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Ball rolls without sliding in a concave bowl

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Ball of mass m and radius r rolls without sliding in a bowl of radius R.
    Express it's angular velocity ω as a function of the angle θ.
    What are the normal force and friction force as a function of θ.
    Snap1.jpg
    2. Relevant equations
    Moment of inertia of a solid ball round it's center: ##I_c=\frac{2}{5}mr^2##
    Sheteiner's theorem: ##I_A=I_c+mr^2##
    Kinetic energy of a rigid body: ##E=\frac{1}{2}I\omega^2##
    Centripetal force: ##F=mr\omega^2##
    Torque and angular acceleration: ##M=I\alpha=I\dot\omega##

    3. The attempt at a solution
    The descending in height is transferred to kinetic energy:
    $$mg(R-r)\sin\theta=\frac{1}{2}\left( \frac{2}{5}mr^2+mr^2 \right)\omega^2\;\rightarrow\; \omega^2=\frac{10}{7}\frac{R-r}{r^2}g\sin\theta$$
    $$\omega=\left( \frac{1}{r}\sqrt{\frac{10}{7}g(R-r)} \right) \cdot (\sin\theta)^{\frac{1}{2}}$$
    $$\dot\omega=\left( \frac{1}{2r}\sqrt{\frac{10}{7}g(R-r)} \right) \frac{\cos\theta}{\sqrt{\sin\theta}}$$
    The normal force is the substarction of the weight's radial component and the centripetal one:
    $$F=mg\sin\theta-\frac{1}{2}I\omega^2=mg\sin\theta-\frac{10}{7}\frac{R-r}{r^2}g\sin\theta=mg\sin\theta\left( \frac{10}{7}\frac{R-r}{r}+1 \right)$$
    The friction force:
    $$M=f\cdot r=I_c\dot\omega\;\rightarrow\; f=\frac{2}{5}mr\cdot\left( \frac{1}{2r}\sqrt{\frac{10}{7}g(R-r)} \right) \frac{\cos\theta}{\sqrt{\sin\theta}}$$
     
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  3. Apr 22, 2016 #2

    haruspex

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    I don't see the connection between the centripetal force and ##-\frac 12 I\omega^2##. I count four things wrong there.
     
  4. Apr 22, 2016 #3
    $$F=mg\sin\theta-mr\omega^2=mg\sin\theta-\frac{2}{5}mr^2\cdot \frac{10}{7}\frac{R-r}{r^2}g\sin\theta=mg\sin\theta\left(1+ \frac{4}{7}\frac{R-r}{r} \right)$$
     
  5. Apr 22, 2016 #4

    haruspex

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    That fixed two problems, two to go.
    Remember that the centripetal force is a component of the resultant of the actual forces. In fact, it is better to renounce the concept of centripetal force and think instead in terms of centripetal acceleration.
    To make sure the signs are right, get into the habit of writing such equations in the standard Fnet=ma form, where the net force is the resultant of the actual forces.
    The other problem concerns the (angular) velocity and radius of the centripetal acceleration. Where is the centre of curvature of the trajectory of the ball's centre?
     
  6. Apr 22, 2016 #5
    $$F=m(R-r)\omega^2-mg\sin\theta=m(R-r)\cdot \frac{10}{7}\frac{R-r}{r^2}g\sin\theta-mg\sin\theta=mg\sin\theta\left(1+ \frac{10}{7}\left( \frac{R-r}{r} \right)^2 \right)$$
     
  7. Apr 22, 2016 #6

    haruspex

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    Still not right.
    I assume you are taking g to have a positive value. If F is the normal force, what is the net radial force on the ball?
    Also, you are confusing ##\omega## with ##\dot\theta##.
     
  8. Apr 22, 2016 #7
    $$F=m\frac{v^2}{R-r}+mg\sin\theta=\frac{m}{R-r}\omega^2r^2+mg\sin\theta=\frac{m}{R-r}\cdot \frac{10}{7}\frac{R-r}{r^2}g\sin\theta r^2+mg\sin\theta=mg\sin\theta\left(1+ \frac{10}{7} \right)$$
     
  9. Apr 22, 2016 #8

    haruspex

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    Yes, that looks right.
     
  10. Apr 23, 2016 #9
    Thank you Haruspex
     
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