Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Determine the speed of the stone

  1. Apr 27, 2004 #1
    Need Help ASAP!!!

    hey can some1 help me with these 2 physics questions..really need it...it's my last hw assigment and i have trouble with this 2...please help it's due tomorrow morning...show what me to do..thankssssssssss

    1. A stone is tied to a string (length = 1.27 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 15% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

    2.Suppose the surface (radius = r) of a cylindrical space station is rotating at 33.8 m/s to provide artificial gravity. What must be the value of r for the astronauts to weigh 1/2 of their earth weight?
    Last edited: Apr 27, 2004
  2. jcsd
  3. Apr 28, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    1) In the horizontal circular motion case, only the string's tension can provide the required centripetal acceleration.
    What forces can provide centripetal acceleration when the circular motion is vertical; in particular, when is the tension of the string at its maximum?
  4. Apr 28, 2004 #3
    1. Tension is max at the bottom of the vertical surface and is given by
    [tex] T_{max} = W +\frac{mv_{0}^2}{L}[/tex]

    2. N is Normal Reaction and it varies as

    [tex] N = W\ cos\theta + \frac{mv^2}{L}[/tex]

    where \theta is the angle of astronaut with the vertical at any instant

    Since its normal rxn is varying with theta , so its weight registered is also varying
    so u can solve it for [tex] \theta =0 [/tex] and N= W/2
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook