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Finding the speed of a block only by using the angle and Radius

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data
    x1aIXFa.gif
    A small block is standing on top of a big smooth sphere. The small block is initially at rest. The block then glides down the sphere. The block is standing a little to the side for it to move down the sphere but it can be assumed that it is starting at the top. The radius of the sphere is known.

    Find an expression for the block's speed after θ degrees.

    3. The attempt at a solution
    I've been thinking a lot about which equations that I can use, unfortunately because I don't know the mass of the block, I can't use Newton's second law. My initial strategy was to use the formula:
    [itex]a_{rad}=\frac{v^2}{R}[/itex]

    However I need to find the acceleration and I can't figure out a way to do this without using the mass of the block.

    Can somebody give me a hint or kickstart me?
     
  2. jcsd
  3. Nov 16, 2013 #2

    TSny

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    My hint would be "energy". Don't worry about not knowing the value of the mass, just call it "m" for now.
     
  4. Nov 16, 2013 #3

    Andrew Mason

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    What force is causing it to fall? What is the force opposing that fall?

    AM
     
  5. Nov 16, 2013 #4
    I thought about the kinetic energy as well, however, that too requires the mass.

    I was thinking about the energy-work theorem.

    If I set point 1 to be the top where it's at rest and point 2 the distance of θ, then I have my nergy-work theorem:
    [itex]W_{tot}=K_2-K_1=K_2-0[/itex] since K1 will be zero because the velocity of the block at point 1 is zero.

    That leaves me with the two unkown variables, the work and the mass.

    For the work, I figured that the formula for a varying force on a curved path could help me:
    [itex]W=\int_{P_1}^{P_2} F \cdot cos(\theta) dl[/itex]

    I want to draw a FBD of the block when it has traveled θ distance however the only force that I can think of is the gravity which points downwards and the normal force perpendicular to the surface of the sphere. It's only the gravity that is causing the block to move down. I don't have a F force.

    Unless it is possible to express my F force in terms of the gravity and normal force?
    FveQA8k.png

    So Newton's first law gives me:
    [itex]n\cdot sin(\theta)-mg=0[/itex]

    But I can't use this in my formula?
     
    Last edited: Nov 16, 2013
  6. Nov 16, 2013 #5
    It is only the gravity (mg) that causes it to fall, right? And the normal force opposing it.
     
  7. Nov 16, 2013 #6

    TSny

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    Right, two forces act on the block: the normal force and the force of gravity. Can you express the work done by each of those forces? Then you can set up ##W_{\rm tot} = K_2-K_1##. Again, don't worry about not knowing the mass.
     
  8. Nov 16, 2013 #7
    I can find the gravitational potential energy. If I set my reference point at y1 at the top of the sphere where the block is at rest, that will leave me with:
    [itex]W_{grav}=mgy_1-mgy_2=0-mgy_2[/itex]

    I'm not sure how exactly to find y2 though. I know that the arc that the block moves from the top to this point is the radius of the sphere multiplied by the length of θ in radians. But that doesn't exactly help me finding the potential height..
     
  9. Nov 16, 2013 #8

    TSny

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    That's good!

    See if it helps to draw an auxiliary line as shown.
     

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  10. Nov 16, 2013 #9
    I can't believe I didn't see that! :)

    So that leaves me with the potential height:
    [itex]y_2=R-(R\cdot cos(\theta))[/itex]

    While going through my book, I think I found a formula that I think would be better to use. The total mechanical energy conserved:
    [itex]K_1+U_1=K_2+U_2[/itex]

    So my K1 was zero because the initial velocity is zero.
    My U1 is zero because my reference point is y1=0 at the top of the sphere.
    My K2 was [itex]\frac{1}{2}\cdot m\cdot v^2[/itex]
    And my U2 was [itex]m\cdot g\cdot R-R\cdot cos(\theta)[/itex]

    That leaves me with:
    [itex]\frac{1}{2}mv^2=-mg(R-Rcos(\theta))[/itex]

    The mass cancels out on both sides and so all that is left is the velocity. :)
    But does the negative sign with the potential energy makes sense? You can't take the squareroot of a negative.
     
  11. Nov 16, 2013 #10

    TSny

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    Yes, good. (But you might need to reconsider the sign of y2.) Setting up conservation of energy like this is equivalent to ##W_{\rm tot} = \Delta K## for this problem. The conservation of energy approach is probably the "best" method for a conservative system like you have here.

    Check your sign for U2 = mgy2. What is the sign of y2?
     
  12. Nov 16, 2013 #11
    Of course, hehe. Since the direction is downwards, it's negative. Thank you! :)
     
  13. Nov 16, 2013 #12
    [tex]mg\cos\theta - N= m\frac{v^2}{R}\\
    \text{It can be observed that }N\cos\theta=mg\\
    \text{Substituting this in yields: }mg\cos\theta - \frac{mg}{\cos\theta} = m\frac{v^2}{R}\\
    \Rightarrow gR\left(\cos\theta+\frac{1}{\cos\theta}\right)=v^2\\
    \Rightarrow v=\sqrt{gR\left(\cos\theta + \sec\theta\right)}[/tex]
     
  14. Nov 16, 2013 #13
    How do you "observe" this? :smile:
    Is the acceleration along the vertical direction equal to zero?

    And another way to check you result: what do you get for θ=0?
    Is this compatible with "initially at rest"?
     
  15. Nov 16, 2013 #14
    Mmmm you're right! [itex]N\cos\theta \neq mg[/itex] as long as the block is past the top of the sphere. But it is still true that [itex]mg\cos\theta-N = m\frac{v^2}{R}[/itex], which makes sense when [itex]\theta = 0[/itex] because the block isn't moving/rotating
     
  16. Nov 16, 2013 #15
    Yes, the first equation is correct.
    Just not necessary here, when only the speed is required.

    You will need this equation if you want to find the angle at which the block leaves the sphere, for example.
     
  17. Nov 16, 2013 #16
    Hey guys, I still need some help if possible! :)

    I need to find an expression for the normal force when it has glided down θ distance and I need to find the angle for when the block leaves the sphere.

    For the expression of the normal force, I drew the FBD:
    FveQA8k.png

    I then tried to use Newton's second law with [itex]a_{rad}=\frac{v^2}{R}[/itex] to find an expression for the normal force.

    Unfortunately Newton's second law includes the mass. So I tried to use Newton's first law in equilibrium and along the y-axis in my FBD and then tried to find an expression for the mass and then insert that expression for the mass into Newton's second law, however that resulted in the normal force canceling each other out. :(

    Any hints? I don't think I'm supposed to use Newton's laws here.

    once I find the expression for the normal force, finding the angle for when the block leaves the sphere should be easy since the normal force would be zero in that situation.
     
    Last edited: Nov 16, 2013
  18. Nov 16, 2013 #17
    Re: angle when it leaves
    I had a problem just like this! I wrote more above, but you want to use Newton's Laws to find an equation involving centripetal acceleration, weight, and the normal. When the blocks loses contact, [itex]N=0[/itex] and that'll give you the result!

    Re: [itex]N(\theta)=?[/itex]
    I'm not sure about finding an expression for [itex]N[/itex] at a given angle, especially if you can't have it be a function of [itex]m[/itex]... Maybe I'm not thinking very thoroughly right now
     
  19. Nov 16, 2013 #18
    Ya, i tried that approach:
    [itex]\sum F_y=m\frac{v^2}{R}[/itex]

    Unfortunately I don't know the mass in the equation. I'm not sure what else I can do.
     
  20. Nov 16, 2013 #19
    The mass will cancel out when N=0 (the sphere loses contact). Be sure you're summing the correct forces, not just the forces in the y direction. *Hint* When done correctly, the sphere will fall after after lowering [itex]\frac{R}{3}[/itex] distance from the top, vertically. Use [itex]N, \; mg\cos\theta, \; \text{and } m\frac{v^2}{R}[/itex]. It'll come together when done correctly.
     
    Last edited: Nov 16, 2013
  21. Nov 17, 2013 #20
    As TSny says, use conservation of energy,
    What is initial energy(taking centre of sphere as reference point) and what's total energy after moving through an angle.
    In the equation, m will be cancelled out.
     
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