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Projectile motion and angular motion

  1. Mar 25, 2005 #1
    Hi.

    I have two questions which I have been pondering on and I just can't figure it out. Maybe someone could be kind enough to answer the questions?

    Q1; Projectile motion: Let's say we dropped a ball from a height called h.
    We achieve a velocity [tex] v=\sqrt{2gh} [/tex]. Let's then say that we hit the ball in differently angle with a paddle. What will the motion equations become?

    Q2: Angular motion: Let's say we have a ball in vaacum. We make this ball rotate. Then this ball achieves angular velocity, the direction of the angular velocity [tex] \omega [/tex] can be found by the right-hand rule. But I am confused, what does the angular velocity's direction tell us? It it where the ball will move towards?

    Thank you.
     
  2. jcsd
  3. Mar 25, 2005 #2

    dextercioby

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    Solve the system

    [tex] m\frac{d^{2}\vec{r}}{dt^{2}}=m\vec{g} [/tex]

    with the initial conditions that u wish to impose...

    Nope.If the ball just rotates around a fix axis (for simplicity),then the direction of [itex] \vec{\omega} [/itex] will be along the rotation axis and,incidentally,the angular momentum [itex] \vec{L} [/itex] will have the same direction.So yes,i the ball doesn't translate,then specifying the modulus,sense & direction of either [itex] \vec{L} [/itex] or [itex] \vec{\omega} [/itex] will completely determine the movement,in case the ball is not acted on by any force (except gravity which would give a zero torque under normal conditions)...

    Daniel.
     
  4. Mar 25, 2005 #3
    I'm sorry to ask again. But I didn't understand your answer. How did you achieve that equation, could you show me? What is [tex]\vec{r}[/tex]? Is it the radius of the ball? Let's say I impose initial conditions as none friction and none air resistance. How will the equation turn out then?

    Thank you for the answer of Q2.
     
  5. Mar 25, 2005 #4

    dextercioby

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    It's the II-nd law of Newton for translation movement...Mass times acceleration is equal to the vector sum of all forces acting on the body.In this case,it's only gravity.That [itex] \vec{r} [/itex] is the position vector for the CM of the body wrt an inertial reference system.

    Daniel.
     
  6. Mar 25, 2005 #5
    Thank you for your help, Daniel. I appreciate it.
     
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