Results 1 to 5 of 5

Math Help - Urgent physics help (rotating vertical bar)

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    2

    Urgent physics help (rotating vertical bar)

    I found this problem very recently on the forum but mine seems to have a 4th and fifth part to it


    The diagram shows a small ball of mass m=0.280kg that is attached to a rotating vertical bar by means of two massless strings of equal lengths L=1.10m. The distance between the points where the strings are attached to the bar is D=1.50m. The rotational speed of the bar is such that both strings are taut and the ball moves in a horizontal circular path of radius R at constant speed v=6.66m/s.


    (a) Determine the magnitude and direction of the net force that is acting on the ball when it is in the position shown in the diagram.
    (b) Determine the magnitude of the tension T1 in the upper string and the magnitude of the tension T2 in the lower string.
    (c) Determine the speed of the ball at which the tension T2 becomes zero.
    (d) Determine the angle of the top string when the velocity is v = 1.48 m/s.
    (e) Determine the tensions T1 and T2 as a function of velocity (2 seperate equations).

    I've already got the answers to a, b, and c, just need d and e.
    Last edited by academy; October 10th 2008 at 02:23 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Trig

    On part (d), can't you just use trig to solve this; namely that \frac{d}{2}=L\cos\theta?

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    2
    no because in the new part D of the problem, they are changing the velocity so that T2 becomes 0 and T1 gets closer to the pole thus Distance isnt exactly half and radius becomes unknown

    thanks for the help though, anything is appreciated as im stumped and quite furious at this point
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by academy View Post
    no because in the new part D of the problem, they are changing the velocity so that T2 becomes 0 and T1 gets closer to the pole thus Distance isnt exactly half and radius becomes unknown

    thanks for the help though, anything is appreciated as im stumped and quite furious at this point
    Yes, that's right. Because I solved for the v when T2 = 0 but the r is still sqrt(1.10^2 -0.75^2) = 0.804674 cm and I found it to be v = 2.9087 cm/sec.
    Since 1.48 cm/sec is less than 2.91 m/sec, then the r should be less than the "taut, maximum" r that is 0.804674 cm.

    The theta of the upper string then is arctan(r/y)
    where
    r = radius
    y = vertical distance of the ball to the top anchor point.

    The lower string is slacked or loose.

    At the FBD of the ball now:
    Let T = T1

    Summation horizontal forces is zero, so,
    T(r/1.10) = (.28)(1.48^2)/r
    T = 0.674643 /r^2

    Summation of vertical forces is zero,so,
    T(y/1.10) = (0.28)(9.8)
    T = 3.0184 /y

    T = T,
    0.674643/r^2 = 3.0184/y
    r^2 = (0.674643*y)/3.0184 = 0.22351*y

    We have 1.10^2 = r^2 +y^2, so,
    r^2 = 1.21 -y^2

    Hence,
    1.21 -y^2 = 0.22351*y
    y^2 +0.22351*y -1.21 = 0By the Quadratic Formula,
    y = 0.9939 m

    So, r = sqrt(1.21 -0.9939^2) = 0.471341 m

    Therefore, theta = arctan(0.471341 /0.9939) = 25.37 degrees -----answer
    Last edited by ticbol; October 10th 2008 at 07:18 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by academy View Post
    I found this problem very recently on the forum but mine seems to have a 4th and fifth part to it


    The diagram shows a small ball of mass m=0.280kg that is attached to a rotating vertical bar by means of two massless strings of equal lengths L=1.10m. The distance between the points where the strings are attached to the bar is D=1.50m. The rotational speed of the bar is such that both strings are taut and the ball moves in a horizontal circular path of radius R at constant speed v=6.66m/s.


    (a) Determine the magnitude and direction of the net force that is acting on the ball when it is in the position shown in the diagram.
    (b) Determine the magnitude of the tension T1 in the upper string and the magnitude of the tension T2 in the lower string.
    (c) Determine the speed of the ball at which the tension T2 becomes zero.
    (d) Determine the angle of the top string when the velocity is v = 1.48 m/s.
    (e) Determine the tensions T1 and T2 as a function of velocity (2 seperate equations).

    I've already got the answers to a, b, and c, just need d and e.
    For the part (e):

    Cut the strings at the points of attachment so that you have a FBD.
    Call the upper attachment point as A; the lower, as B.

    Summation of moments about B is zero.
    [(T1)(0.804674 /1.10)](1.50) = [(0.28)(v^2)/0.804674](0.75) +[(0.28)(9.8)](0.804674)
    (T1)(1.097283) = (0.260975)v^2 +2.208025
    T1 = (0.237837)v^2 +2.012266 .....in joules ------------answer

    For the T2, following the same procedure above, but now taking moments about point A,
    T2 = (0.237837)v^2 -2.012266 ....in joules -------------answer.
    Last edited by ticbol; October 10th 2008 at 07:21 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. URGENT physics question please.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 15th 2008, 02:57 AM
  2. Physics, vertical kinematics, gravity
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: September 18th 2008, 09:25 AM
  3. urgent physics help,
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: June 3rd 2008, 08:38 AM
  4. Physics urgent help plz.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 1st 2008, 04:04 AM
  5. physics help...urgent !
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2008, 02:13 AM

Search Tags


/mathhelpforum @mathhelpforum