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Please clarify questions on forces and energy

  1. Jan 8, 2008 #1
    [SOLVED] Please clarify questions on forces and energy

    Hi everybody,
    I have been doing some reading and practice problems in my AP book lately, so I would appreciate it if people could tell me if the statements below are correct.

    1. A force applied perpendicular to the velocity can only cause a change in direction, not in speed. The force also does no work.

    2. The centripetal acceleration formula, ac = v^2/r, applies even when speed changes.

    3. When a person tosses a ball held by a string in a vertical circle, the speed constantly changes, so the magnitude of the centripetal force and the tension in the string differs at various points as well.

    4. A pendulum is also a case of centripetal acceleration even if it does not go in a complete circle. The centripetal force throughout is equal to T - m*g*cos(theta). m*g*sin(theta) causes the tangential acceleration.
     
  2. jcsd
  3. Jan 8, 2008 #2

    rock.freak667

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    1,2,4:Correct
    Not too sure on 3
     
  4. Jan 8, 2008 #3

    olgranpappy

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    indeed, you can see this by computing the time derivative of [itex]v^2[/itex].

    that formula doesn't apply generally. in that, the actual
    acceleration is not always given by that formula.

    yes

    probably, but it depends on what you mean by theta
     
    Last edited: Jan 8, 2008
  5. Jan 8, 2008 #4
    By theta, I meant the angle that the string makes with the vertical line through its equilibrium position.

    Ah, I see what you're saying. For the speed to change, there has to be some acceleration tangential to the direction of its velocity. The total acceleration would then be the vector sum of the tangential and centripetal acceleration.

    But in cases of circular motion (even when speed is not constant), I assume v^2/r works for finding centripetal acceleration?

    And one last question about the work-energy theorem:
    Wnet = change in KE
    Is this equation always true even in the presence of other forces? If so, is it because any change in KE has to come from a non-zero acceleration due to a non-zero net force?

    Thanks for replying and taking the time to read this.
     
  6. Jan 8, 2008 #5

    olgranpappy

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    what "other" forces? You add up *all* the forces acting on the object and that *is* the net force. The work done by that net force is the net work. If you don't add up all the forces on the object then you don't have the net force.

     
  7. Jan 8, 2008 #6

    olgranpappy

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    So, yeah. the work-KE theorem only works with the *net* force because it is proportional to the acceleration. Any other partial sum of forces just wont do. Again, to see this you can just take a time derivative of v^2.
     
  8. Jan 9, 2008 #7

    Shooting Star

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    That's right. For any motion,

    a = (dv/dt)T + (v^2/r)N, where T is the unit tangent vector and N is the unit principal normal vector.

    Didn't quite understand the question, neither the answer. If you carry something up through an height at constant speed, net work done is not equal to change in KE, but in this case, change in PE.
     
  9. Jan 9, 2008 #8

    olgranpappy

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    Here's what I mean. He was asking about the work kinetic energy theory which states, unequivocally, that: W_net = change in KE

    There's no two ways to slice it, you see:
    [tex]
    \sum_{all i} {\vec F_i}\equiv{\vec F_{net}}= m{\vec a} = m \frac{d \vec v}{dt}
    [/tex]

    thus

    [tex]
    Work_{net} = \int d{\vec x} \cdot {\vec F_{net}}=m \int d{\vec x}\cdot{\vec a}
    [/tex]
    Here, the integral is along the path of the particle (say, \vec x1 to \vec x2). I.e., [itex]d{\vec x}=dt\frac{d{\vec x}}{dt}=dt{\vec v}[/itex]

    thus, changing to an integral in time (limits t1 and t2, corresponding to x1 and x2, respectively)
    [tex]
    Work_{net}= m\int dt {\vec v}\cdot{\vec a}=\frac{m}{2}\int dt \frac{d}{dt}({\vec v}\cdot{\vec v})
    =\frac{m}{2}(v_f^2 - v_i^2)
    [/tex]
    where v_f is the velocity at t2 and v_i is the velocity at t1.

    The net work is alwasy the change in kinetic energy. always. never potential. regardless of whether there is any potential energy of not. For example, if I take an object at rest, and I raise it up to a height 'h' where it again is at rest, the change in kinetic energy is zero. Sure, *I* did work in rasing it up, but the gravitation field did exactly equal and opposite work. The net work is the change in kinetic energy.
     
    Last edited: Jan 9, 2008
  10. Jan 9, 2008 #9

    Shooting Star

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    Now I understand the question, and completely agree with the answer.
     
  11. Jan 9, 2008 #10

    olgranpappy

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    cheers.
     
  12. Jan 9, 2008 #11

    Shooting Star

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    To you, too!
     
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