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Homework Help: Exam in 2 hours. Help!

  1. Nov 19, 2004 #1


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    These might be really silly problems for a lot of you but I'll have an exam in 2 hours and I still don't know how to do these problem ( which's supposedly the types of questions that will show up in the exam). I'll be really appreciate if someone can help me!

    1. An uniform thin rod 80 cm long with mass of 1 kg lies on a frictionless horizontal surface. A second small object of mass 1 kg strikes the rod 20 cm long from one end traveling with a velocity of 10m/s directly perpendicular to the length of the rod. the small object stick to the road after the collision. Compute ( after the collision):
    a, the velocity of the center of mass of the system composed of the rod and the object .
    b, angular velocity of the system about the center of mass.

    2. An ancient military harbor was protected by underwater reef that was 12ft out of the water at low tide and 4 ft under water at high tide. Friendly boats could enter the harbor as long as there was 3 ft of water above the reef. The depth of the water produced by the tides oscillates as if the water level were undergoing simple harmonic motion between high tide and low tide with a period of 12.5 hours. What was the duration of each time interval during which ships could safely enter the harbor.

    is there any hint? Thanks a lot!
  2. jcsd
  3. Nov 19, 2004 #2
    a. The velocity of the center of mass is equal to the weighted average of the mass*velocity of each object.
    b. [tex]L = r \times p = I\omega[/tex]

    2. Convert all that information into a sine wave. You can start by determining the amplitude and vertical shift. The period shouldn't be too hard to incorporate in.
  4. Nov 19, 2004 #3


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    I still don't understand. we have:
    x= 8 cos (omega t + phi)
    omega = 2pi/T
    but how can I find phi?
  5. Nov 19, 2004 #4
    [tex]x = 8cos(\omega t + \phi)[/tex]
    All phi does is shift the graph [tex]-\phi[/tex] on the x axis. Just let it equal 0. Also, be careful to add in the vertical shift.
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