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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 = 3.37×103 kg at P, the center of curvature? (b) What would be the force on the particle if the rod were a complete circle?
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 = 3.37×103 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442784318789:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you 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] = 3.37×103 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 : \[F_y = \frac{GMm}{\pi R^2} \int\limits_{0}^{\color{red}{2\pi}} \, \sin \phi \, d\phi \color{red} {= 0}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442787001551:dw
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