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Review Questions- Need Help

  1. May 15, 2005 #1
    Review Questions- Need Help!!

    A hollow sphere is rolling along a horizontal floor at 5.00 m/s when it comes to a 30.0° incline.
    How far up the incline does it roll before reversing direction?

    I am not sure how to start this problem. I know the answer is 4.25m but do not know/understand how that answer was derived.

    A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 6.00 cm.
    What is the oscillation frequency of the two-block system?

    Here I thought you could say that the 6 cm was the amplitude and derive the answer from that, but my answer was incorrect. I don't know where to go with that not working.

    Thanks for the help
     
  2. jcsd
  3. May 15, 2005 #2

    cepheid

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    For the first question, I'm assuming friction is not a consideration. In that case, use conservation of mechanical energy.
     
  4. May 15, 2005 #3

    OlderDan

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    The energy hint from cepheid is the way you want to go on the first one. Just remember that rolling balls, disks, logs, etc have two contributions to their kinetic energies.

    For the second problem, the information given is sufficient to find the spring constant. It is important to note that the block being added is identical to the one that was already hanging.
     
  5. May 15, 2005 #4
    On the first one I have tried the mechanical energy and have come up empty.
    Here is my work:

    1. .5mv^2 + .5*I*omega^2 = mgh
    2. I then substituted for I and omega to get:
    3. .5mv^2 + 1/3mv^2 = mgh
    4. h= (.8333*v^2)/g

    I get an answer of 2.125. Can you see where I went wrong? Thanks

    On the second one can you give me a little more detail on how to find the spring constant? I can see trying to use the potential energy, but do not know where to get the necessary information.
     
    Last edited: May 15, 2005
  6. May 15, 2005 #5
    what is the inertia of a hollow sphere? [itex]I=\frac{2}{3}mr^2[/itex]?
     
  7. May 15, 2005 #6
    hmmm, lets see:
    [tex]\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=mgh[/tex]
    [tex]\frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{3}mr^2\frac{v^2}{r^2})=mgh[/tex]
    [tex]\frac{1}{2}mv^2+\frac{1}{3}mv^2=mgh[/tex]
    [tex]3mv^2+2mv^2=6mgh[/tex]
    [tex]5v^2=6gh[/tex]
    [tex]\frac{5v^2}{6g}=h[/tex]
    [tex]h=2.125~meters[/tex]
    yea, i got the same thing as cde42003. :confused:
     
  8. May 15, 2005 #7

    cepheid

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    You're both correct. You haven't completed the problem yet though.

    Hint: The question is, how far up the ramp will the ball go?
     
  9. May 15, 2005 #8
    I have figured the first one out now. You had to take the value of h and divide by
    sin(30°) to get the answer.

    Can anyone help me on the second question?
    Thanks
     
  10. May 15, 2005 #9

    OlderDan

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    I did in #3 :smile:

    How can you find the spring constant from the information given, and how do you use it to find the frequency?
     
  11. May 15, 2005 #10
    I am trying to follow you but I cannot think of how to find the spring constant from the given info. I do know how to find the frequency from the spring constant but I cannot get this far. What equation do I need to use for the spring constant? Thanks and I am not trying to get off without doing any work, I am just trying to figure this out before my exam tomorrow.
     
  12. May 15, 2005 #11

    OlderDan

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    Hooke's Law tells you that a spring stretches an amount that is proportional to the force applied. With one mass hanging on the spring, it is stretched by an unspecified amount, but when the second mass is added, it stretches another 6 cm. Any additional force added to a spring stretches it an additional distance dtermined by the same constant. In other words, if

    [tex] F = -kx[/tex]

    then

    [tex] \Delta F = -k\Delta x[/tex]

    You can determine the spring constant from the additional stretching that results from the additional force. You have the total mass of the system after the second one is added, so now you have all you need.

    Good luck on your exam.
     
  13. May 15, 2005 #12
    Ok. So since the the change in force is proportional to the change in x:
    .06N = -k(.06 m)
    When I tried it this way, I got the answer to be 2.03 Hz which was incorrect. What Am I still not getting?
     
  14. May 15, 2005 #13

    OlderDan

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    WAIT!! You can do this!!!

    All you need is the ratio of the mass to the spring constant

    [tex]\Delta F = (\Delta m)g = k\Delta x[/tex]

    Doubling the mass is going to double the stretch so you have

    [tex] 2mg = k (2X_s)[/tex]

    where m is the mass of each of the two masses, and [itex]X_s[/itex] is the stretch from the one mass, or 6 cm. Note that this [itex]X_s[/itex] is not the position of the mass at any particular time during oscillation. It is the amount that one mass stretches the spring when the mass is at equilibrium hanging on the spring.

    [tex] \frac{k}{2m} = \frac{g}{2X_s} [/tex]

    [tex] f = \frac{1}{2\pi}\sqrt{\frac{k}{2m}} = \frac{1}{2\pi}\sqrt{\frac{g}{2X_s}} = \frac{1}{2\pi}\sqrt{\frac{9.8}{.12}} Hz[/tex]
     
    Last edited: May 16, 2005
  15. May 15, 2005 #14
    There really is no other information. That is why I was stuck. I can't see anyway to do the problem. Thanks for the help
     
  16. May 16, 2005 #15

    OlderDan

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    I was wrong about this. Take another look at my previous post.
     
    Last edited: May 16, 2005
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