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!

Question regarding energy

  1. Nov 7, 2006 #1
    Hi there, I have a question regarding energy that I'm not sure if I have solved correctly. I was wondering if anyone here may please have a look. Thanks!

    Question: A 2 kg mass is on the end of a massless rigid rod which pivots freely about one end. The mass is moving in a vertical circle with a speed of 6 m/s at P when it reaches Q, its speed is 8.00 m/s. Find a) the length of the rod b) force exerted by the rod on the mass when the mass is at P.

    Diagram:
    Code (Text):

     P
       \
         \
           \
     30 deg  \
               \
     __________ +
                |*
                |
                | *
                |
                |  *
                |
                |   *
                |40 deg
                |    * Q
     
    The total energy at P: [tex]E_T=\frac{1}{2}(2kg)(6m/s)^2+(2kg)(9.8N/kg)(l\sin 30-l\sin 310)[/tex]
    The total energy at A:
    [tex]E_T=\frac{1}{2}(2kg)(8m/s)^2[/tex]
    The total energy at P and A are equal, so solving for [tex]l[/tex], [tex]\displaystyle l=\left(\frac{64-36}{19.6(\sin 30-\sin 310)}\right)m=2.58m[/tex]
    b) At P: [tex]E_T=F\cdot l\Rightarrow F=\frac{E_T}{l}[/tex]
    Therefore, [tex]\displaystyle F=\left(\frac{64J}{2.58m}\right)=24.8N[/tex]
     
  2. jcsd
  3. Nov 7, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Your approach is correct for part a), but the calculation looks off. For part b) it appears to me you are missing something. The mass is accelerating with a component tangent to the circle and a component toward the center of the circle. The component toward the center is all you need to answer the qustion.
     
  4. Nov 8, 2006 #3
    Hi OlderDan, thank you for checking over my work. For part a), I recalculated and got an answer of [tex]l=1.13m[/tex].

    As for part b), you mentioned that as the mass is being whirled around, the force on the mass is the component towards the center. Does this mean the "centripetal force". If so, we have the information to solve for [tex]F_c=m\frac{v^2}{l}[/tex]
    [tex]F_c=(2)(\frac{6^2}{1.13})=63.8N\approx 64N[/tex]
     
  5. Nov 8, 2006 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    That looks good for part a). And yes to part b). Where does that centripetal force come from?
     
  6. Nov 9, 2006 #5
    Could the centripetal force be from tension force inside the rod and gravitational force on the mass as it is whirled in a circle?
     
  7. Nov 9, 2006 #6

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Yes it could. Tension acts in the direction of the rod. What about gravity?
     
  8. Nov 10, 2006 #7
    When you hurl the rod around, at the top of the circle the force downwards is force of gravity + force of tension in the rod, so that force provides the centripetal force. I think.
     
  9. Nov 10, 2006 #8

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    For any position of the rod, resolve the forces into components parallel and perpendicular to the direction of the rod. What is the relationship between the net parallel-to-rod force and the speed of the mass. What is the perpendicular force doing to the mass?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question regarding energy
Loading...