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anonymous

  • one year ago

A thin rod with mass M = 5.85 kg is bent in a semicircle of radius R = 0.606 m (the figure below). (a) What is its gravitational force (both magnitude and direction) on a particle with mass m = 3.37×10-3 kg at P, the center of curvature? (b) What would be the force on the particle if the rod were a complete circle?

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  1. anonymous
    • one year ago
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  2. IrishBoy123
    • one year ago
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    i can help later

  3. anonymous
    • one year ago
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    thank you @IrishBoy123

  4. IrishBoy123
    • one year ago
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    |dw:1442784318789:dw|

  5. IrishBoy123
    • one year ago
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    you can start by using calculus the force between the central mass M , and the small elemental section of mass dm on the rod follows Newton's Law for gravitation, ie: \(\large |dF| = \frac{GMdm}{R^2}\) the force on dm acts inwardly toward the centre of the circle we can [and need] resolve it in the y direction [in the x direction, the force is symmetrical: overall, left and right, it nets off to zero]. so \(\large dF_y = -\frac{GMdm}{R^2} \, \sin \phi\), with the minus sign because it acts down the y axis we are given the overall mass, m, of the wire so we can say that mass dm of our small element, which subtends an angle \(d\phi\), is \(\large dm = m \times \frac{ d\phi}{\pi}\), which leaves us with : \[dF_y = -\frac{GMm \times \frac{ d\phi}{\pi}}{R^2} \sin \phi = -\frac{GMm}{\pi R^2} \sin \phi \, d\phi\] thus \[F_y = -\frac{GMm}{\pi R^2} \int\limits_{0}^{\pi} \, \sin \phi \, d\phi\] \[= -\frac{GMm}{\pi R^2} \left[ -\cos \phi \right]_{0}^{\pi} = \frac{GMm}{\pi R^2} \left[ \cos \phi \right]_{0}^{\pi} \] \[= -2\frac{GMm}{\pi R^2} \] and, here, we have: m [the rod] = 5.85 kg R [radius of rod] = 0.606 m M [the central particle] = 3.37×10-3 kg remember, the - sign just tells us that the force acts down the y axis if the rod were a complete circle,clearly symmetry would rule the day. the great thing is that we can complement that thought just using the maths. in that event we would have : \[F_y = -\frac{GMm}{\pi R^2} \int\limits_{0}^{\color{red}{2\pi}} \, \sin \phi \, d\phi \color{red} {= 0}\]

  6. IrishBoy123
    • one year ago
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    |dw:1442787001551:dw|

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