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anonymous
 one year ago
A thin rod with mass M = 5.79 kg is bent in a semicircle of radius R = 0.618 m (the figure below). (a) What is its gravitational force (both magnitude and direction) on a particle with mass m = 4.33×103 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 = 5.79 kg is bent in a semicircle of radius R = 0.618 m (the figure below). (a) What is its gravitational force (both magnitude and direction) on a particle with mass m = 4.33×103 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.0dw:1442485921631:dw Given \[M=5.79 \space kg\] \[R=0.618 \space m\] \[m=4.33 \times 10^{3} \space kg\] Let \[\lambda\] be the mass per unit length of the semicircular wire segment(or linear mass density) Then we have \[\lambda=\frac{M}{\pi R}\] consider a small length element dl, having small mass dM Then we have the relation \[dM=\lambda.dl\] \[dM=\frac{M}{\pi R}dl\] Then small force due to small segment having length dl and mass dM is given by \[dF=\frac{Gm.dM}{R^2}=\frac{Gm}{R^2}.\frac{M}{\pi R}dl=\frac{GMm}{\pi R^3}dl\] Now as we vary the small length segments from 0 to pi times R, the force varies from 0 to F(the total force exerted by segment of full pi times R length) Summing up these segments we should be able to find an expression for total force, \[\sum dF=\frac{GMm}{\pi R^3}\sum dl\] Now since we consider infinitely small length segments, our sum becomes an integral with the limits as shown \[\int\limits_{0}^{F}dF=\frac{GMm}{\pi R^3}\int\limits_{0}^{\pi R}dl\] \[[F]_{0}^{F}=\frac{GMm}{\pi R^3}.[l]_{0}^{\pi R}\] Applying fundamental theorem of calculus we get \[F0=\frac{GMm}{\pi R^3}.(\pi R0)\]\[F=\frac{GMm}{\pi R^3}.\pi R \implies F=\frac{GMm}{R^2}\] So the force is simply the same as the force that would be exerted by a particle having as much mass as the rod has What about the direction?? You can tell by symmetry that all the vertical components of the force will cancel out, leaving a purely horizontal force dw:1442486819658:dw From the figure, you can tell that each pair like these will have the vertical components cancelled, leaving horizontal components only So the force can be considered as the same as exerted by a particle of mass M(same as mass of rod) placed at a distance R(radius of the semicircle formed by the rod) from the point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you think by symmetry would happen in case of a full circular loop?
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