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Physics Help

  1. Dec 10, 2004 #1
    A solid sphere of diameter 0.15m and mass 0.5kg is released and rolls without slipping down the ramp. The ramp is 0.81m high and is on a table. The table has a height of 1.40 meters. The balls move freely once it leaves the table.

    a) What is the moment of inertia of the ball about the axis of rotation?
    I = (2/5)(0.5kg)(0.075)^2 = 0.001125 kg*m^2

    b) Find the angular speed of the ball when it reaches the table top.
    Potential energy = Kinetic Energy
    (0.5)(9.81)(0.81) = (1/2)(0.001125)w^2 ---> w=84.04 rad/sec.

    c) Find the initial velocity (magnitude and direction) of the ball when it leaves the table.
    Total Energy = 0 = (1/2)(0.5)(v)^2 - (1/2)(0.001125)(84.04)^2
    velocity = 3.987m/s
    How do I find the direction?

    d) Find the distance d (distance ball travels when it left the table).
    x = vt + (1/2)at^2
    -1.40y + dx = (3.987x)t + (-4.905y)t^2
    x: d = 3.987t
    y: -1.40 = -4.905t^2 ----> t=0.534sec.
    d=3.987(0.534) ----> d=2.13meters

    Are my answers correct? Did I do anything wrong? Please help. Thanks! :smile:
     
    Last edited: Dec 10, 2004
  2. jcsd
  3. Dec 10, 2004 #2

    Pyrrhus

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    Homework Helper

    Remember when both movement (linear and rotational) are combined

    [tex] K = \frac{1}{2}mv^{2}_{cm} + \frac{1}{2}I_{cm} \omega^2 [/tex]

    You can rearrange this to find [itex] \omega [/itex]

    remember [itex] v = \omega r [/itex]

    [tex] K = \frac{1}{2}m \omega^{2} r^{2} + \frac{1}{2}I_{cm} \omega^2 [/tex]
     
  4. Dec 10, 2004 #3
    ok, so
    b) w = 44.92 rad/sec
    c) Would the initial velocity of the ball when it leaves the table still be 3.987m/s?
     
  5. Dec 10, 2004 #4

    Pyrrhus

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    You only considered linear movement kinetic energy, you have an object with linear and rotational motion. Refer to my post for [itex] \Delta K [/itex] You can express the direction by using [itex] \pm v_{x} \hat{i} [/itex]
     
    Last edited: Dec 10, 2004
  6. Dec 10, 2004 #5
    Ok, I redid the problem considering, like you said, the kinetic energy of both linear and rotational movement.

    so,
    (0.5)(9.81)(0.81) = (1/2)(0.5)(0.075^2)(w^2) + (1/2)(0.001125)w^2 ---> w=44.68 rad/sec.

    and velocity when the ball leaves the table is 5.147m/s

    and d = 2.75meters

    right?
     
  7. Dec 11, 2004 #6
    can someone please check my work?
     
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