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(Circular motion) The second problem I really need to solve this weekend

  1. Nov 20, 2005 #1
    A little block with mass m lies still on the top of a frictionless sphere with radius R. Somebody gently hits the block, causing it to slide down the sphere. The mass loses contact with the sphere when the angle between the positionvector and the vertical equals 0c. The drawing below shows the position of the block on two moments in the movement. The velocity caused by the hit in the beginning is negligible.

    http://img453.imageshack.us/img453/6496/sphericalsurface0qu.gif [Broken]

    The question is:

    Find the angle 0c where the block begins losing contact with the surface.

    Unfortunately I may not use energy laws, because my teacher wants me to do it another way, but I just don't know how! Please give me a little start here...
    Questions I've already answered are:

    a) Express the distance s in R and 0.

    s= R * 0 That is when 0= in rad. If 0 is in degree it should be s= R (2pi *0 / 360)

    b) Which forces work on the block if it's sliding down the sphere? Draw them.

    The gravitationforce and the normal force, drawn below:

    http://img173.imageshack.us/img173/4374/sphericalsurface25zz.gif [Broken]

    c) Split the forces in tangential components and perpendicular components on the spherical surface.

    http://img467.imageshack.us/img467/2815/sphericalsurface35tt.gif [Broken]

    d) Express the centripetal acceleration Ac and the tangential acceleration At in the forces found by b.

    If 0 is chosen well in the x-direction (the direction of the tangential acceleration) Fres= Fz,//= Fz sin 0 = m * At. Therefore At= (Fz sin (0))/m

    In the y-direction (the direction of the centripetal acceleration) Fres= Fz,|- Fn= Fz cos 0 - Fn= m* Ac. Therefore Ac= (Fz cos(0) -Fn)/m.

    e) At the point where the mass begins losing contact with the surface there are two conditions:

    Condition 1: |Ac|= g cos(0)
    Condition 2: |Ac|= v^2/R where v= the speed of the block.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 20, 2005 #2


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    Homework Helper

    write your centripetal and tangential accelerations as,

    [tex]A_t = r \ddot \theta[/tex]
    [tex]A_c = r \dot \theta ^2[/tex]


    [tex]r \ddot \theta = gsin\theta \mbox{ -----------------------(1)}[/tex]
    [tex]r \dot {\theta ^2} = gcos\theta - \frac{F_n}{m} \mbox{ --------------(2)}[/tex]

    You want to integrate (1) to get an expression for [tex]r \dot \theta ^2[/tex] which you then substitute into (2)

    I'll start you off,

    Using (1),

    [tex]r\ddot \theta \dot \theta = g \dot \theta sin\theta \mbox{ multiplying both sides by } \dot \theta[/tex]
    [tex]\frac{d \ }{dt}\left(\frac{1}{2}r \dot \theta^2 \right) = gsin\theta\cdot \frac{d\theta}{dt}[/tex]

    Use initial conditions to evaluate your constant of integration.

    The answer is about 48 deg.
    Last edited: Nov 20, 2005
  4. Nov 20, 2005 #3
    Why do you write:

    [tex]A_t = r \ddot \theta[/tex]
    [tex]A_c = r \dot \theta ^2[/tex]

    What do these theta's mean?

    And why do you multiply [tex]r \ddot \theta[/tex] by [tex] \dot \theta [/tex] ?

    And I still don't get what you're doing with integration.... I'm sorry :S
    Last edited: Nov 20, 2005
  5. Nov 20, 2005 #4


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    I'm no expert by any means, and may confuse you more but...

    use conservation of energy

    mgr(top) = 1/2 mv2 + mgrcos(theta) (the angle slides off)

    use your two equations-

    1) Ac = gcos(theta)
    2) Ac = v^2/r

    hence, gcos(theta) = v^2/r... solve for v^2

    3.) v^2 = grcos(theta)
    use this to get rid of the v^2 term in your original equation and solve for theta
  6. Nov 20, 2005 #5


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    Unfortunately, she's not allowed to. :frown:

  7. Nov 20, 2005 #6


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    See http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq for discussion of rotational motion.

