Results 1 to 3 of 3

Math Help - Vertical Circle

  1. #1
    Member
    Joined
    Mar 2007
    From
    England
    Posts
    104

    Vertical Circle

    The question is from Heinemann Modular Mathematics p145

    Q6: A bead of mass m is threaded on a smooth circular wire of radius a and centre O which is fixed in a vertical plane. B is released from rest at the point where OB makes an angle of 30 degrees with the upward vertical.
    b) Given that the speed of B at the lowest point of its path is v, find \<br />
v^2 <br />
\ in terms of a and g. This part I can do and the answer is \<br />
v^2  = ag(2 + \sqrt 3 )<br />
\
    The next part is the bit I'm stuck on:
    c) Find in terms of m and g the magnitude and direction of the reaction of the wire on the bead when OB makes an angle of 60 degrees with the upward vertical.

    I started off by using the conservation of energy, PE lost=KE gained.
    \<br />
2ag\left( {\frac{{\sqrt 3  - 1}}{2}} \right) = a^2 g^2 \left( {2 + \sqrt 3 } \right)^2  - v^2 <br />
\
    which simplifies to
    \<br />
v^2  = a^2 g^2 \left( {2 + \sqrt 3 } \right) - ag\left( {\sqrt 3  - 1} \right)<br />
\

    But I don't know what to do after this. Can someone help please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by free_to_fly View Post
    The question is from Heinemann Modular Mathematics p145

    Q6: A bead of mass m is threaded on a smooth circular wire of radius a and centre O which is fixed in a vertical plane. B is released from rest at the point where OB makes an angle of 30 degrees with the upward vertical.
    b) Given that the speed of B at the lowest point of its path is v, find \<br />
v^2 <br />
\ in terms of a and g. This part I can do and the answer is \<br />
v^2  = ag(2 + \sqrt 3 )<br />
\
    The next part is the bit I'm stuck on:
    c) Find in terms of m and g the magnitude and direction of the reaction of the wire on the bead when OB makes an angle of 60 degrees with the upward vertical.

    I started off by using the conservation of energy, PE lost=KE gained.
    \<br />
2ag\left( {\frac{{\sqrt 3  - 1}}{2}} \right) = a^2 g^2 \left( {2 + \sqrt 3 } \right)^2  - v^2 <br />
\
    which simplifies to
    \<br />
v^2  = a^2 g^2 \left( {2 + \sqrt 3 } \right) - ag\left( {\sqrt 3  - 1} \right)<br />
\

    But I don't know what to do after this. Can someone help please?
    In the absence of friction the net force in the radial direction on the bead is the centripetal force. This will be the sum of the normal force on the bead from the wire plus the component of the weight in the radial direction.
    F_c = \frac{mv^2}{a} = N + w~cos(60^o)
    We can find the magnitude of the normal force from this equation.

    The reaction force on the wire is the same magnitude as the centripetal force, but being a Newton's third law pair with the normal force on the bead, it will be in the opposite direction. So the reaction force is radially outward, rather than radially inward.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by topsquark View Post
    In the absence of friction the net force in the radial direction on the bead is the centripetal force. This will be the sum of the normal force on the bead from the wire plus the component of the weight in the radial direction.
    [snip]
    A proof that for this problem the normal reaction force is radial can be supplied upon request.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vertical Circle ..Bead on a wire problem?
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: December 19th 2011, 06:30 AM
  2. Replies: 6
    Last Post: July 8th 2010, 05:39 PM
  3. Replies: 2
    Last Post: February 6th 2010, 08:31 AM
  4. Motion in a vertical circle
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: December 19th 2009, 06:34 AM
  5. Vertical distance between 2 points on a circle
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2009, 07:00 AM

Search Tags


/mathhelpforum @mathhelpforum