1. Not finding help here? Sign up for a free 30min 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!

Rod rotating round it's end

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A rod of length L and mass m which is free to rotate round it's edge is held, initially, horizontally.
    What are: the angular velocity and acceleration at angle θ
    The force at the nail (the axis)

    2. Relevant equations
    Torque and moment of inertia: ##M=I\alpha##

    3. The attempt at a solution
    Conservation of energy:
    $$mg\frac{L}{2}\cos\theta=\frac{1}{3}mL^2\omega\;\rightarrow \omega=\frac{3}{2}gl\cos\theta$$
    $$M=I\alpha:\; mg\frac{L}{2}\sin\theta=\frac{1}{3}mL^2\alpha\;\rightarrow \alpha=\frac{3g}{2L}\sin\theta$$
    The linear, tangential acceleration ##a_t=\alpha\frac{L}{2}##
    I take forces in the tangential direction, F is the reaction:
    $$mg\frac{L}{2}\sin\theta-F=m\cdot \frac{3g}{2L}\sin\theta \frac{L}{2}\;\rightarrow\; F=\frac{1}{4}mg\sin\theta$$
    But i am not sure.
    The reverse of the axial component of gravity ##mg\cos\theta## is the second component of the reaction in the nail.
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You seem to be mixing energy and angular momentum.
     
  4. Jul 10, 2015 #3
    $$mg\frac{L}{2}\cos\theta=\frac{1}{2} \frac{1}{3}mL^2\omega^2\;\rightarrow \omega^2=\frac{3g\cos\theta}{L}$$
     
  5. Jul 10, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right.
     
  6. Jul 10, 2015 #5
    But what about the main question, the force on the nail?
     
  7. Jul 10, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The force will depend on the rate of rotation (centripetal force), which you seem to have omitted.
     
  8. Jul 10, 2015 #7
    The centripetal force on the COM at the distance L/2: ##m\omega^2\frac{L}{2}## is added:
    $$mg\frac{L}{2}\sin\theta-F=m\cdot \frac{3g}{2L}\sin\theta \frac{L}{2}\;\rightarrow\; F=\frac{1}{4}mg\sin\theta$$
    $$F_{tot axial}=\frac{1}{4}mg\sin\theta+m\omega^2\frac{L}{2}$$
     
  9. Jul 11, 2015 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    F is the component of the axial force parallel to the motion of the mass, i.e. tangential, no?
    You need to find the total radial force (including the centripetal force) and add that vectorially.
     
  10. Jul 11, 2015 #9
    Yes, i made a mistake.
    $$\left\{ \begin{array}{l} F_{tan}=\frac{1}{4}mg\sin\theta \\ F_{rad}=mg\cos\theta+m\omega^2\frac{L}{2} \end{array} \right.$$
     
  11. Jul 11, 2015 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That looks right.
     
  12. Jul 11, 2015 #11
    Thanks
     
  13. Jul 11, 2015 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Karol,

    Should this title actually be "Rod rotating round it's end" ? -- not edge
     
  14. Jul 11, 2015 #13
    Yes, correct, i don't know physics AND english well...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted