal force has the same magnitude as the force which the ball applies onto the wall. At the point of loss of contact, the latter force has magnitude zero, which should therefore also be the magnitude of the normal force.
     
  9. Jan 30, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right. So what ΣF=ma equation(s) can you write? Bear in mind there may be tangential acceleration.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Ball doing a loop
  1. Ball going around loop (Replies: 5)

Loading...