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Easy kinematics question

  1. Jan 13, 2005 #1
    a boy whirls string of length 2 metres around in a horizontal plane. At the end of the string a mass of M2 = 0.1kg is tied while at 1m (midway on the string) another mass M1 = 0.1Kg is fastened. The boy increases angular velocity [tex] \omega [/tex] in the motion of the string. The hand of the boy and the two masses will, when omega is sufficiently large, be almost on a straight line. When the string force reaches 327N the string will break.

    Find the magnitude of the angular velocity when the string breaks and find where the break occurs.

    Since maximum of the tension on the string will be between the boy's hand and the mass m1 the string will break right there. Also since they are almost in a horizontal circle the angular velocity is common.

    when [tex] T = 327 N = m_{1} \omega^2 r_1 + m_2 \omega^2 r_2[/tex]

    which gives [tex] 327 = 0.1 * \omega^2 * 1 + 0.1 * \omega^2 * 2[/tex]

    and this yields 33.02 / s

    Is it possible to whirl the two masses in such a way that the two masses on the string are exactly on a horizontal line?

    I would think that when the two omegas are the same the string will break so is it not possible

    i ned more help with actualyl explaining the second one. Also i need explanations because my prof is like that. Are my explanatuons concise enough or do i need any more detail.

    your suggestions and help are appreciated!
     
  2. jcsd
  3. Jan 13, 2005 #2

    NateTG

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    I would describe part (b) as a poorly phrased question because in the inital text of the problem, you are asked to assume that the masses and string rotate in a horizontal plane. With that assumption, the string could be straight horizontally with [itex]\omega=0[/itex].
    The answer that the book is looking for probably involves ignoring that issue, and, instead looking at all of the forces that are acting on one of the masses. (If you know what a free-body-diagram is, you might want to draw one.) In the spinning reference frame, there are three forces acting on the masses: Gravity, the Centrifugal Force, and the tension in the rope. Since, in this reference frame the net force is zero, gravity acts downward, and the centrifugal force acts outward, the tension must be acting up and in. However, since the tension is always along the same direction as the rope, this means that the rope must also be running up from the mass to the boy's hand.
     
  4. Jan 13, 2005 #3
    awesome explanation!

    i get it now.

    But i was under the impression that centrifugal force was actually centripetal force that acts inward so the tension is actually outward?
     
  5. Jan 13, 2005 #4

    NateTG

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    Any force acting from an object towards the center of rotation is a centripetal force. In practice, it is probably better to avoid the expression "centripetal force" because it is somewhat confusing, and use "centripetal acceleration" instead. In this problem, for example, you might say the tension in the rope is providing the centripetal acceleration.

    The centrifugal force is a force that only exists in rotating reference frames. It acts outward away from the center of rotation What happens is that it is frequently easier to deal with things in rotating reference frames like in this example, but since the rotating frame is accelerated, forces proportional to the inertia of objects need to be added. The centripetal force is called a fictional force by some people because it is the result of an accelerated frame of reference rather than some sort of object to object interaction.

    P.S. "Centripetal force" is not only confusing because of it's similarity and association with centrifugal force, but also because there are situations where different forces act as a centripetal force on an object. For example, when a roller coaster goes through a loop, gravity is a centripetal force at the top, but not at the bottom.
     
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