## 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

2. IrishBoy123

i can help later

3. anonymous

thank you @IrishBoy123

4. IrishBoy123

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5. IrishBoy123

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

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