    [itex]A_t = r\,\alpha = r\,\ddot \theta[/itex] where [itex]\alpha = \ddot \theta[/itex] is the angular acceleration.

    [itex]A_c = r \omega = r \dot \theta ^2[/itex] where [itex]\omega = \dot \theta[/itex] is the angular velocity.

    The block leaves the surface at the point where the centripetal force = the force normal to the surface mg cos[itex]\theta[/itex], or the centripetal acceleration = g cos[itex]\theta[/itex].

    Also [tex]\frac{d \ }{dt}\left(\frac{1}{2}r \dot \theta^2 \right) = r\ddot \theta \dot \theta [/tex]
  8. Nov 20, 2005 #7


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    sorry...that is what happens when you read a post to fast..:yuck:
  9. Nov 20, 2005 #8


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    Those are the expressions for tangential and centripatal acceleration using polar coordinates <r,θ>

    [tex] \theta[/tex] is your angle, or angular displacement. [tex]\dot \theta[/tex] is [tex]\frac{d \theta}{dt}.[/tex] The dot over the [tex]\theta[/tex] simply means the derivative wrt time. Similarly [tex]\ddot \theta \mbox{ is } \frac{d^2 \theta}{d \theta^2}.\mbox{ Also, } \omega = \frac{d\theta}{dt} = \dot \theta \mbox{ and } \alpha = \frac{d^2 \theta}{d \theta^2} = \ddot \theta[/tex]

    That multiplication was just a "trick" to simplify the intgration. You do this all the time when you multiply a DE by an integrating factor.
    I'll continue with the integration.

    Using (1),

    [tex]r\ddot \theta \dot \theta = g \dot \theta sin\theta \mbox{ ,multiplying both sides by } \dot \theta[/tex]
    [tex]\frac{d \ }{dt}\left(\frac{1}{2}r \dot \theta^2 \right) = gsin\theta\cdot \frac{d\theta}{dt}[/tex]
    Integrate both sides wrt time.

    [tex]\frac{1}{2}r \dot \theta^2 = \int gsin\theta\cdot \frac{d\theta}{dt}\cdot dt[/tex]
    [tex]\frac{1}{2}r \dot \theta^2 = \int gsin\theta\ d\theta[/tex]
    [tex]\frac{1}{2}r \dot \theta^2 = -gcos\theta + c[/tex]

    [tex]\mbox{at } t=0,\ \theta = 0,\ \dot \theta = 0,\ \mbox{giving } c = g,[/tex]

    [tex]r \dot \theta^2 = 2g(1 - cos\theta)[/tex]

    now substitute for [tex]r \dot \theta^2[/tex] into (2)
    Last edited: Nov 20, 2005
  10. Nov 20, 2005 #9
    Wow thanks for your explanation! Our book didn't explain anything about [tex]\omega[/tex] and [tex]\alpha[/tex] but I understand it now fine! I only found 86.10 degrees as an answer..... Please tell me what I'm doing wrong... what I did was substituting
    [tex]r \dot \theta^2[/tex] = 2g (1-cos[tex]\theta[/tex]) into
    [tex]r \dot \theta^2[/tex] = g cos[tex]\theta[/tex] - (Fn/m).


    2g (1-cos[tex]\theta[/tex]= g cos[tex]\theta[/tex] - (Fn/m).
    On the moment the block loses contact with the sphere Fn= 0, leaving:
    2g- 2g cos[tex]\theta[/tex]= g cos[tex]\theta[/tex]
    2g= 3g cos[tex]\theta[/tex]
    1= 1.5g cos[tex]\theta[/tex]
    cos[tex]\theta[/tex]= 1/1.5g= 2/3g= 0.0679578661
    [tex]\theta[/tex]= arccos 0.0679578661= 86.10 degrees
  11. Nov 20, 2005 #10


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    The g cancels on both sides :smile:

    cos theta = 2/3
  12. Nov 20, 2005 #11
    Awwww of course! I feel so stupid right now :D Thank you!!!
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